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. . . . . . C26:What should be themaximumdiameter of awire of length 1metre and density6 × 103 kgm 3which can float horizontally on the surface of a liquid of surface tension 80 × 10 3N/m(neglec t the buoyancy effect) Sol. Here 2 d L g T 2L 2 .. . . . . .. .. . 3 3 d 8T 8 80 10 1.86 10 metre. g 22 6 10 9.8 7 . . . . . . . .. . . . C27: Explain the following : Athinsteelneedle floats onwater but whena little soa p solution is carefullymixed with thewater the needle sinks. Sol. When a needle is placed gently on the surface ofwater, as no part of it is submerged inwater, it is not buoyed up byArchemedes principle.There is ofcourse onwayfor steel to float byArchimedes principle.Since its density is greater than that ofwater so it will sink if submerged. mg T A T It is actuallykept afloat by the surface tension ofwater.When it is placed on the surface ofwater, it depresses the surface ofwater slightlydue to itsweigh t.The stretchedsurface due to surface tension exerts a restoring force, the vertical component ofwhich canm aintainequilibriumwith the weight ofthe needle. However,when a detergent is added towater its surface tensionwillsuddelydecrease and the force due to surface tension is no longer sufficient to support theweight of the needle and s o it will sink. Example 49. Aring is cut froma platinumtube of 8.5 cminternal and 8.7 cmexternaldiameter. It is supported horizontallyfroma pan of a balance so that it comes incontact withthewater ina glass vessel. What is the surface tension ofwater of an extra 3.97 g weight is required to pull it away fromwater ? (g = 980 cm/s2) F T T F Cross Section Sol. The ring is in contact with water along its inner and outer circumference; sowhenpulledout thetotalforce onit due to surface tensionwillbe F = T (2.r1 + 2.r2) So . . 1 2 T mg 2 r r . . . [as F = mg] i.e., T = . . 3.97 980 3.14 8.5 8.7 . . . = 72.13 dyne/cm
description
Physics
Transcript of ref phy 18
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. . . . C26:What should be themaximumdiameter of awire of length 1metre and density6 × 103 kgm�3which can float horizontally on the surface of a liquid of surface tension 80 × 10�3N/m(neglect the buoyancy effect) Sol. Here 2 d L g T 2L 2 .. . . . . .. .. . 3 3 d 8T 8 80 10 1.86 10 metre. g 22 6 10 9.8 7 . . . . . . . .. . . . C27: Explain the following : �Athinsteelneedle floats onwater but whena little soap solution is carefullymixed with thewater the needle sinks.� Sol. When a needle is placed gently on the surface ofwater, as no part of it is submerged inwater, it is not buoyed up byArchemedes�principle.There is ofcourse onwayfor steel to float byArchimedes�principle.Since its density is greater than that ofwater so it will sink if submerged. mg T A T It is actuallykept afloat by the surface tension ofwater.When it is placed on the surface ofwater, it depresses the surface ofwater slightlydue to itsweight.The stretchedsurface due to surface tension exerts a restoring force, the vertical component ofwhich canmaintainequilibriumwith the weight ofthe needle. However,when a detergent is added towater its surface tensionwillsuddelydecrease and the force due to surface tension is no longer sufficient to support theweight of the needle and so it will sink. Example 49. Aring is cut froma platinumtube of 8.5 cminternal and 8.7 cmexternaldiameter. It is supported horizontallyfroma pan of a balance so that it comes incontact withthewater ina glass vessel. What is the surface tension ofwater of an extra 3.97 g weight is required to pull it away fromwater ? (g = 980 cm/s2) F T T F Cross Section Sol. The ring is in contact with water along its inner and outer circumference; sowhenpulledout thetotalforce onit due to surface tensionwillbe F = T (2.r1 + 2.r2) So . . 1 2 T mg 2 r r . . . [as F = mg] i.e., T = . . 3.97 980 3.14 8.5 8.7 . . . = 72.13 dyne/cm
FLUID MECHANICS www.physicsashok.in 51 Surface Energy Liquid surface in contact with vapour is usually called a free surface. It is in fact an interface of liquid and vapour.The surface in contact withsolid is also aninterface of solid and liquid.Themolecules inthe surface interact with both similar and dissimilar molecules. The forces between similarmolecules use cohesive forces and those between dissimilar molecules are adhesive forces. If one removes a molecule froma interface to infinity, one has to spend an energy ofvaporization, equal to work done against the attractive forces ofcohesion and adhesion. What happenswhen amolecule inthe body of liquid comes to the surface?To see it let us inquire into its surround in the two states. Themolecular force varieswith distance and vanishes if distance betweenmolecule (r) is large (see figure). A r0 r B R r Fr ThemoleculeAis fixed at the origin and distance ofmoleculeB from F Ais r.At r = r0 themilecules are at equilibriumdistance havingminimum potential energyofmolecular force. For r < r0, repulsion and for r > r0 attraction results.We can drawa sphere aroundAofradiusRsuch that r > Rmeans negligible force. This sphere is called sphere of influence.Amolecule within the sphere of influence interactswithAwhile that outside it does not. Amolecule in the body of liquid (well belowthe surface) is surrounded by almost equaldensity of similar molecules.The resultant cohesive force is almost zero.This is the rate of thatmolecule alsowhose sphere of influence is almost touching the free surface [Fig (a) and (b)]. (a) (b) As themilecule rises towards the surface, the sphere of influences rises above the surface. The number of molecules in the upper hemisphere becomes less than those is the lower hemisphere [Fig (c) and (d)]. The molecule feels aninward force against which it has to rise. (c) (d) It iswill lose kinetic energyand average kinetic energyper particlewill fall; coolingwill occur.Thus ifwe increase the free surface area ofa given volume ofliquid,molecules in thebodyare brought to surface losing the kinetic energyin the formofenergyofpotential formcalled surface energy.This increment in free surface area is attended by fall in internal kinetic energy (and hence cooling) and increase in surface energy. The surface energy is the result of intermolecular forces and depends upon the nature ofmolecules (host and foreign) and their spacing (temperature). It is associatedwith free surface of solids also. Inthe figuremolecule Aofmedium�1 has a stored energylike themoleculeBofmedium�2. Whenconsideringphenomena onthe interfacebetweenvariousmedia, × × × × × × × × × × × × × × × × × × × × ×
× × × × × × × × × × × × × ×× × A r1 1 B r2 2 × × × × × × wemust keep inmind the surface energy of a liquid aswell as solidliquid and solid-gas interfaces.
FLUID MECHANICS www.physicsashok.in 52 C28: PQRS represents a framewhich is kept vertically.Over this frame a rodMN of length 20 cmcan slide as shown. The whole system is dipped in a soap solution and thenwithdrawn.What should be themass of the bloockCwhich has to be suspended fromthe wire in order to keep it in equilibrium? [S.T. of soap solution = 30 × 10�3 N/m; neglect friction] P S M N Q R C Sol. Here the force exerted by the S.T. = 2TL For equilibriumof the rod, Mg = 2TL WhereMis the mass of the block C. . 3 M 2TL 2 30 10 0.2 1.22 10 3 kg g 9.8. . . . . . . . . Example 50. Adrop of radius 4 cmis broken into 125 equal small drops. Calculate thework done if surface tension ofwater is 75 × 10�3N/m. Sol. Let us suppose that, R = Radius of the big drop, and r = radius of the small drop. Since volume remains constant, V= nV1 4 R3 n 4 r3 3 3 . . . . 1R . n3r Hence, increase in surface area = 4.r2n � 4.R2 = 4. 2 2/3 R n n � 4.R2 = 4.R2 . . n1/3 .1 Thereforework done . 4.R2 .n1/ 3 .1.T . 4.R2 .(53 )1/3 .1.T = 4.R2[5 � 1]T = 16 .R2T C29: Calculate thework done in breaking awater drop of radius 4 cminto 64 small drops, surfae tension of water is 75 × 10�3 N/m. Sol. W . 4.R2 .n1/ 3 .1.T . 4.R2 .(43 )1/ 3 .1.T 2 12 R2T 12 4 75 10 3 4.52 10 3 joule 100 . . . . . . .. .. . . . . . . . . C30: Calculate thework done in breaking a soap-bubble of radius 10 cminto 512 small bubbles; S.T. of soap solution is 30 × 10�3N/m. Sol. W . 2 . 4.R2 .n1/ 3 .1.T W . 2 . 4.R2 ..83 .1/3 .1.T . 56 .R2T
FLUID MECHANICS www.physicsashok.in 53 C31: Calculate the surface energy released if eight small drops ofwater each of radius 6 cmcombines to form a big drop. Sol. Let r and R be the radius of small drop and big drop respectivelythen as the volume remains constant, 4 R2 n 4 r3 3 3 . . . . R . n1/3r Total surface area of n small drops = 4. r2n and surface area of big drop = 4.R2 Total release of surface energy = [4.r2n � 4.R2]T = 4.[r2n � n2/3r2]T = 4.r2[n � n2/3]T = 4.r2[8 � (23)2/3]T = 4.r2[8 � 4]T = 16 .r2T Example 51.The surface tension ofwater is 0.073Nm�1. 8 drops ofwater, eachof radius 1mm,merge to form a single drop.Determine the change in internal energyofthe drops. Sol. Surface energylost due tomerging goes into the formof change in internal energy. Surface energy befoce themerging = 8 × 4. (1mm)2 × 0.073 (Nm�1). The eight dropsmerge to forma single drop. Hence, 8 × 43 . (1 mm)3 = 43 .r3 r = 2 mm New surface area = 4..(2mm)2. Surface energyafter themerger = 4..(2mm)2 × 0.073 Nm�1 Lost surface energy= [8 × 4. (1 mm)2 × 0.073 (Nm�1)] � [4. × (2 mm)2 × 0.073 Nm�1] = 4. × 0.073 [8 × 106 � 4 × 10�6] J = 4. × 0.073 × 4 × 10�6 J = 3.67 × 10�6 J The change in internal energyis, thus, 3.67 × 10�3 J EXCESS PRESSURE IN A LIQUID DROP liquid drop is sphericalwhen effect of gravity ismasked by surface tension effect, or when it is in gravity free space. Let P be the pressure inside the drop and P0 outside the drop. P P0 Liquid Drop P R2 P R2 0 Sdl P0 ds We consider equilibriumof half drop (hemisphere) under the forces listed below: (a) The liquid on the lift exerts normal contact force on the right part, equal to P × .R2.
FLUID MECHANICS www.physicsashok.in 54 (b) Air on the right exerts normal contact force P0 ds ... at each area element ds, whose resultant is P0 × .R2, acting leftward. (c) The free surface of left hemisphere pulls the free surface of right hemisphere at the contact circle with leftward force S dl over length dlof the circlewhose sumover thewhole circle is S × 2 .R. Thus, balancing these forces leads to S × 2.R + P0 .R2 = P .R2 2S/R + P0 = P. We see that pressure P is greater than outside pressure.The excess pressure P� P0 is given by P � P0 = 2S/R This is true also for an air drop enclosed in a liquid. BUBBLE Abubble has two free surfaces, externaland internal. Thickness ofliquid filmis negligible relative to radius of the bubble.Hence we takeR as radius of both the outer and the inner surfaces. P0 inner free surface P R2 Sdl P0 ds P Liquid outer free surface Pds The equilibriumofhalf bubble is under the forces� (a) Resultant of Pds�forces, which equals P.R2 (right) (b) Resultant of P0ds�forces, which equals P0 × .R2 (left) (c) Resultant of S dl�forces on two free surfaces,which equals 2 × 2.rS. Balancing these P × .R2 = P0 × .R2 + 4.R2S P � P0 = 4S/R The exces pressure is twice that in a drop. C32: Compare the state of a soap-bubble with that of a rubber balloon in following respects : (a) Has each surface tension ? (b) Does the surface tension depend on area ? (c) isHooke�s lawapplicable ? Sol. (a) The soap bubble (having two free surface of a liquid) has surface tensionwhile the balloon beingmade of rubber has no surface tension but tension due to elasticity. (b) Surface tensionis independent of the area of the soap filmwhile the tensionin the balloon is proportional to stretch, i.e., change in area. (c) As in case of bubble surface tension is independent of stretch, Hooke�s lawis not applicable. However, in case of balloon as tensionarises due to elasticityand is proportional to stretch,Hooke�s lawis applicable. C33: Calculate excess pressure inside a soap bubble of radius 4 cm; surface tension of soap solution is 30 dyne/cm. Sol. Excess pressure P 4T 4 30
r 4 . . . = 30 dyne/cm2.
FLUID MECHANICS www.physicsashok.in 55 Example 52. The lower end of a capillary tube of diameter 2.00mmis dipped 8.00 cmbelowthe surface of water in a beaker. Find the pressure required in the tube in order to blowa hemispherical bubble at its end in order to blow a hemispherical bubble at its end inwater.Also calculate the excess pressure. Surface tension ofwater is 7.30 × 10�2 N/mand atmospheric pressure is 1.01 × 105 Pa. Sol. The excess pressure in a bubble of gas in a liquid is 2S/r, where S is the surface tension of the liquid-gas interface and r is the radius of the bubble. The pressure outside the bubble equals the atmospheric pressure plus the pressure due to 8.00 cmofwater column : pout = 1.01 × 105 + 0.08 × 1000 × 9.8 pout = 1.01784 × 105 Pa The pressure inside the bubble is pin = pout + 2S/r Pin = 1.01784 × 105 + (2 × 7.3 × 10�2/10�7) Pin = (1.01784 + 0.00146) × 105 Pin = 1.02 × 105 Pa. Here the radius of the bubble is taken to be equal to the radius of the capillary, since the bubble is hemispherical. Excess pressure inside the bubble = 0.00146 × 105 Pa = 146 Pa. Example 53. Amercury drop of radius 1 cm is sprayed into 106 drops of equal size. Calculate the energy expended. Surface tension ofmercury is 35 × 10�5N/m. Sol. Let R be the radius of the big drop and r be the radius of a small drop. Since the total volume remains conserved on the formation of small drops. 4 r3 106 4 R3 3 3 . . . . or r = R × 10�2 = 0.01 cm Energy expended = Surface tension × Increase in surface area = T × 4.(R2 � 106 r2) = 35 × 10�3 × 4 × 3.14 [106 (0.01 × 10�2)2 � (10�2)2] = 4.35 × 10�3 J Example 54. Aglass plate of length 10 cm, breath 1.54 cmand thickness 0.20 cmweights 8.2 g in air. It is held verticallywith the long side horizontal and the lower half underwater. Find the apparent weight of the plate. Surface tension ofwater is 7.3 × 10�2N/m. Sol. Apparent weight =Trueweight + Force due to surface tension �Buoyant force True weight = mg = 8.2 × 10�3 kg�wt Force due to surface tension = 7.3 × 10�2 × 2 (10 + 0.2) × 10�2 = 14.892 × 10�3 N = 14.892 10 3 9.8 . . = 1.5196 kg�wt = 1.52 kg wt Buoyant force =Volume of plate immersed inwater ×Densityofwater × g
FLUID MECHANICS www.physicsashok.in 56 = 10 × 0.2 × 1.54 2 × 10�6 × 103 × 9.8 N = 1.54 × 10�3 kg�wt . Apparent weight= (8.2 + 1.5196 � 1.54) × 10�3 = 8.2 kg�wt. Example 55. What is the excess pressure inside a bubble of radius 5.00 mm jormea of a soap solution of surface tension 2.50 × 10�2N/m. If an air bubble of the same dimensionwere formed at a depth of 40.0 cm inside a container containing the soap solution (or relative density 1.20), what be the pressure inside the bubble ? (1 atm= 1.01 × 105 Pa) Sol. Excess pressure inside the soap bubble. 2 3 4S 2 2.5 10 P r 5 10 . . . . . . . Excess pressure inside the air bubble in soap solution p´ 2S 10 Pa r . . Total pressure inside the bubble =Atmospheric pressure + Hydrostatic pressure due to soap solution of height 40 cm+ Excess pressure due to surface tension = 1.01 × 105 + 0.4 × 1200 × 9.8 + 10 = 101000 + 4704 + 10 = 105714 Pa = 1.06 × 105 Pa Example 56. The limbs of a monometer consist of uniform capillary tubes of radii 1.4 × 10�3 m and 7.2 × 10�4 m. Find out the correct pressure difference if the level of the liquid (density 103 kg/m3, surface tension 72 × 10�3 N/m) in narrower tube stands 0.2mabove that in the broader tube. Sol. If p1 and p2 are the pressures in the broader and narrower tubes of radii r1 and r2 respectively, the pressure just belowthemeniscus in the respective tubeswill be 1 1 p 2T r . and 2 2 p 2T r . So that 1 2 1 2 p 2T p 2T h g r r
. . . .
. . . . . . . . .
. . . . h p2 p1 B A or p1 � p2 = h.g � 2T 2 1 1 1 r r . . . . . . . Assuming the angle of contact to be zero, i.e., radius ofmeniscus equal to that of capillary, p1 � p2 = 0.2 × 103 × 9.8 � 2 × 72 × 10�3 4 4 1 1 7.2 10. 14 10. . . . .. . . .. or p1 � p2 = 1960 � 97 = 1863 Pa
FLUID MECHANICS www.physicsashok.in 57 Example 57. Two separate air bubbles (radii 0.02mand 0.004m) formed of the same liquid (surface tension 0.07N/m) come together to forma double bubble. Find the radius and the sense of curvature ofthe internal filmsurface common to both the bubbles. Sol. If r1 and r2 are the radiiof smaller and larger bubbles and p0 is the atmospheric pressure, the pressure inside themwillbe p1 = p0 + 1 4T r and p2 = p0 + 2 4T r ...(1) Nowas the pressure inside the smaller bubblewill be more than inside the larger bubble, so for interface, p = p1 � p2 ...(2) Nowas excess pressure acts fromconcave to convex side, the interface willbe concave towards smaller bubble and convex towards larger bubble (as shown in Fig.) and ifRis the radius of interface, p = (4T/R) ...(3) So substituting Eqns. (1) and (3) in (2), we get p2 p1 r1 p R r2 1 2 4T 4T 1 1 R r r . . . . . . . . , i.e., . . 1 2 2 1 R r r r r . . So here R 0.002 0.004 0.004 m 0.004 � 0.002 . . . Example 58. Under isothermal condition two soap bubbles of radii a and b coalesce to forma single bubble of radius c. If the external pressure is p0 showthat surface tension, 3 3 3 0 2 2 2 T p (c a b ) 4(a b c ) . . .
. . Sol. As excess pressure for a soap bubble is (4T/r) and external pressure p0, pi = p0 + (4T/r) so a 0 b 0 p p 4T , p p 4T a b . . . . . . . . .. .. .. .. and c 0 p p 4T c . . . . .. .. ...(1) and 3 a V 4 a 3 . . , 3 b V 4 b 3 . . and 3 c V 4 c 3 . . ...(2) Nowasmass is conserved, µa + µb= µc i.e., a a b b c c a b c p V p V p V RT RT RT . . as PV µRT, i.e., µ pV RT . . . . .. .. As temp. is constant, i.e., Ta = Tb= Tc, so the above expression reduces to paVa + pbVb = pcVc
FLUID MECHANICS www.physicsashok.in 58 Which in the light ofEqn. (1) and (2) becomes 3 3 3 0 0 0 p 4T 4 a p 4T 4 b p 4T 4 c a 3 b 3 c 3 . . . . . . . . . . . . . . . . . . . . .. .. .. .. .. .. .. .. .. .. .. .. i.e., 4T(a2 + b2 � c2) = p0(c3 � a3 � b3) i.e., . . . . 3 3 3 0 2 2 2 p c a b T 4 a b c . . . . . ANGLE OF CONTACT The free surface of a liquid in contact with a solidwallmeets at some angle. The angle between tangents drawn in the free surface and solid,measured inside the liquid, is known as angle of contact. < 90º 90º > 90º glass-water silver-water glass-mercury In the figure, . is the angle of contact. The angle of contact may be acute, right angle or obtuse. Tangent in the free surface drawn through contact Tangent along the solid Howdoes angle of contact arise ? In the figure three substances, namely, a liquid (i), a solid (2), and a gas (3), are in direct contact with one another. They produce such an orientation of free surface at contact line as to haveminimumof the total energy. this is required for stability. This happens when the line of contact of the three substances are oriented on the surface of the solid in such away that the sumof the surface tension forces acting on its length element equals zero. 1 3 2 S13 S23 Solid S12 1 3 2 S13 S23 S12 Let S12, S13 and S23 be the surface tensions at the solid-liquid, liquid-gas and solid-gas interfaces.
Balancing forces along the solid surface gives S23 = S13 cos . + S12 23 12 13 cos S S S. . .
FLUID MECHANICS www.physicsashok.in 59 The angle ..ismeasured inside the liqiud between tangents to the surface ofthe solid and the surface of the liquid and is called the contact angle. Since | cos ..| ..1, the contact angle is obtained if 23 12 13 | S S | 1 S. . If this condition is not satisfied, equilibriumcannot set in at any value of .; the contact linewill accelerate. Wetting and Non-wetting Waterwets glass butmercurrydoes not.At water-air-glass contour, the angle of contact is tending to zero andwetting takes place. the contact angle is zero in completewetting. The inward adhesive force by solid ismore thannormal cohesive force. For any value of ...close to ., the liquid-solid interface tends to contract into a point, and the liquid separates fromthe surface of the solid. Complete non-wetting takes placewhen the contact angle is ..The adhesive force is smaller than normal cohesive force. Capillarity The Latin capillesmeans hair. Literally, capillary is a �tube as thin as a hair�. Usual capillariesmay have larger bore thanhair. If such a tube of glass be immersed inwater, water rises in it. If this tube be immersed inmercury, the levelis depressed. Suchactions are knownas capillarityormeniscus effect.Wemayunderstand capillaryaction inthe followingway. The existence of the contact angle leads to curvature of the surface of a liquid near thewalls of the vessel containing it. In a narrowtube (capillary) or ina narrowgap between twowalls, the entire surface is curved. If the liquidwets thewalls, the surface is concave, and if it does not wet them, the surface is convex. Such curved surfaces of a liquid are called meniscuses. For a narrowtube it maybe assumed spherical. h P0 P0 R Meniscus P0 � 2S R < 2 ; h > 0 Meniscus h > 2 ; h < 0 Using pressure concept,we have high pressure on concave side.Tomeet this requirement hydrostatically, themeniscus rises (or falls) by suitable height (or depth). A difference h sets in between the level of a liquid in a capillary and in the broad vessel such that the hydrostatics pressure .gh is compensating it bythe capillary pressure : 0 0 P 2S gh P
R . . . . gh 2S R . . In this euation, S is the surface tension on the liquid-gas interface, andRis the radius of curvature of the meniscus. The latter (R) can be expressed through the contact angle ..and the radius of the capillary r.
FLUID MECHANICS www.physicsashok.in 60 Indeed, figure shows that R = r/cos.. Using this value inwe arrive at the equation h 2Scos gr . . . This is known as Jurin�s law. Ifwe consider effect of liquid between lower level ofmeniscus and the free surface, h is replaced byh+r/3 for sphericalmeniscus. Inaccordancewith the fact that awetting liquid rises in a capillary, while a non-wetting liquid lowers in it, gives a positive h for . < ./2 (because cos .> 0)and negative h for . > ./2 (because cos . < 0) the radius rmust bemeasured at the position of themeniscus. [Note :We assumed here that themeniscus has a spherical shape. The equation for h canalso be obtained on the basis of energyconsiderations, and there is no need to make a special assumption on the shape of the meniscus. The equilibriumposition of themeniscuswill correspond to aminimumenergyEof the liquidcapillary system. This energyE is the sumofthe surface energy of the liquid-wall, liquid-gas andwall-gas interfaces, and also of the potentialenergy of the liquid in the field ofthe Earth�s gravitation.] Surface Tension events Even Informations Completewetting, capillaryrise ..= 0 wetting, capillaryrise ..< 2. non-wetting, no capilarity . = 2. non-wetting, capillarydepression . > 2. Complete non-wetting, capillarydepression . = . equilibriumstates sg SL g S S cos S. . . l ; | cos ..| ..1 non-equilibriumstates cos . > 1 Capillary rise 2Scos h r rg 3 . . . . LiquidDrop P � P0 = 2S R Bubble P � P0 =
4S R Variation of surface tension : stronger the intermolecular force of a liquid larger is the surface tension. It is dependent upon temperature (decreases on rising temperature) and contamination (decreases bymixing detergent inwater)
FLUID MECHANICS www.physicsashok.in 61 Surface Tension events Substance Surface Tension (Jm�2) Water 7.29 × 10�2 Mercury 4.6 × 10�1 Benzene 2.89 × 10�2 Ethanol 2.23 × 10�2 Glycerol 6.34 × 10�2 Oxygen (�183ºC) 1.32 × 10�2 Helium(�270ºC) 2.39 × 10�4 C34: Calculate the rise ofwater in a capillary tube of radius 2mm, assuming the angle of contact to be 0º; S.T. of water is 72 × 10�3 N/m. Sol. Here 2.r T cos . = .r2h.g If . = 0º, cos . = 1 then, 3 3 3 3 h 2T 2 72 10 7.34 10 metre. r g 2 10 10 9.8 . . . . . . . . . . . . C35: Acapillary tube of radius 2 mmis dipped in mercury kept in a vessel. Calculate the depression of the mercuryin the capillarytubewith respect to the level inthe vessel[angle ofcontact ofHgwithglass is 135º, and S.T. ofHg is 465 dyne/cm.] Sol. Here, h 2Tcos 2T cos135º 2T cos(90º 45º ) r g r g r g . . . . . . . . 2Tcos 45º 2 455 1/ 2 h 24.67 cm r g 0.2 13.6 9.7 . . . . . . . . . . C36: In a vessel equalmass of alcohal (sp. gravity0.8) andwater aremixed together.Acapillary tube of radius 1mmis dipped vertically in it. If themixture rises to a height 5 cm. in the capillarytube, then calculate the S.T. of themixture; (assuming angle of contact to be 0º). Sol. 2.rTcos . = .r2h.g . T = rh g 2 cos . . but 1 2 3 1 2 2 2 0.8 1 1.6 8 gm/ cm 1 0.8 1.8 9
. . . .
. . . . .
. . . .
. T = 0.1 5 8 / 9 980 2 . . . = 217.9 dyne/cm. Example 59. Aglass capillarysealed at the upper end is of length 0.11mand internaldiameter 2 × 10�5m. The tube is immersed verticallyinto a liquid of surface tension5.06 × 10�2N/m.Towhat length has the capillary to be immersed so that the liquid level inside and outside the capillarybecomes the same ?Whatwillhappen to thewater level inside the capillaryif the seal is nowbroken ?
FLUID MECHANICS www.physicsashok.in 62 Sol. IfAis the cross-sectional area of the tube and Lits length, the initial volume of air inside it will beV1=AL while pressure p1 = p0 = atmospheric pressure. Nowwhen the tube is immersed inwater with its length x inwater, the level ofwater inside and outside is same; so the volume of air in the tubewill beV2 =A(L� x). Further if p2 is the pressure of gas in the tube, 2 0 p 2T p r . . , i.e., 2 0 p p 2Tr . . Nowif temperature is constant, .............. ....... .. .. .............. ...... . ... . ................... .............. . .. ... . ... . ...... . ... . ...... . ... . ................... .. . ........... .......... p(L�x) 2 x p0 p2� 2Tr P1V1 = P2V2 0 0 p AL p 2T A(L x) r . . . . . .. .. or 0 x 1 rp L 2T . . . . .. .. i.e., 5 5 2 x 1 1.012 10 1 10 0.11 2 5.06 10 . . . . . . . . . . . . . . . or x 0.11 0.01m 11 . . If the sealis broken the pressure inside the capillarywill become atmospheric, i.e., p0while capillaritywill take place and the risewill be 2 5 3 h 2T 2 5.06 10 1.03m r g 10 10 9.8 . .
. .
. . .
. . . However, the lengthof the tube outside thewater is 0.11 � 0.001 = 0.1m; so the tubewillbe of insufficient length and so the liquidwill rise to the top of the tube andwill stay therewithradius ofmeniscus, 5 hR 1.03 10 4 r 1.03 10 m L 0.1 . . . . . . . Example 60. Aconical glass capillary tube of length 0.1 mhas diameters 10�3 and 5 × 10�4 m at the ends. When it is just immersed in a liquid at 0ºCwith larger diameter in contact with it, the liquid rises to 8 × 10� 2min the tube. If another cylindrical glass capillarytubeB is immersed in the same liquid at 0º C, the liquid rises to 6 × 10�2 mheight. The rise of liquid in the tube B is only 5.5 × 10�2 mwhen the liquid is at 50ºC. Find the rate at which the surface tension changeswith temperature considering the change to be linear.The densityof the liquid is (1/14) × 104 kg/m3 and angle of contact is zero. Effect of temperature on density of liquid and glass is negligible. Sol. If r is the radius of themeniscus in the conical tube, then as shown in Fig. 1 2 1 tan r r r r L h L . . . . . . i.e., r 2.5 10 4 (5 2.5) 10 4 0.1 0.08 0.1 . . . . . . . . r r1 r2 LA B h i.e., r × 104 � 2.5 = 0.2 × 2.5 i.e., r = 3 × 10�4 m Nowas capillarity is independent of the shape of tube so at same temp. . = 0º C.
FLUID MECHANICS www.physicsashok.in 63 hArA = hBrB = (2T0/.g) = constant. so rB = (0.08 × 3 × 10�4)/(6 × 10�2) = 4 × 10�4 m Nowas fromh= (2T/r.g) for cylindrical tube, 0 2 4 4 0 h gr 1 1 T 6 10 10 9.8 4 10 2 2 14 . . . . . . . . . . . . . . . . . T0 = 8.4 × 10�2 N/m Nowas for a given tube and liquid T . h (as T = h.gr/2) 50 50 0 0 T h T h . So, 2 50 2 5.5 10 T 6 10 . . . . . × 8.4 × 10�2 = 7.7 × 10�2 N/m So rate ofchange of surface tensionwith temperature assuming linearity, 2 50 0 T T T (7.7 8.4) 10 50 0 50 . . . . . . . .. . = �1.4 × 10�2 N/mºC Ans. Negative sign shows that with rise in temperature surface tension decreases.7 Example 61. Aconical glass capillary tubeAof length 0.1mhas diameters 10�3 mand 5 × 10�4mat the ends. When it is just immersed in a liquid at 0ºC with larger diameter in contact with it, the liquid rises by 8 × 10�2 min the tube. In another cylindrical glass capillarytube B, when immersed in the same liquid at 0ºC, the liquid rises by 6 × 10�2 m. The rise of liquid in tube B is only 5.5 × 10�2 mwhen the liquid is at 50ºC. Find the rate at which the surface tension chargeswith temperature considering the change to be linear.The densityof the liquid is (1/14) × 104 kg/m3 and the angle of contact is zero.Effect of temperature on the density of liquid and glass is negligible. Sol. Let r1 and r2 be the radii of the lower and the upper ends of the capillary. The radius r at height h at the position of themeniscus is given by 1 1 2 r r r r ( tan ) h. . . . . l r = r1 � (r1 � r2) (h / l)
r = 0.5 × 10�3 � (0.25 0.5) 0.25 . × 0.8 × 10�1 r = 0.3 × 10�3 m r r2B r1 h l A Surface tension of the liquid at 0ºC �3 2 4 0 rh g 0.3 10 8 10 10 9.8 S 2cos 2 14 1 . . . . . . . . . . . . . (. . = 0) S0 = 0.084 N/m If r´ is the radius of the cylindrical tube B, then at 0º C 0 0 S r´h g 2. .
FLUID MECHANICS www.physicsashok.in 64 and at 50ºC 50 50 S r´h g 2 . . . 2 50 50 2 0 0 S h 5.5 10 11 S h 6 10 12 . . . . . . . or 50 0 S 11 S 11 0.084 0.077 N /m 12 12 . . . . . Rate of change of surface tensionwith temperature, assumed constant for S (.) as a linear function,willbe 50 0 1 �1 S S 0.077 0.084 Nm K 50 50 . . . . = �1.4 × 10�4 Nm�1 Cº�1 Thus, the surface tension decreases at the rate of 1.4 × 10�4 N/mper degree celsius rise in temperature.
FLUID MECHANICS www.physicsashok.in 65 ELASTICITY ELASTIC BEHAVIOUR The length of a steelwire canbe increased by applying forces at its ends. Inthe figure a steelwire of length l is fitted with a vernier scale V by the side ofmain scaleM. By putting a loadW, the length l becomes l + .l as indicated by reading of the vernier scale. l l + l V M W If loadWis decreased, the value of .l also decreases.Whenwbecomes zero, .l also goes to zero. The wire has returned to its original shape and size. There are severalbodies around us, which behave like a steelwirewe described just now. This behaviour is known as elastic behaviour or elasticity.The property of certainmaterials to regain their shape and size when the cause of deformation is removed, is known as elasticity. Thematerial showing elasticity is called elastic. In deformation, the particles of the body are diplaced fromtheir equilbriumpositions to newpositions. These newpositions give rise to newshape and size of the body. The forces between particles tryto bring themback to original positions. The tendency is decided by howthemolecules act an each other during deformation. Thesemolecular forces are called internal forces that oppose the applied forces andwant to restore the old shape and ize (see figure). When the cause of deformation ceases, there are backmotions of displaced particles. If the particles return back reversibly, the deformation is categorised as elastic. Amodel ofmolecules under spring-like forces of othermolecules is as shown here for simplicity sake. If atoms (black balls) are displaced and released, the springs force themto come back. Such forces are at the root ofelastic behaviour. It shows that it is amolecular (atomic) levelphenomenon, electromagnetic forces governing thewhole game. Condon andMorse proposed a force derivable frompotential. U repulsion attraction r0 r m n U A B r r . . (A, B, m, n are suitable constants) This potentialhas both repulsive and attractive parts, giving spring-like action. In certainmaterials, particles are pushed to new arrangement of equilibriumduring deformation. Hence upon removal of cause of deformation, the body does not return to previous shape and size fully. Such
bodies are called plastic. If you press a putty or mud, and remove your force, it stays in the �pressed position�. Thus it is plastic. The deformation is irreversible.
FLUID MECHANICS www.physicsashok.in 66 We can observe a mixture of elasticity and plasticity in the same body. It depends of upon the extent of deformation. For smalldeformation elastic behaviour is observedwhile amixture atmore deformation and purely plastic at large deformation.Wemust performexperiments to knowthe actual elastic behaviour of bulkmatter. STRESS AND STRAIN Elastic behaviour is described by answering two questions : (a) Howmuch is deformation produced. (b) Howmuch is opposition to the deformation. There are two terms��stress� and �strain��is answer such questions. �Strain� is related to tirst questionwhile stress to the second. Stress : If a body is deformed, forces arise that want to restore the equilibriumconfiguration.These are known as internalrestoring forces or elastic forces. Ifwe consider a surface in the bodyit divides the bodyinto two parts, one on one side of the surface and the other on the other side. The two parts exert on each other elastic forces.An areaAofthe surfacemayhave a force Fdistributed onit.The internalelastic force per unit area of a surface is known as stress.We denote it by .. Then F . A . . ASurface A F F A The value of . depends upon the orientation of surface.Hence itmayhappen that (i) F . is normal to tbe surface. In that casewe call . �normal stress�, denoted by .n. (ii) F . is not normal to the surface. In that casewe can resolveF . into two components : Fn, normal to surface and Ft along the surface, Normal stress is defined as normal internal elastic force per unit area. That is, n n F . A . . A FnF Ft Normal stressmaybe arising due to tension in the body. In that case it is called tensile stress.On the other hand, if it arises due to �compression�, it is called compressive stress. It produces a change in size. Tangential Stress is defined as tangential internal elastic force per unit area.That is, t t F . A . .
It is also called shaearing stress. It tries to move the surface tangentially causing a change in shape. In hydrostatics, the normal force per unit area of a surface is pressure. The value of pressure at a point does not depend upon choice of the orientation of the surface. The normal stress however, is not pressure because its value depends upon choice of orientation of the surface. It may in same cases coincidewith pressure. Thus, stress, although has the same unit and deminsions as the pressure, is not a scalar, nor it is a vector. The complete specification of �stress�requires description of both normal and tangential stresses, which depend upon orientation of surface and are known to behave as tensors.
FLUID MECHANICS www.physicsashok.in 67 There are situationswhere interaction of one part of a bodyon the other part is not only a force but also by a couple. Such couplemoments are distributed in the body, for example in a polar dielectric in electric field or in amagnet in amagnetic field. In those cases couplemovement per unit area is also needed to describe the stress. Stress is distributed in the volume the body.Wemust consider the particular point and force at that point for finding out stress at that point of the body. Example 62. Athick book likeRamcharitManas is placed on a rough table. It is pushed slant byhand as in the figure.We find that its base remains fixed but upper layers shift.Also the thickness is slightlyreduced due to pressing. Consider ahand areaof A = 5 cm2 in contact and applying a force of F = 10Nat 60ºwith normal. Calculate normal stress and tangential stress. Ft Fn Ft Fn 60º A rough Sol. Normal stress n n FA . . Here Fn = F cos 60º Fn = 10 × 1/2 = 5 N A = 5 cm2 = 5 × 10�4 m2 . n 4 2 5N 5 10. m . . . .n = 104 Nm�2 .n = 104 Pa (1 pascal = 1 Pa = 1 Nm�2) Shearing stress, t t FA . . Here Ft = F sin 60º = 5 3 N . .t = 3 × 104 Pa = 1.7 × 104 Pa. Example 63. Ascissor fulcrumis a nail of radius r. It is squeezedwithforce F as in the figure byputting a strip to cut. Inequilibrium, calculate the shearing stress developed byblades on the nail. b a FF Sol. Consideringmoment of forces about Oin the figure, O Ft F Ft
Ft f f N a b F (a + b) = Ft b
FLUID MECHANICS www.physicsashok.in 68 t F F 1 ab . . . . .. .. Blades exert forces Ft on area .r2. Hence shearing stress t t 2 2 F F(a b) r r b . . . . . . . Example 64.Awire ofcross-sectional areaAand lengthLis hanging fromceiling.The density of itsmaterial is p. Thegravityis pulling it downwardwhile ceilingis preventing it fromfall. So it is in stressed state. Calculate stress at a cross section located at height x fromlower end. (The force is axial, the stress is tensile). Sol. Here the elastic internal force is tensionforce. Let T be the tension in thewire at a cross section located at height x above the lower end. The tensionT is balancing theweight ofwire of length x.Thisweight is x weight T w= volume × density× g w = xA. . . g . T = xA..g. The stress is given by TA . . ..= x .g. C37:Abeamis placed on two nails PandQ.Aload hung is fromthe centre. There is no friction.Which part of the beamis in tensile stress andwhich part isn compressive stress ? P Q P Q Sol. The part abovemiddle horizontal layer (surface) is �shortened� and is under compressionwhile the part belowthemiddle surface is elongated and undre tension. thus, upper half is in �Compresve stress�while lower half is in �tensile stress�. (Themiddle surface is neutral). Strain :When the particles of a body are displaced fromtheir equilibriumpositions under the intermolecular forces, the bodyis said to be strained.You squeez a rubber ball, it is strained.Youmay pull apart the ends of awire, it is strained. l + l l We can deforma body in three ways : (a) Changing its length. (b) Chang it volume. (c) Twisting it (changing its shape). Longitudinalstrainisdefined as change inlinear dimension per unit ofits originalvalue.The linear dimension may be length of awire, diameter of a cylinder, width of a rectangular bar, etc.
It l be ameasure of linear
FLUID MECHANICS www.physicsashok.in 69 dimention and .l be the change (positive or negative), the ratio . l l is known as longitudinal strain. If the length is decreased due to deformation, .l is negative. Then we call . l l compressive strain. If .l is positive, the length is increased and . l l is tensile strain. If D be dimater of a cylinder and .D be the change in it due to deformation, D D . is known as lateral strain. The ratio of lateral strain to longitudinal strain is defined as Poisson ratio. D µ D . . . .. .. . . . . . .. .. l l Its theoreticalvalue can be �0.5 to 0.5 but practicalvalue is found between zero and half. Theminus sign is put to define µ as positive, .Dand .l are of opposite signs. l l + l D + D D h + h hl b b + b l + l (.l > 0, .D < 0); b µ b . . . . . . .. . . . . . . . . l l D µ D . . . . . . .. . . . . . . . . l l h µ h . . . . . . .. . . . . . . . . l l Volume Strain or, hydraulic strain, is defined as the change in volume per unit originalvolume, due to deformation. IfVbe the volume before the deformation andV+ .Vbe the volume due to the deformation. Then V V . is the volume strain.
V + V V An air bubble at some depth in water has volume V. It is rising up. The pressure is less at less depth. Hence the bubble is relaxed a bit, its volume becomesV+ .V. There is a volume strain of .V/V, which is positive. V + V, V V < 0 We can squeez a rubber ball in which case .V is negative and volume strain is .V/Vis negative.
FLUID MECHANICS www.physicsashok.in 70 Example 65. Ametalwire of length l is stretched by dl. Its Poisson ratio is µ.Determine its volume strain. Sol. The initial volume of thewire isV= .r2l. Here dV = .d (r2l) dV = .(2r dr . l + r2 dl) 2 2 2 dV 2r dr r d V r r . . . .. . . . . . . l l l l dV 2 dr d V r . . l l Nowthe Poisson ratio µ relates dr r to d l l as dr µ d dr dD D r r D D . . . . . . . . . .. . l l . dV 2µ d d V . . . . . .. .. l l l l dV .1 2µ. d V . . l l Volume strainofwire is (1 � 2µ) times the longitudinal strain. (It willbe zero ifµ = 1/2). The Shearing Strain Shear is the kind of deformation inwhich all the flat layers of a body, parallel to certainplane, undergo rigid displacement parallel to one another. During this, the size of layers does not change. In the figure shear of a cubeABCD is shown. The shear plane isAB. Alayer DC is parallel to it. It is displaced to C´D´. There is no change in its size (C´D´ = CD). A B D C A B D D´´A´ B´D´ C´ Ft Ft We observe that the ratio of relative shift in a layer relative to some layer, to the separation between these
two layers is the same for anypair of parallel layers and is called shearing strain. In the figure. DD´ = shift of layer relative toAB. D´´D´= shift of layer relative toA´B´. DD´ D´´D´ tan AD A´B´ . . . . . This ratio is known as �shearing strain�. Let us denote it by .t. C38:Awire of length l and radius ris clamped at one end. It is twisted at the other end byanangle . about its axis.Write the shearing strainproduced, in a small area lying in the cross-section at the end at a distance x from the axis. l
FLUID MECHANICS www.physicsashok.in 71 Sol. The figure shows an area A at position x fromaxis which is parallel to area P at the clamped end. Due to twisting, A goes toA´ bysuffering a displacementAA´ . x . relative to P. The separation between the parallel area P and A is l. Hence shearing strain, t x . . . . l = t A A´ x P Relationship Between Stress and Strain To see the relationship between stress and strain, wemust performexperiment on the given body. V M Aslotted weight Let us take ametalwire,measure its length and diameter, and suspend fromceiling.Nowe attacha vernierscale andmain scale by a sidewire. Suspend a hanger onwhich slottedweight could be put.Adjust the vernier to read zero; this remove calculation for effect ofweight of hanger. Let a loadw be put on the hanger. Then stress is 2 W ( D / 4) . . . The strain is calculated by noting the much (.l) the vernier descends and dividing it by the length ofwire above vernier.This is . . . l l l . IncreasingWin small steps,we get several pairs of . and t . Agraph is plotted between ..and . for various values ofWand corresponding values of . l l . This graph shows elastic behaviour of thematerial. It is dependent uponmaterial.We discuss typicalgraphs for steel wire, rubber band and aorta. The vernier-method described is good for small stretching. Formore sensetive stepwise stretching and measuring tension a tensometer is used. Steel wire : Acommon graph of stress . against stream... for a wire ofmetal has the features shown in the figure. (i) Proportionate region : For lowstress, the stress-strain graph is a straight passing through origin. l A B C D E F
= W ( D ) 2 4 O = l We have stress = (slope) × strain. For the straight. Part the slope is constant Hence stress . strain. This behaviour is called proportionalityofstress to strain. The regionOAis, therefore, called proportionate region.
FLUID MECHANICS www.physicsashok.in 72 (ii) Elastic region : If the cause of deformation be removed, the body returns reversibly (along to the same states throughwhich it went on deforming) to original shape and size. This behaviour occurs upto a stress called elastic limit. Inthe figureOAandABare suchregions,Bis calledelastic limit andAis called proportional limt. In the region between proportional limt and elastic limit, the graph is curved and hence stress is not proportionalto strain, although it growswith growing strain. If the stress is further increased, the plastic and elastic behaviourmix up.Asmaterialyields to plasticity (agrees to become slightlyplastic also), the elastic limit is also known as yield point. The stress at yield point is known as yield strength, .y. (iii) Plastic region : As the material is deformed beyond yield point (elastic limit), there is a few degree of structural change in equilibriumpositions ofits particles causing plasticity. If the stress is lowered, the body returns towards initialconfiguration but not totally.When stress is zero, a permanent deformationstayswith the body, called permanent set (OF). In the graph,BDis the regionof plasticity.The stress increaseswith strain upto anultimate value known as ultimate tensile strength, .u (SeeTable ). Table : Some Elasticity Data Material Ultimate Strength Yield Strength (GPa) (GPa) Aluminium 110 95 Copper 400 200 Iron(Wrought) 330 170 Steel 400 250 (iv) Fluid region : Beyond the point where stress is equal to ultimate tensile stress, .u, thematerial undergoes deformation even for smaller value of stress than .u. It acts as if it is flowing. Thewire thickness goes on reducing, called neck formation.At a certain point, it breaks. This point is called fracture point (E) The corresponding stress is breaking stress, .b. The important offluid regionis that it classifiesmaterialinto�brittle� and�ductile�. If the point of ultimate tensile stress and fracture point are veryclass to each other thematerial is brittle, like brass, cast iron. If these points are for apart, thematerial is ductile.That is, it canbe drawn into wire form. If youwant to test the proportionate behaviour of a steelwire in laboratory, the allowed load is upto half of that causing fracture. Elastomers : If you stretch a steelwire, a large force will cause a small strain.However, a rubber wire can be stretched by a small force to have a large strain. There are several such substances. The vessels carrying blood fromheart, called aorta, have tissues that showlarge reversible strain for smallstresses,without well defined plasticity. Suchsubstances are called elastomers. 0.5 (x102 Pa)
0.5 1.0
FLUID MECHANICS www.physicsashok.in 73 The stress�strain graphs of elastomers are not straight lines, yet they are elastic. The tissue of aorta has a graph of stress against strain as inthe figure (a); the figue (b) shows that for a rubber. Governing the elastic behaviour : The elastic behaviour depends upon the following factors : (a)Temperature (b) presence of notches (c) time rate of loading (d) presence ofembrittling agent like hydrogen (e) loadingand unloading (f) structure ofmaterial. HOOKE�S LAW Robert Hookewas the first person to notice the propertyof elasticity. Based on experiments on springs and springybodies, he announced, �Ut tensio sic vis�. (The stretching of a spring body is as the applied force). This law has been given more precise form now and is known as Hooke�s law :Within elastic limit stress is directly proportionalto strain. If stress be . and strain be . then ..... or, ..= E .. Here the proportionalityconstant is E which depends upon thematerial and the type of stress.We call it modulus ofelasticity. Hooke�s lawneed not be obeyed for a body to be elastic.We knowthat up to elastic limit ametalwire remains elastic.WhileHooke�s lawis valid onlywithin proportionate region.Also there are severalmaterials which are elastic but have non-linear stress strain graphs. The elastic materials obeyingHooke�s law are known asHookean and those not obeying it are known asHencky. For Henckymaterialswe have . = a .n where a and n are material constants. Modulus of elasticity for such bodies are defined by slopes of tangent and secant of strain-stress graph.We shall not consider these further. DifferentModuli of Elasticity : There are three types of stress�longitudinal, bulk and shearing and there are threemodulidefined for these. (a)Young�smodulus,Y, is the ratio of longitudinal stress to longitudinal strain. Ifelastic normal force Fn acts on areasAand . l l be strain, then . . n F / A Y / . . l l (b) Bulkmodulus, B, is the ratio of volume stress change in pressure to volume strain. If a change in pressure dP causes a change in volumeVby dV, the strain is (dV/V) and bulkmodulus B is given by . . B dP dV/ V . .
whereminus signin put to keepBa positive quantity. Inverse of bulkmodulus is known as compressibility, K. That is, K dV VdP . .
FLUID MECHANICS www.physicsashok.in 74 (c) Rigiditymodulus,G, is the ratio of shear stress to the shear strain. If elastic internal force Ft acts parallel to areaA, and . be the angular shift in the layer,we get G . Ft /A . Table shows elastic constants of somematerials. Units and dimensionsofElasticmoduli : AsE = Stress Strain , and strain is unitless and dimensionless, Ehas the same unit and dimensions as stress,which is the ratio of force and area. Thus, the SI unit of modulus of elasticity is Nm�2 or Pa (passcal). The dimensional formula is [ML�1 T�2] Example 66. The following four wires are made of the same material.Which of these will have the largest extensionwhen the same tension is applied ? (A) length= 50 cm, diameter = 0.5mm (B) length= 100 cm, diameter = 1mm (C) length = 200 cm, diameter = 2mm (D) length= 300 cm, diameter = 3mm Sol. [A] Y T / A/ . .l l . T T A Y Y A . . . . . . l l l Here TY is constant. Therefore A .l . l . Al is largest in the first case. MEANING OF A MORE ELASTIC BODY In commonman�s languagemore elastic bodies are easily elongated to larger extent. However in physics, elasticity ismeasure of opposition of to a deformation. Harder to deforma body,more elastic is the body. Thusmore elastic bodieswillhave higher value ofmodulus of elasticity. This iswaywe say that steel (Y= 200 GPa) ismore elastic than rubber (Y = 2.4 × 10�3GPa). Relationship among elastic constants : Let Y, B and G be Young�s modulus, bulk modulus and rigidity modulus ofmaterialof a solid bodyand µ bePoison ratio.The relationship among thesemaybe summerised as below.Derivation of these results is out of scope of the book. (i) Y = 3B (1 � 2µ) = 2G(1 + µ) (ii) 3B 2G 2G 6B . . .
. (iii) 9 1 3 Y B G . . Elastic Fatigue Due to repeated deformation and restoration, the elasticityfalls and delayed recovery is observed. This is known as elastic fatigue. If the bodyis allowed to take rest for time, the original elastic propertyis restored.
FLUID MECHANICS www.physicsashok.in 75 Elastic after-effect The delayed recoveryof original configurationbysome elasticmaterialswhen the cause of deformation is removed is known as elastic after effect. This is a type of viscoelastic behaviour. In some materials the recoveryoforiginal configuration is quicker (quartz, phosphor bronze, etc.)while in some it is considerably large (glass fibre). Strain�hardening As the deformation becomes plastic,material needs larger and larger stress (becomes harder to deform). This is known as strain�hardening. Example 67. Acopper bar has a compression force of 2kN distributed over its cross section of diameter 20 mm. Its length is 2mandYoung�smodulus isY= 110GNm�2. Calculate the compression .l of the bar. Sol. UsingHooke�s law, 2 F y ( D / 4) . . . l l 2 F Y (D / 4) . . . l l 2 2 (2kN) (2m) ( / 4)(110 GN m. ) (20 mm) . . . l 3 9 3 2 (2 10 ) 2 (3.14 / 4)(110 10 ) (20 10. ) . . . . . . l .l = 1.116 × 10�4 m .l = 1.12 × 10�4 m. Example 68. Aload of 4.0 kg is suspended froma ceiling through a steelwire of length 2.0 mand radius 2.0 mm. It is found that the length of thewire increases by0.031mmas equilibriumis achieved. FindYoung�s modulus of steel. (Take g = 3.1 .m/s2) Sol. Longitudinal stress= 3 2 (4.0) (3.1 ) (2.0 10. ) . . . = 3.1 × 106 N/m2
Longitudinal strain= 0.031 10 3 2.0 . . = 0.0155 × 10�3 Thus, Y = 6 3 3.1 10 0.0155 10. .. = 2.0 × 1011 N/m2. Example 69.Determine the elongation of the steelbar 1mlong and 1.5 cm2 cross-sectionalareawhen subjected to a pull of 1.5 × 104 N. TakeY= 2.0 × 1011 N/m2. Sol. Y F /A/ . . l l
FLUID MECHANICS www.physicsashok.in 76 . F AY .l . l Substituting the values, 4 4 11 (1.5 10 )(1.0) (1.5 10. )(2 10 ) . . . . . l .l = 0.5 × 10�3 m = 0.5 mm Example 70. Abar ofmassmand length l is hanging frompoint Aas shown in figure. Find the increase in its length due to its ownweight.TheYoung�smodulus of elasticityofthewire is Yand area of cross-section of thewire isA. B A Sol. Consider a small section dx of the bar at a distance x fromB. Theweight of the bar for a length xis, W . . mg. x .. l .. Elongation in sectiondxwillbe B Adx W x d dx AY . . . .. .. l d mg x dx AY . . . . . . . l l Total elongation in the bar can be obtained by integrating this expression for x = 0 to x = l . . x x 0 d . . . . . l l l 0 mg x dx AY . . . . . . . . .l l l or mg 2AY .l . l Example 71. Two wires ofmetalsAand B have unloaded lengths lA and lB and equal
cross sectional area .. They are loaded as in the figurewithweightsW1 andW2. If theYoung�smoduli areYA forAandYB for B, calculate elongations inAandB due to loading. W1 W2 AB Sol. The free body diagrams are shown for wireA, loadsW1 andW2, and wire B.We see that TB = W2 ...(i) TA =W1 + TB ...(ii) The stress inwireAis A A . . T.1 B A W . T . . . W1 TB TA TA TA TB TB W2 TB A B 1 2 A W .W . . . ...(iii)
FLUID MECHANICS www.physicsashok.in 77 Now A A A A Y . . . l l A A A A Y. .l . l 1 2 A A A (W W ) Y . . . . l l [using (iii)] The stress inwire B is B B . . T. 2 B . . W. [using (i)] Hence B B B B Y . . . . . . . . . l l 2 B B B WY . . . l l Example 72. A wire of length l = 1.0 m and cross-sectional area . = 5.00 × 10�4 cm2 is stretched by load atmid point with amass m= 100 g.Assuming stretchingwithinelastic limit, calculate the depression ofwire at themid point. TheYoung�smodulus of elasticity is 1.9 × 1011 Pa. m Sol. Let the depression be x. Then tan x . ( / 2) . . l Now W= T cos . + T cos . T W Wsec 2 cos 2
. . .
. UsingHooke�s law W l/2 T T T x T Y z ( / 2) . . l where z is elongation in the half length (l /2). . W sec 2Y z 2 . . . l Now 2 2 z x2 2x 2 2 . . . . . . .. .. l l l 2 x2 2 sec x 2 . . . .. .. . . . l l
FLUID MECHANICS www.physicsashok.in 78 . W 2x2 2x 4 y . . .l l l W 3 1/3 x 16 y . . . . . . . . l Putting the value 3 3 4 4 11 x 100 10 10 1 0.10 m 16 5 10 10 1.9 10 . . . . . . . . . . . . . . . . . . . Example 73. Amassmis tied to an elatic band of length l andwhirled in a vertical circle.The angular speed of themass at the lowest point of the path is .. IfYbeYoung�smodulus, calculate elongationat this point. Sol. Let x be the elongation.Using dynamics of circularmotionm.2 (l + x) = T �mg T = m.2 (l + x) + mg ...(i) UsingHooke�s law, T . Y x . l l + x T mg V T Y x . . l ...(ii) From(i) and (ii), Y x m 2 ( u) mg . . . l . . l Y m 2 x m( 2 g) .. . . . . . . . . . . l l 2 2 m( g) x Y m . . . . . . . . . . . . l
l Example 74. The breaking stress of amaterial is .B = 106 Nm�2 and density is . = 5 × 103 kgm�3.Determine themaximumlength that can be hungwithout breaking. Sol. Let l be the length hung. Theweight will be w= l .., where ..is cross-sectional. The stress due to weight ismaximumat the top end. Hence we expect fracture fromthe top.We avoid break of a top bywriting B w . . . B .p . . . l B . . . l Themaximumlength is, then, B .. , i.e., and across-sectional area 6 min 3 10 200 m 5 10 . . . l
FLUID MECHANICS www.physicsashok.in 79 Example 75. TwometalwiresAandBhave the same length l under no loading. TheYoung�smoduli areYA andYB, cross-sectionalareas are . and 2. for A and B respectively. The rod PQ is light.Nowa loadWis applied. (a) Write extension in thewiresAand B as a function of x. L Wx AP B Q (b) Write the value of x for equal stress in thewires. (c) Write the value of x for equal strain in thewires. Sol. Let tension be TA inAand TB inB.The equilibriumof rod suggests thatmoment ofwand TB about Pmust be balanced. HenceWx = TB L B T W xL . ...(i) W L TA x TB As TA + TB =W A T W 1 xL . . . . .. .. ...(ii) The stress in thewireAis given by a A A Y . . . l l A A A T Y . . . l l A W 1 x Y L . . . . . . .. .. l l The stress in thewire Bis given by B B B Y . . . l l B B B T Y
2 . . . l l B W x 2 L . . . . . .. .. l l (b) Equal stresses in the two wiresmeans A B T T x 2 . . B a T T2 . Putting the values ofTA andTB, Wx W 1 x 2L L . . . . .. ..
FLUID MECHANICS www.physicsashok.in 80 x x 1 2L L . . x 2 L 3 . (c) Equal strains inthe twowiresmeans A B . . l . l l l A B A B T T Y 2 Y . . . [ . stress =Y. strain] A B A B T T Y 2Y . Putting the values ofTA andTB, A B W 1 x Wx Y L 2LY . . . . .. ..AB x Y x 1 L Y 2L . . . . . . . . AB x Y 1 1 L 2Y . . . . . . . . A B x L (1 Y / 2Y ) . . Example 76. Asolid cube of steel of volume 1m3 is immersed inwater at a depth of 1 km. Find the decrease in its volume due to volume stress. The densityofwater is . = 1000 kgm�3 and the bulkmodulus of steel is B = 135 GPa. Sol. UsingHooke�s law, .P = � B V V . . . .. .. Here volume stress is .P = .gh .P = 1000 × 9.81 × 1000 Pa .P = 9.81 × 106 Pa . .V = � V (.P)
.V = � (1 m3) × (9.81 × 106 Pa)/ (135 GPa)
.V = � 0.072 × 10�3 m3
.V = � 7.2 × 10�5 m3 Theminus sign shows that the volume has decreased. Example 77. Awire of length l and radius r is clamped at one end. Its rigiditymodulus isG. Find the torque about its axis needed to produce a twist by angle . at the end.
FLUID MECHANICS www.physicsashok.in 81 Sol. Let d. be the torque acting along the axis due to forces of share along the ring surface of area 2. x dx. The sheart strain is x. l . The stress is dF 2. x dx . Using Hooke�s law dF G x 2 x dx . . . l . x d dF 2 G x2dx l . . . . l Torque about the cylinder axis is given by d. = x dF d 2 Gx3 dx . . . . l The total torque needed is given by r 3 0 2 G x dx . . . . . l 2 G r4 4 . . . . . . . . l . . r4 G 2 . . . . . . . . . l . Here ./. is constant. This constant is called torsional constant, c. c = r4G . 2 . l ELASTIC ENERGY The energystored in deformed bodyper unit volume is equal to half of the product of stress and strain. It is called elastic energydensity. If u be the elastic energyper unit volume, u = 12 × stress × strain. We consider the Longitudinal case for deriving the above result (without loss ofgenerality). Let there be awire of length l and cross sectional areaAwhich is extended by x using a force F. Then, usingHooke�s law, F Y x
A . l xdx l F . Y Ax F l Let thewire be further extended by dx. Thework done byF is dW.Here dW = F dx
FLUID MECHANICS www.physicsashok.in 82 dW . YA x dx l If thewire is stretched fromx = 0 to x = .l, 0 YA w x dx . . .l l . .2 w YA 2 . . l l w 1 YA 2 . . . . . . . . . . l l l w = 12 × Force × extension. Thiswork is stored in thewire as elastic energyU. Thus 2 U YA 1 Y A 2 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . l l l l l l l l This gives U = 12 × stress × strain × volume . u U volume . u = 12 × stress × strain. This is true for all the three types of deformations. Example 78. Ametalwire of length 2mis supporting a 20 kg ball. Suddenly the ball is detached. Calculate the thermal energy developed in thewire ifYoung�smodulus of itsmaterialis 200GPa. Sol. The elastic energy stored in thewire in presence of load is U = 12 × stress × strain × volume U 1 Y A 1 Y A 2 2 . . . . . . . . . . . . . . . . l l l l l l l l Now Y A mg . l .
l . 1 mg 1 (mg)2 U mg . 2 YA 2 YA . . l l 2 9 6 1 (200) 2 U 0.04 J 2 200 10 5 10 . . . . . . This energyis converted into thermal energy.
FLUID MECHANICS www.physicsashok.in 83 Example 79. Awire of area of cross-section 3.0 mm2 and natural length 50 cmis fixed at one end and amass of 2.1 kg is hung fromthe other end. Find the elastic potential energy stored in the wire in steady state. Young�smodulus of thematerial of the wire = 1.9 × 1011 N/m2. Take g = 10 m/s2. Sol. Volume ofthewire is V = (3.0 × 10�6) (0.50) = 1.5 × 10�6 m3 Tension in thewire is T =mg T = (2.1) (10) = 21 N Stress = T/A Stress = 6 21 3.0 .10. = 7.0 × 106 N/m2 Strain= Stress/Y Strain = 6 11 7.0 10 1.9 10 . . = 3.7 × 10�5 The elastic potential energy of thewire is U= 12 (stress) (strain) (volume) U = 12 (7.0 × 106) (3.7 × 10�5) (1.5 × 10�6) U = 1.9 × 10�4 J Determination of Young�s modulus of the material of a wire in laboratory Two identicalwires (sayof steel) are hung fromthe ceiling.Ais auxiliarywire and B is experimentalwire whoseYis to be determined.Amain scaleMis fitted in auxiliarywireAand vernier scaleVin experimental wire. Identical pans H (or hangers) are attached and loaded to remove kinks.Measure the diameter of wiresBbya screwgaugevernier is adjusted to zero.Nowallowed load is calculated fromtable of constants (half of breaking stress).We have to put a load not greater than this. Now we start froma minimumload. Put the load on right pan and take the reading on vernier scale. Calculate the value of (.l / l) for given load. Plot a graph between stressW/.r2 and strain (.l / l). It turns out to be a straight line. Hooke�s lawgives 2 W Y r . . . . . . . . . l l . M l A l B l H H V
The value ofYcan be calculated : (W/ r2 ) Y ( / ) . . . l l It can also be graphically calculated fromthe slope of stress � strain graph. x y Stress Strain l l ( ) (W/ r 2)
FLUID MECHANICS www.physicsashok.in 84 Slope = tan . = yx =Young�smodulus. Why dowe takewireAalongwithB ?This is to compensate for any temperature effect on strain. IMPORTANT FEATURES 1. Moduli of elasticityare three, viz,Y, Band .while elastic constants are four, viz,Y, B, ..and .. Poisson�s ratio ..is notmodulus of elasticityas it is the ratio of two strains and not of stress to strain, Elastic constants are found to depend on each other through the relations : Y= 3 B (1 � 2.) and Y = 2..(1 + .) Eliminating ..orYbetween these,we get Y 9B 3B . . . . and 3B 2 6B 2 . . . . . . 2. Table : Some Elastic Constans of SelectedMaterials ofEngineering andMedical Interest Material Young�sModulus ShearModulus Bulk Modulus (Y) (1011 Pa) (S) (1011 Pa) (B) (1011 Pa) Aluminium 0.70 0.30 0.70 Brass 0.91 0.36 0.61 Copper 1.1 0.42 1.4 Glass 0.55 0.23 0.37 Iron 1.9 0.70 1.0 Steel 2.0 0.84 1.6 Tungsten 3.6 1.5 2.0 3. Gases have two bulk-moduli, namely isothermal elasticityE. and adiabatic elasticityE.. It has been found that at a given pressure P, E. = P and E. = .P So that PV E C 1 E C .. . . . . , i.e., E. > E. i.e., adiabatic elasticityis greater thanisothermal elasticity. 4. In case of compression of a fulid density = . = mV , so V V V .. . . . But bydefinitionof bulkmodulus, i.e., B V P
V. . . . or V P V B . . . . . P B .. . . . or ´ P B . .. . . . [as .. = .´ � .]
FLUID MECHANICS www.physicsashok.in 85 or P 1 ´ 1 [1 C P] as C B B . . . . . . . . . .. . . . . . . . . . . . . . . . . . 5. In case of bending of a beamof length l, breadth b and thickness d, bya loadMg at themiddle, depression . is given by 3 3 Mg 4bd Y . . l and for a beamof circular cross-section of radius r and length l. 3 4 Mg 12 r Y . . . l 6. In case of twisting of a cylinder (or wire) of length l, and radius r, elastic restoring couple per unit twist is given by 4 C 2 .. . ll when . ismodulus of rigidityof thematerialofwire. 7. In case of a rod of length l and radius r fixed at one end, angle ofshear . is related to angle of twist ....by the relation. l . = r ..
FLUID MECHANICS www.physicsashok.in 86 THINKING PROBLEMS FLUID MECHANICS 1. Awooden cylinder floats inwater in a vesselwith its axis vertical. Howwill the level ofwater in the vessel change ifthe cylinder floatswithits axis horizontal? 2. Avessel filledwithwater is supported on a knife-edge.Will the equalibriumbe disturbed if a small board carrying aweight is placed on the surface of thewater? 3. Aboat floats in a swimming pool.Water fromawell is pumbed into the pool.Whenwillthe pumps perform movework: when a big boat floats in the pool, or a small toy-boat ? 4. Avessel ofwater is placed on the floor of an elevator.Will the pressure at the bottomof the vessel change if the elevator goes upwith uniformacceleration a ? 5. Awooden cylinder floats inwater in a vessel placed on the floor of an elevator. The length of the cylinger outsidewater is l. Ifthe elevator goes downwith uniformacceleration a,will l change? 6. Mercury is poured into a verticalU-tube, andwater is poured in above it. The level ofwater is the same in both arms.Will the levelofthewater and themercurybe the same if a piece ofwood is dropped into one arm and somewater equal inweight to this piece is added to the other ? 7. An air mattress is filled with air to a pressure greater than the atmospheric pressure. When will the air pressure in themattress be greater: when aman stands on it or when he lies on it? 8. Aman carries a tyre tube and decides tomake it lighter bymaking use of the buoyancyof air. In order to do this, he inflates the tube, thus increasing its volume.Willhis aimbe achieved? 9. Atank containingwater is placed ona spring balance.Astone ofweight wis hung and lowered into thewater without touching the sides and the bottomof the tank.Explain howthe readingwill change. 10.Aballfloats on the surface ofwater in a countainer exposed to the atomsphere.Willthe ballreemain immersed at its former depthorwillit sink or rise somewhat if (a) the container is covered and the air is removed, (b) the container is covered and the air is compressed? 11.A solid cylinder is placed erect in a container in contact with the base. When liquid is poured into the container, none of it goes beneath the solid, which remain closelyin contact with the base. Is there a buoyant force onthe solid?Explain. 12.An open bucket ofwater is on a smooth inclined plane, forming an angle .with the horizontal.Howwill the level rest relative to the planewhen the bucket is allowed to slide down the plane? 13.Explainwhya uniformwooden stickwhichwill float horizontallyif it is not loaded, but will float verticallyif enoughweight is added at one end. 14.It is found that a liquidwillflowfaster andmore smoothlyfroma sealed canwhen two holes are punctured in the can thanwhen one hole ismade. Explain. 15.Two rowboatsmoving parallel to one another in the same directionare pulled towards one another. Explain. 16.Can the action of a parachute in retarding free fallbe explained byBernoulli�s
theorem? 17.Can you explainwhy an object falling froma great height reaches a steadyterminalvelocity? 18.While taking offwould it be better for an airplane to move into thewind or with thewind?While landing? 19.Does the difference in pressure between the lower and upper surfaces of an airplance wing depend on the altitude of themoving plane?Explain. 20.If p is the pressure of gas inside the exhaust chamber of a rocket and p0 is the pressure of the gas outside the chamber, the forward thrust on the rocket is 2a (p � p0) instead of a (p � p0),where a is the area of the orifice. Why does the factor �2� appear ? 21.The destructive effect of a tornado (twister) is greater near the centre of the disturbance than near the edge. Explain. 22.Whywill a light ping-pong balldance over a streamof gas or water issuing at high speed froma tube?
level. The temperature of water is slowly raised from 0ºC to 20ºC. How will the volume V change with the rise in temperature? 27.Abodyimmersed in a liquid is balanced on a scale. Will the reading on the scale be d if the liquid is heated together with the body? 28. A glass bulb is balanced by a brass weight in a sensitive beam balance. State what will happen when the balance is covered by a bell jar which is the evacuated. Explain. 29.Ablock ofice is floating ina liquid ofspecific gravity 1.2 contained ina beaker. When the ice melts completely, level. The temperature of water is slowly raised from 0ºC to 20ºC. How will the volume V change with the rise in temperature? 27.Abodyimmersed in a liquid is balanced on a scale. Will the reading on the scale be d if the liquid is heated together with the body? 28. A glass bulb is balanced by a brass weight in a sensitive beam balance. State what will happen when the balance is covered by a bell jar which is the evacuated. Explain. 29.Ablock ofice is floating ina liquid ofspecific gravity 1.2 contained ina beaker. When the ice melts completely, caused. Yet when a hydraulic press, in which the pressure is much higher, blows up, the damage caused is not very great. Why ? 32.A hollow sphere and a solid sphere of the same radius and the same material fall through air from the same height. Which one arrives on the ground first ? (Neglect viscous effect of the air.) 33.Do rain drops of different sizes reach the earth with the same speed? Explain. FLUID MECHANICS 23.During storms, the strong winds tear off the roofs of thatched houses along the ridge C if the roofisfastenedmorefirmlyattheedgesAandB(seefigure)thanat theridge. Ontheother hand, if the roof is secured more firmly at the ridge C than at the edges, the wind will first lift theroofupandthencarryit away.Explain.
24.Adevice(seefigure) consistsoftwodiscsAandB.Thelower discB hangsonthreepins fixedtoAalongwhichit canfreelymoveupanddown.Theupper discAisprovidedwith apipeat itscentre. Ifacompressedair streamispassedthroughthepipe, thelower disc begins to knock the upper one. Explain.
25.Aballoonfilledwithairisweighedsothat itbarelyfloatsinwater asshowninthefigure. Explainwhyit sinksto thebottomwhenit issubmergedmorebyashort distance. 26.A block os wood is floating on water at 0ºC with a certain volume V above the water will the level in the beaker change ? 30.It is the practice of masons, while laying and levelling the foundations of buildings, to use a long, transparent, plastic pipe. The theory is that water, seeking a common level, will be at the same level in both halves of the pipe and thus help to obtain levelling. What happens if a bubble of air is left in the pipe? 31.When a steam boiler in which the pressure of steam is 10 to 15 atm blows up,
considerable damage is 34.Asphericalbobmadeofcorkfloatshalf-submergedinapotofteaat restontheearth.Willthecorkfloat or
move in the beginning and after a short time? (see figure)
sink aboard a spaceship coasting in free space? On the surface of Jupiter? 35.Why is it easier to curve the flight of a tennis ball than it is to curve that of a baseball? 36.Explain why it is dangerous to stand near a speeding train. 37.Very often a sinking ship turns over as it gets immersed in water. Why? 38.What is the pressure at a depth h below the free surface of a liquid falling freely? 39.A ship gets a large hole O in its underwater portion. In what direction willit begin to 40.The vessel shown in Figure is entirelyfilled with water. What will happen if the tap S is opened?
41.A sailor found a small hole in the hole of his vessel, through which water was entering into it. He tried to stop the flow with a plank but the stream of water pushed the plank away. He managed to bring the plank close to the hole with the aid of another sailor, and then found that he could hold the plank alone. Explainwhythepressureontheplankisdifferent inthetwo cases. www.physicsashok.in 87
FLUID MECHANICS 42.A vessel with a side-cock is filled with water and placed on the platform of a spring balance. Will the equilibrium change if the cock is opened? The outflowing water falls on to the same pan on which the vessel is placed. 43.Explain why one has to blow over a piece of paper rather than under it, to keep it horizontal. 44.Why does a flag flutter in a strong breeze ? FLUID MECHANICS 42.A vessel with a side-cock is filled with water and placed on the platform of a spring balance. Will the equilibrium change if the cock is opened? The outflowing water falls on to the same pan on which the vessel is placed. 43.Explain why one has to blow over a piece of paper rather than under it, to keep it horizontal. 44.Why does a flag flutter in a strong breeze ? 45.Why do water jets taper when the tap is almost closed ?
SURFACE TENSION
1. Water can rise to a height h in a certain capillarytube. Suppose that this tube is immersed in water so that only a height h/2 is above the surface. Will there be a fountain? Explain. 2. Why is moisture retained longer in the soil if it is harrowed? 3. Why are raindrops spherical in shape ? 4. Why does the end of a glass rod become round on being heated strongly? 5. Explain how an iron can be used to remove greasyspots from clothing. 6. Why do drops of water appear at the end of a piece of firewood when it is being dried in the sun while its other end is in the shadow ? 7. Acapillary tube is dipped in water vertically. It is long enough for the water to rise to the maximumheight h in the tube. The length of the portion immersed in water is l < h. The lower end ofthe tube is closed and then the tube is taken out and opened again. Will all the water flow out of the tube? Explain. 8. Two capillary tubes A and B are immersed in water � One is straight and the other is in the form of a rectangular U-tube. The tubeAis sufficientlylong. The lower end of the bent tube is at a depth H. What form will the meniscus take and will there be any flow of the water? Consider the following five cases : (a) H > h (b) H = h (c) 0 < H < h (d) H = 0 (e) H < 0 9. Explain how detergents clean dirty clothes. 10.Two soap bubblesAand B of different diameters are blown at the two ends of a bent tube. By opening the stop cock S, the two bubbles are put in communication. What will happen? ELASTICITY 1. Stress and pressure are both forces per unit area. Then in what respect does stress differe from pressure ? 2. Is elastic limit a property of the material of the wire ? 3. Which is more elastic, steel or rubber ? 4. Which one has a greater force constant � a steel wire or a rubber wire of the same length and radius ? 5. There are two springs of the same material and length but the area of cross-section of one is double that of the other. Which one will have a greater force constant ?
6. Among solids, liquids and gases, which one can have all the moduli of elasticity ? 7. Among solids, liquids and gases, which possesses the greatest bulk modulus ? THINKING PROBLEMS SOLUTIONS
FLUID MECHANICS
1. There will be no change in the level of water because in both the positions the cylinder displaces the same volume of water. 2. No. The equilibriumwillnot be disturbed, since according to Pascal�s law, the pressure on the bottomwillbe the same at every point. 3. Inboththecasesthepumpsperformthesamework,sincethesameamountofwater pumpedinrisestothe same level. 4. Considering the upward motion of a column ofliquid of depth www.physicsashok.in 88
FLUID MECHANICS www.physicsashok.in 89 h, pA � p0A� (Ah.)g = (Ah.) a . p = p0 + h. (g + a). When the elevator is at rest, pi = p0 + h.g . .p = p � pi = h.a. Thus the pressure at the bottomincreases. 5. When there is nomotionAL.g = (L� l)A..g, where L= total length of cylinder,A= area of cross-section, ..= density of cylinder, .. = densityofwater, ..L. = (L� l) .. When the elevator goes downwithacceleration a, the effective acceleration due to gravityon both cylinder andwater is (g � a). . AL. (g � a) = (L� l�)A.. (g � a) ..L. = (L� l�) .. . l = l�. So the cylinderwill neither sink nor rise. 6. Dropping the piece ofwood is exactlythe same as adding the amount ofwater displaced byit,which is equal to theweight of the body.Hence, if the cross-sections ofthe arms are the same, the levels ofwater in both the armswill stand at the same horizontal level. 7. If themanstands on themattress, hisweight willbe distributed over a smaller area, that is, the area of his feet thanwhen he lies down. Hence, the air pressure in themattresswill be greater when themass stands than when he lies on it. 8. No, his aimwill not be achieved. The increased buoyancy of air will bemade up for byweight of air blown in. Since the densityof compressed air is generallygreater than the densityof atomspheric air, hewill achieve the opposite of his aim. 9. Make free-bodydiagrams of the bodies separatelyand consider their equilibrium.Like all other forces, buoyancyis also exerted equallyon the two bodies in contact. Hence if thewater exerts a buoyant force, say, B on the stone upward, the stone exerts the same force on tehwater downward. The forces acting one the �water + container�systemare :W, weight of the system downward, B, buoyant force of the stone downward; and the force of the stone downward; and the forceR of the spring in the upward direction. For equilibriumR =W+ B. Thus the reading of the spring scalewillincrease byan amount equal to theweight of the liquid displaced, that is, byan amount equal to the buoyant force. Note : The weight of the stone will not be included in the free-body diagram of the �water + container� system because the weight is the action and reaction between the stone and the earth and not between the stone and the water. 10.Let v be the volume of the ball above the level ofwater andVits totalvolume.W=weight of the ball. Then remembering that, strictlyspeaking, air also exerts buoyant force W= v.0g + (V� v) ..g where .0= density of atmopheric air, ...= densityofwater. Let v� be the volume abovewater levelwhen density of air is changed. thenW= v�.g + (V� v�) ..g where . = density of air. . v.0g + (V � v) ..g = v�.g + (V� v�) ..g or w 0 w 0 w w v' v or v '
v . . . . . . . . . . . . . . (a)When air is removed .= 0 . w 0 w v' v . . . . . . v�< v. So the ball sinks. (b)When air is compressed
FLUID MECHANICS www.physicsashok.in 90 ..> .0 ....� ..< .. � .0 ... w 0 w 1 . . . . . . . . v' 1 v . or v�> v. So the ball rises. 11.No. There is no buoyant force on the solid because the liquid is not in contact at the bottomand so it exerts no upward thrust on it. 12.The levelwill remain parallel to the plane during the journeydown the plane.The component of the forcemg down the plane,whichismsin . produces acceleration down the plane. The surface is subjected to the force mg cos . perpendicular to the plane.Aliquid surface keeps itself at right angles to the force to which it is subjected. So the level rests parallel to the plane during the downward journey ofthe bucket. 13.Afloating body float s in stable equilibriumwhen its centre of gravityiswell belowthe centre of buoyancy, that is, the centre of gravityof the displaced liquid.When loaded byenoughweight the centre ofgravitygoes well belowthe centre of buoyancy. 14.It is found that a liquidwillflowfaster andmore smoothlyfroma sealed canwhen two holes are punctured in the can thenwhen one hole ismade. Explain. 15.When they are close to each other, the velocity of the water between themincreases, resulting in a fall of pressure there according to Bernoulli�s theorem. The pressures fromthe sides push themtogether. 16.When the parachute opens out, the pressure of the air above drops and so an upward thrust is called into play to balance theweight of the parachutist. 17.The viscous force on a bodydepends on its velocity.The greater the velocitythe greater is the viscous force. When a bodyfalls froma sufficient height, it acquires enoughvelocityto produce a viscous force that balances itsweight.The resultant force on the bodybeing zero, the bodymoveswith uniformvelocity, called terminal velocity. 18.While taking off an airplane needsmaximumdynamic lift, which is associatedwith an unsymmetrical set of streamlines relativelyclose together ontheupper side andrelativelyfar apart below.Theunsymmetricdistribution of streamlines can be obtained better bymoving into thewind.While landing, it shouldmovewith thewind when the streamlineswill be uniformlydistributed on the two sides of the airplane and so no liftwill arise and the planewillland under gravity. 19.Yes, because 2 1 2 2 1 2 1 v p 1 v p
2 2 . . . . . . . 2 2 . 2 1 1 2 p p p 1 v v 2 . . . . . . At high altitudes the densityis considerablylowand so the pressure differencewilldepend on the altitude of the airplane. 20.The formula �thrust = pressure difference× area� holds for fluids at rest. This does not hold for fluid inmotion. ByBernoulli�s theorem(a theoremapplicable to fluids in streamlinemotion) 2 0 2 p 1 p 1 V v 2 2 . . . . . andAV= avwhereA= area of the chamber, a = area of the orifice. SinceA>> a, v >>V, henceV2maybe neglected in comparisonto v2. . . . 2 0 p p v 2 . . . Nowthrust on the rocket v dMdt . But themass (dM) of gas flowing out in time dt is (a v . dt). . dM av dt . .
FLUID MECHANICS www.physicsashok.in 91 . thrust = av2. = a. × . 0 . . . 0 2 p p 2a p p . . . . 21.Near the edge the velocity is zero and the pressure is the normal atmospheric pressure.At the centre, the velocity ismaximumand so there ismaximumdrop of pressure at the centre. Thus, the pressure difference between the centralregion and the surrounding air being themaximum, the destructive power ofa tornado is maximumat the centre. 22.Since the streamhas a high velocity, the pressure inside the streamis belowthe atmospheric pressure. The ball is supported fromthe bottomby the thrust of the streamand kept in position by the lateral pressure difference. 23.The pressure of the air streaming over the roof is less than that of the stationary air belowthe roof. This is according to Bernoulii�s theorem.The surplus pressure of the stationaryair of the roomacts normallyon the two inclined halves of the roof. When the roof is secured at the edgesAand B, the thrusts on the two halveswill have turning effects about the edges and so the roof is likelyto tear along the ridge.Onthe other hend ifthe roof is secured rigidlyalong the ridge, the thrustswill thenlift the roof against gravityand it willbe carried away bythewind. 24.When air is blown through the pipe, the air between the discs is set inmotion, resulting in a heavy drop of pressure. The surplus pressure frombelowpushes the disc Band knocks it against the upper disc. 25.When it is submerged more by a short distance, the air inside the balloon is compressed resulting in a reduction ofthe buoyant force on it. Thus, downward force (weight of the sinker attached +weight of air in the balloon) being in excess of the buoyant force, it sinks to the bottom. 26.Let V� be the total volume of the block of wood and Wbe the weight of the block. Then by the law of flotation (V� � V) × .tg =W where .t= density ofwater at tºC. Sincewood has a negligible coefficient of expansionV�maybe considered constant. . V = V� �W/.tg As the temperature is gradually increased from0ºC, .t increases and so V increases up to 4ºC, when the density ofwater becomesmaximum.Above 4ºC, .tdecreases and soVdecreases continuously. 27.Reading on the scale =weight of liquid and beaker + buoyant force =W= V.tg, whereW=weight of liquid + beaker V= volume of the solid at tºC and .t= density of liquid at tºC . Reading . . 0 0 W V 1 t g 1 ' t
. . .
.
.
.
. whereV0 =volume of the solid at 0ºC, . =coefficient of expansion ofthe solid, .� = coefficient of expansion ofthe liquid . Reading =W+ V0.0 {1 � (.� � .)t}g As t is increased, {1 � (.� � .)t} decreases and so the reading also decreases becauseWis a constant. 28.In air Wg � Fg = Wb � Fb where Wg = weight of glass bulb, Fg = buoyant force on the bulb, Wb = weight of brass weights, Fb = buoyant force on the brassweights Since the volume of the bulbis greater than the brassweights Fb < Fg or Wb � Wg + Fg < Fg or Wb < Wg Thuswhen air is removed the bulbwillbecome heavier than the brassweights. 29.Let Hbe the height of the liquid in the beaker andH� be the height of the liquid after all the icemelts. LetAbe the area of cross-section of the beaker.
FLUID MECHANICS www.physicsashok.in 92 Let Vbe the volume of the ice block andV� the volume of it inside the liquid. Then V.ice × g = V� × 1.2 × 1000 × g ice V' V 1200 . . Volume of liquid in the beaker =AH �V�=AH ice V1200 . . When all the icemelts, it is converted into water of the samemass but of volume, say,V��. . V�� × 1000 = V.ice ice V'' V1000 . . Therefore, totalvolume of liquid after icemelts ice ice AH V V AH' 1200 1000 . . . . . . .. .. . . . . . ice A H' H V 1 1 ve 1000 1200 . . . . . . . .. .. . . H� >H. So the level of the liquidwill rise. 30.Let h1 and h2 be the heights above the horizontal line. Let p1 be the pressure of the entrapped air. Then p1 � p0 = .gh2 and also and p1 � p0 = .gh1 . .gh1 = .gh2 . h1 = h2 So the entrapped air will not affect the levelling test. 31.Steamis highlycompressible.When a steamboiler blows up, steamexpands violently causing considerable damagewhilewater being incompressible expands nominallywith release of pressure. So it does not cause much damage. 32.Since theyare of the same radius, the buoyant force on each of themis the same. . b W F W a g . . . whereWis theweight of the hollowsphere and b W' F W' a ' g . . . whereW� =weight of solid sphere . b W 1 a F W' 1 a ' g g . . . . . . . . . . . . . . . . But W� > W .
1 a 1 a ' a a ' g g . . . . . Hence the solid sphere arrives at the ground earlier. 33.No. Raindrops of different sizes reach the earthwith different speeds. The terminal velocity acquired by a drop is proportional to the square of its radius. At the final stage, upward viscous force = downward driving force . 3 . . 2 6 rv 4 r g v r 3 . . . . . .. . . 34.For flotation Weight of the body=weight of the liquid displaced V.g =V�.�g whereV= volume of the body,V�= volume of the submerged portion of the body. . V�/V= ./.�
FLUID MECHANICS www.physicsashok.in 93 Thus, the ratio of the submerged volume to the totalvolume is independent of the accelerationdue to gravity. Hence in free space and on the surface of Jupiter the corkwill neither sinkmore nor risemore. 35.The pressure difference on the two sides ofa tennis balldue to the spinimparted to it is greater than it is in the case of a baseball because the former is roughter. Hence, it is easier tomake the flight a tennis ball curve. 36.Due to the high speed of the train, the pressure ofthe air between the train and themanmay fall considerably and the pressure differencemaybe sufficientlyhigh to push theman towards the speeding train. 37.To pass on to the stable equilibriumposition. 38.p = p0 (atmospheric pressure). In general p = p0 + . (g + a)h. Here a = �g. Therefore p = p0. 39.Initially, the shipwillmove to the right because the in-flowingwater (flowing to the left)willexert a force 2pA to the right,where p is the pressure at a depth h of the hole, andAis its area.As soon as the streamofwater reaches the oppositewall, thiswallwillbe acted upon by the leftward forceF = .Av2,where v is the velocity of the streamwith respect to the ship.As a result, themotionwillbegin to retard. 40.Siphonic actionwillstart andwaterwill flowto the reservoir. 41.ByBernoulli�s theoremthe velocityof flowis v = 2gh where h is the height ofwater above the hole. The force on the plank = .v2S = 2.ghS.Once the hole is covered, the velocityof flowis zero and so the force on the plank = .ghS which is just half as great. 42.In the beginningwhen the streamhas not yet reached the pan, equilibriumwill be disturbed. The panwill swing upwards since thewater flowing out of the vesselwillnot longer exert force on its bottomand hence on the pan. That is, the reading of the balancewill decrease. Themoment the streamreaches the pan, equilibriumwill be restored. On each element gravity imparts a downward impulse mg 2h m 2gh g . . . (impulse=force×time) andonreaching the panthe same element ofthe liquid experiences on upwardimpulse .m 2gh (impulse=change inmomentum).Thus, eachelement experiences equal and opposite impulse. since the streamis continuous, the upward and downward impulses of elements cancelout and o the equilibriumis restored. Themoment when the streamstops flowing, the panwill swing down, since the last elements of the liquid falling on the pan act on it with a force that exceeds theweight of the elements. 43.When air is blown over the paper there is a drop in pressure and the atmospheric pressure from below balances theweight of the paper. 44.Due towind on the two sides of the flag, the difference of pressure produces �folds� in the flag in compliance withBernoulli�s theorem. 45.This is in compliancewith the equation of continuity.V(rate of volume flow) a1v1 = a2v2 i.e., a .1/v.At the
lower end the velocity is greater and so the are of cross-section is less. SURFACE TENSION 1. No, therewillnot be a fountain. The curvature of the free liquid surface at the top of the tubewill change till the upward force 2.rTcos. balances the gravitationalpull ofthe column of liquid of height h/2. 2. When the soilis not harrowed, there are large capillaries in it.Water in the soil rises up the capillary holes to the surface fromwhere it evaporates continuously.Thus the soilcontinuouslyloseswater.Whenharrowed, all these capillaries are destroyed, and so capillarysuction stops altogether.Thus, water is retained longer in the soil. 3. Because ofsurface tension ofwater.Due to surface tension the surface of a small volume ofwater tends to occupytheminimumarea. The area of a given volume ofwater has theminimumvalue for a spherical shape. 4. The end of a glass rod becomes round on being heated because of the surface tension ofmolten glass.
FLUID MECHANICS www.physicsashok.in 94 5. The grease melts, and capillary forces carry it to the surface of the cold fabric placed under the clothing, where it is soaked. 6. The end inthe shadowis colder than the end in the sun. The capillaryforces drive thewater in this direction. 7. No, thewaterwill not flowout.When the tube is taken out, a convexmeniscus is formed and a force due to surface tension is called into playin the upward directionwhich keeps thewater in the tube.Thus the length ofwater column remaining in the tubewillbe l + h. 8. In a capillarytube the height of the liquid that can be retained is given by .r2h.g = 2.rT cos. where . is the angle at which the liquidmeets the solid. h.gr = 2T cos. (a) SinceH> h, surface tension cannot sustain the column ofwater and sowater flows out. (b) SinceH= h, thewater does not flowout. Themeniscus is convex. (c) Thewater does not flowout. Themeniscus is convex and is less curved than in the second case. (d) Thewater does not flowout. Themeniscus is flat. (e) Thewater does not flowout. Themeniscus is concave. 9. Water cannot remove grease stains. This is because water does not wet a greasy spot.Themolecules ofa detergent are hair pinshaped, experiencing different amounts of forces at the ends due to watermolecules and greasemolecules. Themolecules are as if pinned to smallglobes of greasydirt and forman interface betweenwater and greasy dirt globes. The greasydirt is thus dislodged fromthe clothes and is carried awaywith runningwater. 10.The excess pressure inside a soap bubble is inversely proportional to its radius. Hence, the pressure insideAis greater thanthe pressure inside B.Airwill flowfrom Aand B. Therefore,Awill become smaller and Bwillbecome larger. ELASTICITY 1. Pressure is the external force per unit area, while stress is the internal force called into play fromwithin a strained body acting transverselyper unit area of the body. 2. No. It depends on the radius of thewire aswell. 3. Steel. Elasticityismeasured bythe stress per unit strain. For the same amount of strain,much greater stress is produced in steel than in rubber. 4. The steelwire has a greater force constant. 5. The onewith greater areawill have a smaller force constant. 6. Only solids.Liquids and gases have onlybulkmodulus. 7. Solids.
FLUID MECHANICS www.physicsashok.in 95 ASSERTION-REASON 1. Statement-1 : Imagine holding two identical bricks underwater. BrickA is completely submerged just below the surface ofwater, while Brick B is at a greater depth. The magnitude of force exerted by the person (on the brick) to hold brickB inplace is the same asmagnitude of force exerted bythe personon the brick) the hold brickAin place. Statement-2 :Themagnitudeofbuoyant force onabrick comletelysubmergedinwater is equaltomagnitude ofweight ofwater it displaces and does not depend on depth of brick inwater. 2. Statement-1 :Upto elastic limit ofa stress-straincurve the steelwire tends to regainits originalshapewhen stress is removed. Statement-2 :Within elastic limit thewire followsHook�s law. 3. Statement-1 :When anideal fluid flows through a horizontal tube of variable cross-section, the pressure becomes different as different points. Statement-2 : Raindrops falling froma great height reach the groundwith a relativelysmallvelocity.This phenomena involves the viscosityof air. 4. Statement-1 :When a drop of ink falls on a newspaper, it spreads on it. Statement-2 :Adhesive force betweenink and paper isgreater than cohesive force between inkmolecules. 5. Statement-1 : In the diagramshown, a cube is floating inwater in tilted position, in shown situation, the cube is in stable position i.e., cubewon�t sink. C Statement-2 :The point ofapplicationofgravitationalforce and buoyancyforce are passing through the same line in above described case. MATCH THE COLUMN 1. Bernoulli�s equation canbewritten in the folliwing different forms (column-I). Column-II lists certain units each ofwhich pertains to one of the possible forms of the equation.Matchthe unit associatedwith each of the equations : Column-I Column-II (A) v2 p 2g g . . + z = constant (P) Total energyper unitmass (B) V2 2 . + P + .gz = constant (Q) Total energyper unitweight (C) V2 P 2 . . + gz = constant (R) Total energyper unit volume 2. InColumn-I, a uniformbar of uniformcross-section area under the application of forces is shown in the figure and inColumn-II, some effects/phenomena are given.Match the entries ofColumn-Iwiththe entries ofColumn-II.
Column-I Column-II (A) F F (P) Uniformstresses developed in the rod
(C) Spraying of liquid (R) Surface energy increases (S) Surface energy decreases 4. Column-II shows five systems in which two objects are labelled as X and Y. Also in each case a point P is shown. Column-I gives some statements about X and/or Y. Match these statements to the appropriate system(s) from Column-II. [JEE, 09] Column-I Column-II (A) The force exerted by X on Y (P) Block Y of mass M left has a magnitude Mg. on a fixed inclined plane X, slides on it with a Y X P constant velocity. (B) The gravitationalpotentialenergy (Q) Two ring magnetsY and Z, of X is continuously increasing. each of mass M, are kept in frictionless vertrical plastic stand so that they repel each other. Y rests on the base X Y P Z Xand Z hangs in air in equilibrium. P is the topmost point of the stand on the common axis of the two rings. The whole system is in a lift that is going up with a constant velocity. (C) Mechanical energy of the system (R) ApulleyY of mass m0 is fixed to (C) Spraying of liquid (R) Surface energy increases (S) Surface energy decreases 4. Column-II shows five systems in which two objects are labelled as X and Y. Also in each case a point P is shown. Column-I gives some statements about X and/or Y. Match these statements to the appropriate system(s) from Column-II. [JEE, 09] Column-I Column-II (A) The force exerted by X on Y (P) Block Y of mass M left has a magnitude Mg. on a fixed inclined plane X, slides on it with a Y X P constant velocity. (B) The gravitationalpotentialenergy (Q) Two ring magnetsY and Z, of X is continuously increasing. each of mass M, are kept in frictionless vertrical plastic stand so that they repel each other. Y rests on the base X Y P Z Xand Z hangs in air in equilibrium. P is the topmost point of the stand on the common axis of the two rings. The whole system is in a lift that is going up with a constant velocity. (C) Mechanical energy of the system (R) ApulleyY of mass m0 is fixed to FLUID MECHANICS (C) Smooth
(D) Smooth F (B) F F (Q) Non-uniformstressesdevelopedintherod
F (R) Compressive stresses developed
(S) Tensile stresses developed 3. Matchthefollowing : Column-I Column-II Temperature increases Temperature decreases (A) Splittingofbiggerdropintosmalldroplets (P) (B) Formation of bigger drop from small droplets (Q) X + Yis continuously decreasing. a table through a clamp X. A block of mass M hangs from a string that goes over the pulley and is fixed at point P of the table. The whole system is kept in a lift that is going down with a constant velocity.
X YP (D) ThetorqueoftheweightofYabout (S) AsphereYofmassMisputina point P is zero. nonviscous liquid X kept in a container at rest. The sphere is
X
released and it moves down in
P
the liquid.
(T) AsphereYofmassMisfalling with its terminalvelocityin a viscous liquid X kept in a container. X
P
Y Y www.physicsashok.in 96
(D) The reading will be less or more depending the upon position of piece of iron in the liquid Which is based on an application of Bernoulli�s equation for fluid flow ? (A) Capillary rise (B) Dynamic lift of an aero plane (C) Viscosity meter (D) Hydraulic press (D) The reading will be less or more depending the upon position of piece of iron in the liquid Which is based on an application of Bernoulli�s equation for fluid flow ? (A) Capillary rise (B) Dynamic lift of an aero plane (C) Viscosity meter (D) Hydraulic press FLUID MECHANICS LEVEL � 1
(A) height of the liquid above the bottom (C) densityoftheliquid 4. (B) The balance will indicate more weight Asample of a metal weights 210 g in air, 180 g in water and 120 g in an unknown liquid. Then (A) the densityof the metal is four times the density ofthe unknown liquid (B) the density of the metal is 3 g/cm3 (C) the density of the metal is 7 g/cm3 (D) the metal will float on water 5. Abeaker containing water is put on the platform ofa spring balance.Apiece of iron suspended from a string is immersed in the water without touching the sides or the bottom of the beaker. How will the reading on the spring balance be affected, assuming that no water flows out of the beaker ? (A) The balance will indicate less weight (C) The reading will not change 6. 7. Asteel ballof mass m falls in a viscous liquid with a terminalvelocity 4 cm s�1. Another steel ballof mass 64 mwillfallthrough the same liquid with a terminalvelocityof (A)4cms�1 (B)16cms�1 (C)8cms�1 (D)64cms�1 1. The force due to pressure in a liquid at a given depth below the free surface (A) is always exerted downward of orienting the surface (B) is the same in all directions of orienting the surface (C) equals the total weight of liquid above that depth (D)dependsupontheamount ofliquidbelowthat depth 2. InaU-tubeexperiment,acolumnPQofwaterisbalancedbyacolumn h1 h2SR(B)(h1�h2)/h1(D)2h2/(h1+h2) RS of(some liquid see figure). The relative densityof the liquid is (A) h2/h1 (C) 2 h1/h2 3. Thepressureatthebottomofaliquidtankisnotproportionaltothe (B) accelerationduetogravity (D) areaoftheliquidsurfaceat thetop 8. In the apparatus shown, initially the stop-cock A is closed and B is open. Now B is connected to a vacuum pump so that pressure is reduced and the liquid rises up to some point C above A. B is closed at this position of liquid and A is opened. Then (A) liquidinthetubewillfalltillitslevelissameasoutside (B) liquid willflow out ofA (C) liquid willrise further
(D) liquid willnot flow out ofA C
A
B P Q
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FLUID MECHANICS www.physicsashok.in 98 9. In a hydraulic press, f and Fare the forces acting on the smallpiston and the large piston having diameters d andD respectively.Then f/F is (A) 2 d/D (B) 2 D2/d2 (C) d2/D2 (D) D/ d 10. Two capillary tubes of the same radius and lengths l1 and l2 are fitted horizontallyside byside to the bottom of a vessel containingwater. The length of a single tube of the some radius that can replace the two tubes such that the rate of steadyflowthrough this tube equals the combined rate of flowthrough the two tubes, is(A) 1 2 1 2 2.l l l l (B) l1 + l2 (C) 1 2 1 2 . l l l l (D) 1 2 2 l . l 11. Two capillary tubes of the same length and radii r1 and r2 are fitted horizontallyside by side to the bottom of a vessel containingwater. The radius of a single tube that can replace the two tubes such that the rate of steady flowthrough this tube equals the combined rate offlowthrough the two tubes, is (A) 1 2 r r (B) (r12 + r22)1/2 (C) r1 + r2 (D) (r14 + r24)1/4 12. Acapillarytube is dipped ina beaker containing a liquid. The angle ofcontact of the capillarywith the liquid is 90º.The liquid in the tubewill (A) fall (B) rise (C) neither rise nor fall (D)mayrise or falldepending on the densityfthe liquid 13. An oildrop is placed on the surface ofwater. It will (A) partlybe as a sphericaldroplets and partly as thin film (B) remain on the surface as a sphere (C) spread as a thin layer (D) remain on the surface as distorted drop 14. If thework done in blowing a bubble of volumeVisW, then thework done in blowing a bubble of volume 2Vwill be (A) (4)1/3 W (B) 2W (C) 8 W (D) (2)1/3W 15. Whichoneofthe following curves shows correctlythe variationofvelocityvwithtime t for a smallspherical bodyfalling verticallyin a long column ofviscous liquid ? (A) v t (B) v t (C)
v t (D) v t 16. Acylindricalvesselfilledwithwater is released on an inclined surface of angle . as shown in figure. The friction coefficient of surfacewith vessel is µ (< tan .). Then the constant anglemade bythe surface ofwaterwith the incline willbe : Fixed (A) tan�1 µ (B) . � tan�1 µ (C) . + tan�1 µ (D) cot�1 µ
FLUID MECHANICS www.physicsashok.in 99 17. Acylindrical container of radiusRand height h is completely filledwith a liquid. Two horizontalL shaped pipes of small cross-section area a are connected to the cylinder as shown in the figure. Nowthe two pipes are opened and fluid starts coming out of the pipes horizontally in opposite directions. Then the torque due to ejected liquid on the systemis : R R h/2 h/2 2R (A) 4 agh.R (B) 8 agh.R (C) 2 agh.R (D) none of these 18. In the figure shownwater is filled in a symmetricalcontainer. Four pistons of equal area A are used at the four opening to keep the water in equilibrium. Nowan additional force Fis applied at each piston. The increase in the pressure at the centre of the container due to this addition is F1 F4 F3 F2 (A) F/A (B) 2F/A (C) 4F/A (D) zero 19. Anarrow tube completely filled with a liquid is lying on a series of cylinders as showninfigure.Assuming no slidingbetweenanysurfaces, the valueof accelerationofthe cylinders forwhichliquidwillnot come out ofthe tube fromanywhere is given by open to atmosphere H L a (A) gH 2L (B) gH L (C) 2gH L (D) gH 2 L 20. Asquare box ofwater has a smallhole located in one of the bottomcorner.When the boxis full and sitting ona level surface, complete opening of the hole results in a flowofwater with a speed v0, as showninfigure (1).Whenthe boxisstillhalfempty, it is tilted by 45º so that the hole is at the lowest point.Nowthewaterwillflowoutwitha speed of V0 Fig. (1) Initial V0 Later-on (A) V0 (B) V0/2 (C) 0 V / 2 (D) 4 0 V / 2 21. Amosquitowith 8 legs stands onwater surface and each legmakes depression of radius a. If the surface tension and angle of contact are T and zero respectivelythen theweight ofmosquito is : (A) 8T . a (B) 16 .Ta (C) Ta/8 (D) Ta/16 . 22. In determinationof youngmodulus ofelasticity ofwire, a force is applied
and extension is recorded. Initial length ofwire is 1m. the curve between extension and stress is depicted thenyoungmodulus ofwirewillbe : 4 mm Extension 2 mm 4000 KN/m2 8000 KN/m2 Stress (KN/m2) (A) 2 × 109 N/m2 (B) 1 × 109 N/m2 (C) 2 × 1010 N/m2 (D) 1 × 1010 N/m2
FLUID MECHANICS www.physicsashok.in 100 23. Ablock ofmassMarea of cross-sectionA&length l is placed on smooth horizontal floor.Aforce F is applied on the block as shown. If y is young modulus ofmaterial, then total extensionin the blockwillbe : F Area �A� l (A) Fl /Ay (B) Fl / 2Ay (C) Fl / 3Ay (D) cannot extend 24. Auniformrod ofmassmandlengthl is rotatingwithconstant angular velocity. about anaxiswhichpasses through its one end and perpendicular to the length of rod.The area of cross section of the rod isAand its young�smodulus isY.Neglect gravity.The streamat themid point of the rod is : (A) m 2 8AY . l (B) 3m 2 8AY . l (C) 3m 2 4AY . l (D) m 2 4AY . l 25. Aballofmass 10 kg anddensity1 gm/cm3 is attached to the baseofa container having a liquid ofdensity1.1 gm/cm3,withthe help of a spring as shown in the figure. The container is going up with an acceleration 2 m/s2. If the spring constant of the spring is 200N/m, the elongation inthe spring is 2 m/s2 (A) 2 cm (B) 4 cm (C) 6 cm (D) 8 cm 26. In the figure shown, the heavy cylinder (radius R) resting on a smooth surface separates two liquids of densitieis 2. and 3.. The height h for the equilibriumofcylindermust be h 2 R 3 R (A) 3R2 (B) R 32 (C) R 2 (D) R 34 27. Aconical flask ofmass 10 kg and base area 103 cm2 is floating in liquid of specific gravity 1.2 as shown in the figure. The force that liquid exerts on curved surface of conical flask is (g = 10m/s2) 10 cm (A) 20Nin downward direction (B) 40Nin downward direction (C) 20Nin upward direction (D) 40Nin upward direction 28. In the figure shown, forces of equalmagnitude are applied to the two ends of a uniformrod.ConsiderAas the cross-section area of the rod. For this situation, mark out the incorrect statement(s). F F
(A) The rod is in compressive stress. (B) The numericalvalue of stress developed in the rod is equal to F/A. (C) The stress is defined as internal force developed at any cross-section per unit area. (D) None of the above.
FLUID MECHANICS www.physicsashok.in 101 29. For the block shown in the figure, the Poisson�s ratio is .. The body is under state of compression. Inthis situation c b a F F (A) the volume of the stressed body decreases (B) the decrease in length ismore, than compensating for increase in area (C) the volume of the stressed body increase (D) both (A) and (B) are correct. 30. In above question, if the fractional change in length of block due to application of force F is 0.001 and corresponding change involume is 0.0005, find the value of Poisson�s ratio for thematerialof the blocks? (A) 1/2 (B) 1/10 (C) 1/4 (D)Not possible to determine the value of Poisson�s ratio fromthe given data 31. The space between two larger horizontal metal plates, 6 cm apart, is filled with a liquid of viscosity 0.8N-s/m2.Athin plate of surface area 0.01m2 ismoved parallel to the length of the plate such that plate is at a distance of 2 cmfromone of the plates and 4 cmfromthe other. If the platemoveswith a constant speed of 1m/s, then (A) the layer of the fluidwhich is havingmaximumvelocity is lyingmid-waybetween the plates. (B) the layers of the fluidwhichis in contract with themoving plate is having themaximumvelocity. (C) the layer of the fluidwhich is in contactwiththemoving plate and is onthe side of farther plate ismoving withthemaximumvelocity. (D) the layer ofthe fluidwhich is in contactwith themoving plate and is on the side of nearer plate ismoving withthemaximumvelocity. 32. The coefficient of restitution for collisionof two bodies (A) depends onYoung�smodulius of both the bodies (B) does not depend onYoung�smodulius of both the bodies (C) depends onYoung�smodulus of the lighter body (D)mayormay not depend uponYoung�smodulus 33. Aheavy block ofmass 150 kg hangs with the help of three vertical wires of equal length and equal cross-section area as shown in the figure. x x I II III Y1 Y2 Y3 150 kg Wire II is attached to themidpoint (centre ofmass) of block. Take Y2 = 2Y1. For this arrangement mark out the correct statement(s) (A)Thewire I and III should have sameYoung�smodulus. (B) Tension in I and III would be always equal. (C) Tension in I and IIIwould be different. (D) Tension in Ii is 75 g. 34. Ametalwire of lengthL, cross-section areaAandYoung�smodulusYis stretched bya variable force F.F is varying in suchawaythat Fis always slightlygreater thanthe elastic forces ofresistance in thewire.When the elongation inthewire is l, upto this instant (A) the work done by Fis
YA 2 2Ll (B) the work done by F is YA 2 Ll (C) the elastic potential energy stored inwire is YA 2 2Ll (D) no energyis lost during elongation
FLUID MECHANICS www.physicsashok.in 102 35. Asolid floats in a liquid is in partiallydipped position, (A) the solid exerts a force equal to itsweight on the liquid (B) the solid doesn�t exert any force on the liquid (C) the solid exerts a force equal to buoyancy force on the liquid (D) the liquid exerts a force of buoyancy on the solidwhichis equal to theweight of solid 36. Allthe fourwires in the optionsbelowaremade up ofthe samematerial.Which of thesewillhave the largest extension,when the same tension is applied ? (A) Length 200 cmand diameter 2mm (B) Length 100 cmand diameter 0.5mm (C) Length 300 cmand diameter 1mm (D) Length 50 cmand diamter 0.5mm 37. Auniformplank of length L andYoung�s modulus Y is pushed over a smooth horizontal surface by a constant horizontal force F0. F0 smooth The areaofcross-sectionofthe plankisA.The compressive strain (.L/L) of the plank in the direction of the force is : (A) 0 3F AY (B) 0 F AY (C) 0 F 2AY (D) 0 2F AY 38. Two wires of same material and length but cross-sectional area in the ratio 1 : 3 are used to suspend the same loads. The extensions in themwill be in the ratio : (A) 1 : 3 (B) 3 : 1 (C) 4 : 1 (D) 1 : 4 39. Awire elongates by l mmwhen a loadWnewton is hang fromit. If thewire goes over a pulleyand two weightsWeach are hung at the two ends, the elongation of thewire (inmm)willbe : (A) zero (B) l (C) l / 2 (D) 2l 40. Awork of 2 × 10�2 J is done on awire of length 50 cmand area of cross-section 0.5 mm2. If theYoung�s modulus of thematerial of thewire is 2 × 1010N/m2, then thewiremust be : (A) elongated to 50.1414 cm (B) stretched by 0.707mm (C) contracted by 2.0mm (D) none of these 41. The value of Poisson�s ratio for all practical bodies lies between : (A) 0 to 0.5 (B) �1 to 1.5 (C) 0.5 to 1 (D) �1 to 1 42. Two bodies ofmasses 1 kg and 2 kg are connected by a metalwire shown in figure.Aforce of 10Nis applied on the body ofmass 2 kg. F = 10N 1 kg 2 kg The breaking stress ofmetalwire is 2 × 109 N/m2.What should be Smooth surface minimumradius of thewire used, if it is not to break ? (A) 0.23 × 10�4 m (B) 4 × 10�4 m (C) 5 × 10�4 m (D) 5.2 × 10�4 m 43. Two wires, nemode of coper and other of stealare joined end to end. (as shown in figure). The area of cross-section of copperwire is twice that of steelwire. F Cooper Steel F They are placed under compressive force ofmagnitudes F. The ratio of their lengths suchthat change in lengths of bothwires are same is : (YS = 2 × 1011 N/m2 and YC = 1.1 × 1011 N/m2) (A) 2.1 (B) 1.1 (C) 1.2 (D) 2
FLUID MECHANICS www.physicsashok.in 103 44. The curve in figure represents potential energy(U) in betweentwo atoms in a diatomic molecules as a function of distance x between atoms. The atoms are : U y x A B C (A) attracted when x lies betweenA and B and repelledwhen x lies between B and C (B) attractedwhen x lies betweenB and Cand repelledwhen x lies betweenAand B (C) attractedwhen they reachB (D) repelledwhen they reachB 45. one end of a steelwire is fixed to ceiling of an elevator moving up with an acceleration 2m/s2 and a load of 10 kg hangs fromother end.Area of crosssection of thewire is 2 cm2. The longitudinal strain in thewire is : (Take g = 10 m/s2 andY= 2 × 1011 N/m2) a = 2m/s2 0 (A) 4 × 1011 (B) 3 × 10�6 (C) 8 × 10�6 (D) 2 × 10�6 46. Two bodies of masses 1 kg and 2 kg are connected by a steel wire of crosssection 2 cm2 going over a smooth pulley (as shown in figure). The longitudinal strain in the wire is : (Take g = 10 m/s2,Y= 2 × 1011N/m2) 1 kg (A) 3.3 × 10�7 (B) 3.3 × 10�6 (C) 2 × 10�6 (D) 4 × 10�6 2 kg 47. Water froma tap emerges vertically downwadswith an initla speed of 1.0ms�1. The cross-sectional area of the tap is 10�4m2.Assume that the pressure is constant throughout the streamofwater, and that the flow is steady. The cross-sectional area of the stream0.15 mbelowthe tap is [JEE, 98] (A) 5.0 × 10�4 m2 (B) 1.0 × 10�5 m2 (C) 5.0 × 10�5 m2 (D) 2.0 × 10�5 m2 48. Alarge opentank has two holes in thewall.One is a square hole of sideLat a depth y fromthe top and the other is a circular hole of radiusRat a depth 4yfromthe top.When the tank is completely filledwithwater, the quantities ofwater flowing out per sectionfrombothholes are the same. then,Ris equalto :[JEE,2000] (A) L/ 2. (B) 2.L (C) L (D) L/2. 49. Ahemisphericalportionof radiusRis removed fromthe bottomof a cylinder of radiusR. The volume of the remaining cylinder isVand itsmass isM. It is suspended by a string ina liquid of density.where it stays vertical.The upper surface of the cylinder is at a depth h belowthe liquid surface. The force on the bottomof the cylinder bythe liquid is [JEE, 01] h 2r (A) Mg (B) Mg � v.g (C) Mg + .R2h.g (D) .g (V + .R2h) 50. Awooden block, with a coin placed on its top, floats in water as shown in figure. The distances l and h are shown there.After some time the coin falls into thewater. Then [JEE, 02] h l coin (A) l decreases and h increases (B) l increases and h decreases (C) both l and h increase (D) both l and h decrease 51. Water is filled in a container upto height 3m.Asmallhole of area �a�is punched in thewallof the container at a height 52.5 cmfromthe bottom. The cross sectional area of the container isA. If a/A= 0.1 then v2 is
(where v is the velocity ofwater coming out of the hole) [JEE, 05] (A) 48 (B) 51 (C) 50 (D) 51.5
FLUID MECHANICS www.physicsashok.in 104 PASSAGE Comprehension � 1 Two cylindrical tanks of radii r and 2rwith their bases at the same level containa liquid of density. to heightsHand 3H, respectively as shown in figure. The tanks are linked through a pipe of very small cross-sectionalareaA.Due to pressure difference liquid starts flowing fromnarrower vessel to broader vessel to equalize the pressure. r 2r 3H H Based onabove information, answer the following questions : A 1. The final commonlevel of liquid inboth vessels is (A) 2H (B) 7H/5 (C) 3H/2 (D) 5H/2 2. The time takenfor the liquid levels to become equal frominitial levels is (A) 5H r2 gA . . (B) 5 r2 H A g . (C) r2 H A g . . (D) 8 r2 H 5A g . . 3. Thework done by gravityin equalising these levels is (A) .r2 × .gH2 (B) 41 10 × .r2 ×.gH2 (C) 85 × .r2 × .gH2 (D) 15 2 × .r2 ×.gH2 Comprehension � 2 Awooden cylinder of diameter 4r, height h and density ./3 is kept on a hole of diameter 2r of a tank, filledwithwater of density..as shownin the figure. The height of the base of cylinder fromthe base oftank isH. h2 h1 h /3 2r H 4r 4. If level of liquid starts decreasing slowlywhen the level of liquid is at a height h1 above the cylinder, the block just startsmoving up.Then, value of h1 is (A) 2h/3 (B) 5h/4 (C) 5h/3 (D) 5h/2 [JEE, 06] 5. Let the cylinder is prevented frommoving up, by applying a force and water level is further decreases. Then, height ofwater level(h2 infigure) forwhich the cylinder remains inoriginal
positionwithout application of force is [JEE, 06] (A) h/3 (B) 4h/9 (C) 2h/3 (D) h 6. If height h2 ofwater level is further decreased, then [JEE, 06] (A) cylinderwillnotmove up and remains at its originalpotition. (B) for h2 = h/3, cylinder again startsmoving up (C) for h2 = h/4, cylinder again startsmoving up (D) for h2 = h/5 cylinder again startsmoving up
FLUID MECHANICS www.physicsashok.in 105 LEVEL � 2 1. The solid ballin the figure hangs froma spring balanceB1 and submerged ina liquid contained in a beaker placed on a spring balance B2. Themass of the beaker is 1.0 kg and that of the liquid is 2.5 kg. Balance B1 reads 3.5 and B2 reads 8.5 kg. The volume of ball is 0.005m3. B1 B2 (a)Write the buoyant force. (b) Find the density of the liquid. (c)What will the balances B1 and B2 read if the ball is pulled up out of the liquid ? 2. Auniformrod of length Lm, specific gravity0.5 andmassmis hinged at one end at a distance of 1/2 Lmbelowwater surface. M (a)What massMmust be attached to the other end of the rod so that 5/6 Lmof the rod gets submerged? (b) Find themagnitude and direction of the force exerted by the hinge on the rod. 3. AU-tube, placed in a verticalplane, is partially filledwith liquidX.AliquidY, immisciblewithX, is next poured into one side untilXrises onthe other side by25 cm. If the density ofYrelative toXis 0.8, bywhat distance theY-levelwill stand higher than theX-level ? 4. Avessel contains a liquidX(density 0.8 g/cm3) over another liquidY(density 13.6 g/cm3), notmixing together.Ahomogeneous sphere floatswith half its X volume immersed inupper and the other half in lower liquid. Find the density Y of thematerialof the sphere. 5. Aliquid flows out of two smallhlesXandYin thewallof a tank.The two streams strike the ground at the same point. If the hole X is at a height h above the ground and the level ofwater stands at a height Habove the ground, find theheight ofY. H XYh 6. Two copper (density .) ballsAand B are released in a liquid of viscosity .. They are connected by a narrow light rod. How does the tension in the rod varywithtime ? A B r2r 7. Water rises to a height of4 cmin a capillarytube. If the area of cross-section ofthe tube is reduced to 1/16 of the former value, find howhighwillthe level rise. 8. Asoap bubbleAofradius r1 and another soap bubble B ofradius r2 (> r1) are brought together so that the combined bubble has a common interface. Find the radius of the interface. 9. Find the work that must be done to get n small equal size spherical drops froma bigger spherical drop (radius = R) of a liquid having surface tension S.Amercury drop breaks into 8 equal drops. The surface energy increases by a fact or n. Find n.
FLUID MECHANICS www.physicsashok.in 106 10. Calculate thework done against surface tension in blowing a soap bubble froma radius of 10 cmto 20 cm if the surfacetension of soap solution is 2.5 × 10�2 Jm�2. 11. A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between thewire and a light slider supports aweight ofW= 1.5 × 10�2N (which includes the smallweight of the slider). The length of the slider is 30 cm. What is the surface of the liquid forming the film? 12. Atube Tofradius r is connected to a bubble of a liquid having surface tension S. The radius of the bubble isRat time t = 0. Showthat it will shrink to zero in time t = 2.lR4/Sr4, where . is the co-efficient of viscosityof air. l R T 13. Length of horizontal armof a uniformcross-sectionU-tube is l = 21 cmand ends of both of the vertical arms are open to surrounding of pressure 10500 N/m2.Aliquid of density. = 103 kg/m3 is poured into the tube such that liquid just fills the horizontal part of the tube. Now one of the open ends is sealed and the tube is then rotated about a vertical axis passing through the other vertical armwith angular velocity .0 = 10 rad/sec. 21 cm 6 cm If length ofeach vertical armbe a = 6 cm.Calculate the length ofair column in the sealed arm. [g = 10 m/sec2] 14. Asimple accelerometer (ann instrument formeasuring acceleration) can be made in the form of a tube filled with a liquid and bent as shown. During motion, the levelofthe liquid inthe left armwillbe at a height h1, and inthe ring armat a height h2. h1 h2 /4 /4 Determine the accelerationa ofa carriage inwhichthe accelerometeris installed, assuming that the diameter of the tube ismuch smaller than h1 and h2. 15. Ahemispherical tank of diameter 4mis filledwithwater.Averysmall hole is punched at 1mabove the bottomas shown in figure. Find x (distance at which water strikes the surface). 1m x ................................................. ................................................ ........................................ ........................... . . . . 16. Two arms of a tube have unequaldiameters d1 = 1.0mmand d2 =1.0 cm. Ifwater (surface poured into the tube held in the vertical position, find the difference of level in the tube.Assure the angle of contact to be zero. T = 0.07N/m. 17. A block mass 0.5 kg is attached to free end of 80 cm long aluminium wire of diamter 0.7 mmand suspended vertically. It is now given circularmotion into a horizontalplane at a rate suchthat thewiremakes an angle of 85ºwith the vertical. Find the increase in length of thewire. 85º (sin 85º = 0.9962, cos 85º = 0.0872,Young�smodulus of aluminium= 7 × 1010Nm�2) 18. Ametalwire of cross section 4mm2 and length 5mis suspended vertically froma rigid support.Abob of mass 100 kg is now attached at the lower end. If the is bob gets snapped, calculate the thermal energy generated in thewire. TheYoung�smodulus is 210 PGa and densityis 7800 kg/m3.
FLUID MECHANICS www.physicsashok.in 107 19. The graphof stress . against strain (.) for amaterialis shown inthe figure. The arrow shows the case of increasing stress. How much is the elastic energydensity inthematerialwhen the strain is : (a) 0.1 (b) 0.3 and (c) 0.4 ? (GPa) 2.0 1.5 1.0 0.5 0.1 0.2 0.3 0.4 20. An elastic band of length l is attached to a block ofmassMwhich is held by a person at rest so that the band is unstretched. The person suddenly releases the block. (a) Find themaximumelastic energystored in the band as the block falls. l (b) If the block finally comes at rest, howmuch is thermal energy generated in the bank ?The Young�smodulus of the band isY, cross-sectional area is . andmass is negligible relative to that of the block. 21. Acolumn ofmercuryof 10 cmlength is contained inthemiddle of a narrowhorizontal1mlong tubewhich is closed at both the ends.Both the halves ofthe tube contain air at a pressure of76 cmofmercury.Bywhat distancewill be column ofmercury be displaced if the tube is held vertically ? 22. Asinker ofweight w0 has anapparent weight w1whenweighted in a liquid at a temperature t1 andw2when weight in the same liquid at temperature t2. The coefficient of cubicalexpansion ofthematerialofsinker is ..What is the coefficient of volume expansionof the liquid. 23. Apointmassmis suspended at the end ofamasslesswireof length l andcross sectionA. IfYis theYoung�s modulus for thewire, obtain the frequecyofoscillationfor the simple harmonicmotionalongthe verticalline. 24. Acube ofwood supporting 200 gmmass just floats inwater.When themass is removed, the cube ruses by 2 cm.What is the size of the cube ? 25. Abeaker containing water is placed on the pan of balance which shows a reading ofMgms.Alump of sugar ofmassmgms and volume Vcc. in now suspended by a thread in such away that it is completely immersed inwaterwithout touching the beaker andwithout anyoverflowofwater.What willbe the reading of the balance just when the lump of sugar is immersed ?Howwill the reading change as the time passes on? 26. Aboat floating in awater tank is carrying a number of large stones. If the stones are unloaded into water, what willhappen to thewater level ? 27. Two idendical cylindricalvesselswith their bases at the same level each contain a liquid of density . (rho). The height of the liquid in one vessel is h1 and in other is h2. The area of either base isA.What is thework done bygravity in equalizing the levelswhen the two vessels are connected. 28. Awooden plank of length 1mand uniformcross-section in hinged at one end to the bottomofa tank as shownin fig. The tank is filledwithwater upto a height 0.5 m. The specific gravity of the plank is 0.5. Find the anlge . that the plank makes
with the vertical in the equilibriumposition. (Exclude the case .= 0º) 29. Athin tube of uniformcross-section is sealed at both ends. It lies horizontally, themiddle 5 cmcontaining mercury and the two equal and containing air at the same pressure P.When the tube is held at anangle of 60ºwiththe verticaldirection, the length of the air column above and belowthemercurycolumn are 46 cm and 44.5 cmrespectively. Calculate the pressure P in centimeters ofmercury. (The temperature of the systemis kept at 30ºC).
FLUID MECHANICS www.physicsashok.in 108 LEVEL � 3 1. Acontainer of large uniformcross-sectional areaAresting on a horizontal surface, holds two immiscible, non-viscous&incompressible liquids of densities d&2d, each ofheight H/2 as shown infigure. The lower density loquid is open to the atmosphere having pressure P0. (a) Ahomogeneous solidcylinder oflengthL(L<H/2) cross-sectional areaA/5 is immersed such that it floatswith its axis vertical at the liquid-liquid interface with the length L/4 in the denser liquid. Determine : (i) The densityDof the solid and (ii) The totalpressure at the bottomof the container. H/2 H/2 d2d x h (b) The cylinder is removed and the originalarrangement is restored. Atinyhole of area s(s<<A) is punched on the vertical side of the container at a height h(h <H/2). Determine : (i) The initial speed of efflux of the liquid at the hole; (ii) The horizontaldistance x travelled by the liquid initiallyand (iii) The height hm at which the hold should be punched so that the liquid travels the maximumdistance xm initially.Also calculate xm. [Neglect the air resistance inthese calculations]. [JEE, 95] 2. Acylindrical tank 1min radius rests on a platform5mhigh. Initially the tank is filledwithwater to a height of 5 m. Aplug whose area is 10�4 m2 is removed from an orifice on the side of the tank at the bottom. Calculate the following : (i) initial speedwithwhich thewater flows fromthe orifice; [REE, 95] (ii) initial speedwithwhich thewater strikes the ground and (iii) time taken to emptythe tank to halfits originalvalue. 3. Athin rod of lengthL&area of cross-section S is pivoted at its lowest point P inside a stationary, homogeneous&non-viscous liquid (Figure).The rod is free to rotate ina verticalplane about a horizontalaxis passing throughP.The density d1 of thematerialof the rod is smaller than the dencity d2 of the liquid. The rod is displaced bya small angle . fromits equilibriumpositionand then released. d d P 2 1 Showthat themotion of the rod is simple harmonic and determine its angular frequencyin terms of the given parameters.[JEE, 96] 4. Alarge open top container of negligiblemass&uniformcross-sectional areaAhas a small hole of crosssectional areaA/100 in its sidewallnear the bottom.The container is kept on a smooth horizontalfloor and contains a liquid of density. andmassm0.Assuming that the liquid starts flowing out horizontallythrough the hole at t = 0, calculate (i) the accelerationof the container and; (ii) its velocitywhen 75%of the liquid has drained out. [JEE, 97]
5. Anonviscous liquid of constant density 1000 kg/m3 flows in a streamline motion along a tube of variable cross section. The tube is kept inclined in the virtical plane as shown in the figure. The area of cross section of the tube at two points P and Q at heights of 2 meters and 5 meters are respectively 4 × 10�3 m2 and 8 × 10�3 m3. P2m 5mQ The velocity of the liquid at point P is 1 m/s. Find thework done per unit volume by the pressure and the gravityforces as the fluid flows frompoint PtoQ. [JEE, 97]
FLUID MECHANICS www.physicsashok.in 109 6. Awooden stick of length l, and radius R and density . has a smallmetal piece ofmassm(of negligible volume) attached to its one end. Find theminimumvalue for themassm(interms of givenparameters) that wouldmake the stick float verticallyin equilibriumin a liquid of density. (> .). [JEE, 99] 7. A uniform solid cylinder of density 0.8 gm/cm3 floats in equilibrium in a combinationoftwo nonmixingliquidsAandBwithitsaxisvertical.Thedensities of the liquidsAandB are 0.7 gm/cm3 and 1.2 g/cm3, respectively.The height of liquidA is hA = 1.2 cm. The length of the part of the cylinder immersed in liquid B is hB = 0.8 cm. air AB hhhAB (a) Find the total force exerted byliquidAonthe cylinder. (b) Find h, the lengthof the part of the cylinder in air. (c) The cylinder is depressed in such awaythat its top surface is just belowthe upper surface of liquidAand is then released. Find the acceleration of the cylinder immediately after it is released. [JEE, 02] 8. Consider a horizontallyoriented syringe containingwater located at a height of 1.25mabove the ground.The diameter ofthe plunger is 8 mmand the diameter of the nozzle is 2 mm. The plunger is pushedwitha constant speed of0.25m/s. Find the horizontalrange ofwater streamon the ground. Take g = 10 m/s2. [JEE, 04] 1.25 m D=8mm d=2mm Ground 9. Asolid sphere of radiusRis floating ina liquid of density.withhalf of its volume submerged. Ifthe sphere is slightlypushed and released, it starts performing simple harmonicmotion. Find the frequency of these oscillations. [JEE, 04] 10. Ahorizontalpipe line carrieswater ina streamline flow.At a point along the pipewhere the cross-sectional are is 10 cm2, thewater velocityis 1ms�1&the pressure is 2000 Pa.The pressure ofwater at another point where the cross sectional area is 5 cm2, is __________ pa. [Density ofwater = 103 kgm�3] [JEE, 04] 11. A U tube is rotated about one of it�s limbs with an angular velocity .. Find the difference in height H of the liquid (density .) level,where diameter of the tube d << L. [JEE, 05] H L 12. Acylindricalvessel of height 500mmhas an orifice (small hole) at its bottom. The orifice is initialyclosed andwater is filled in it up to height H. Nowthe top is completely sealedwith a cap and the orifice at the bottomis opened. Somewater comes out fromthe orifice and thewater level in the vessel becomes steady
withheight ofwater columnbeing 200mm. Find the fallinheight (inmm) ofwater leveldue to opening ofthe orifice. [Take atmospheric pressure = 1.0 × 105N/m2, density ofwater = 1000 kg/m3 and g = 10m/s2. Neglect any effect of surface tension.] [JEE, 09] 13. Two soap bubbles Aand B are kept in a closed chamber where the air ismaintained at pressure 8 N/m2. the radii of bubbles Aand B are 2 cmand 4 cm, respectively. Surface tension of the soap-water used to make bubbles is 0.04N/m. Find the ratio nB/nA, where nA and nB are the number ofmoles ofair in bubbles AandB, respectively. [Neglect the effect of gravity.] [JEE, 09]
FLUID MECHANICS FLUID MECHANICS Answer Key
ASSERTION-REASON TYPE
Q. 1 2 3 4 5 Ans. A C B A C
MATCH THE COLUMN 1. [(A�Q), (B�R), (C�P)] 2. [(A�PR), (B�PS), (C�QS), (D�QR)] 3. [(A�QR), (B�PS), (C�QR)] 4. [(A�PT), (B�QST), (C�PRT), (D�Q)] LEVEL�1 Q . 1 2 3 4 5 6 7 8 9 A n s. B A D C B B D D C Q . 10 11 12 13 14 15 16 17 18 A n s. C D C C A D A A A Q . 19 20 21 22 23 24 25 26 27 A n s. A D B A B B C B A Q . 28 29 30 31 32 33 34 35 36 A n s. D D C B A A BD A C D A CD B Q . 37 38 39 40 41 42 43 44 45 A n s. C B B A A A B B B Q . 46 47 48 49 50 51 A n s. A C A D D C PASSAGE
Q. 1 2 3 4 5 6 Ans. B D C C B A
www.physicsashok.in 110
FLUID MECHANICS www.physicsashok.in 111 LEVEL�2 1. (a) 50 N, (b) 1.7 × 104 kg m�3, (c) 8.5 kg (B1); 3.5 kg (B2) 2. (a) 7/8 m, (b) 5/4 mg 3. 12.50 cm 4. 7.2 g cm�3 5. H � h 6. . . 3 6 rg 4 r g 1 e t where 3 3 4 r .. . .. . . . . . . . . . . . . . . 7. 16 cm 8. 1 2 2 1 r r r . r 9. 4. SR2 (n1/3 � 1) ; 2 10. 6. mJ 11. 2.5 × 10�2 Jm�2 13. 5 cm 14. 2 1 2 1 g(h h ) a h h . . . 15. x = 2.81 m 16. 25.2 mm 17. 1.67 mm 18. 2.98 J 19. (a) 0.025 Jm�3, (b) 0.225 Jm�3, (c) 0.4 Jm�3 20.(a) 2M2g2 AY l (b) M2g2 2AYl 21. 3 cm 22. 2 1 0 1 0 2 2 1 0 2 w w (w w ) (w w )(t t ) (w w ) . . . . . . l . . . 23. 1 YA 2. mL 24. 10 c 25. Readingwill increasing. 26. The levelofwater falls. 27. . .2 g 1 2 W Ag h h 4 . . . 28. . = 45º 29. 75.4 cm LEVEL�3 1. (a)(i)D= 54 d, (ii) p = P0+ 14 (6H+L)dg ; (b) (i) v = g (3H 4h) 2 . , (ii) x = h(3H . 4h) , (iii) xmax= 3
4 H 2. (i) 10m/s, (ii) 14.1m/s, (iii) 2.5 hr 3. w = 2 1 1 3g d d 2L d . . . .. .. 4. (i) 0.2 m/s2, (ii) 0 m 2g A. 5. +29625 J/m3, � 30000 J/m3 6. mmin = .r2l ( .. ..) ; if tilted then it�s axis should become vertical. C.M. should be lower than centre of bouyancy. 7. (a) 0, (b) h = 0.25 cm, (c) a = g/6 (upward) 8. x = 2 m 9. f = 1 3g 2. 2R 10.500 Pa 11. H = L2 2 2g . 12. 6mm 13. 6 *******
ROTATIONAL MECHANICS
DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 1 A RIGID BODY There are severalbodies around uswhose shape and size remains the same because the distance between any two points of the body remains the same. Such a bodyis called �rigid�.An iron rod, steel ball, stone etc. are almost rigid bodies. The rigid bodies are so onlyfor small forces.Alarge force can be applied to a steelrod to elongate or bend it. So a �rigid body� is an �ideal case� when small changes in the shape or size of the body can be just neglected. We shall consider only such ideal bodies unless stated otherwise. C1:A(rigid) rod PQis sliding at the corner ofwalls as in the figure. The velocity of P is downward.Write the velocity ofQrightwardwhen PQmakes an angle .with floor. \\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ . Q u Pv Sol: As the rod is rigid, the distance between P andQwillnot change. For this, the approach velocity of P towardsQmust be zero. Nowapproach velocity of P towardsQ(see figure) can be written as vPx � vQx. Hence vcos(90º � .) � ucos. = 0 . Q u P v 90 � . 90º x . vsin. = ucos. . u = v sin cos . . = vtan. Anothermethod: Setting up axesXYandwriting the co-ordinates of P and Q, we have . Q u P vO (x, 0) x yy (0, v)x PQ = .x2 . .y2 l2 = x2 . y2 Squaring both side, l2 = x2 + y2
Differentiating relative to time, d dt (l2) = d dt (x2) + d dt (y2) = d dx (x2) × dx dt + d dy (y2) dy dt [chainruleofdifferentiation] . 0 = 2x dx dt + 2y dy dt Now, Pis coming down so y is decreasing: dy dt = �v Qismoving to the right, so x is increasing: dx dt = u . 0 = 2x(u) + 2y(�v) . u = yx v = (tan.)v
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 2 A rigid body in motion Themotion of a rigid bodyis controlled bythe condition that approach velocityof anyof its point towards its any other point must be zero; only then the disance between two points of rigid bodywill remain constant duringmotion. We consider nowthe very important type of rigid bodymotion- rotation.Therewill arise two aspects of this motion-howdoes the body rotate andwhy does the bodyrotate. In the first casewe say that we are studying kinematics ofrotation.We define angular position, angular displacement, angular velocity, angular acceleration, and find their relationships. In discussing the second case, we turn to dynamics of rotation.We here define torque, angularmomentum,moment of inertia and recast Newton�s laws ofmotionfor rotating bodies. Note: Although we like to present the things in the above order, you may opt your own order of studies. Kinematics of rotation It is better to locate a rigid body by fixing XYZ-axis in the body (hence called body-axes) and measuring anglesmade by these body-axeswith axesX0Y0Z0 fixed in laboratory. Anglesmade bybody-axeswith respective axes of lab-frame are sometimes known asEulerian angles denoted by., . and ..: Y0 Z0 X0 Y Z X O X � X0 : . Y�Y0 : . Z � Z0 : . These specifythe angular position ofthe body.During rotationthe angular positions changewith time. Rotation: If a bodymoves in such awaythat either a point, or a line ofthe bodyremains stationary, themotionis called rotation. centre of rotation O O axis of rotation xx. In the figure point Ois at rest. It is called centre of rotation. If a line likeXX. be at rest the line is called axis of rotation. If a rigid body rotates about of point, the points not lying on the centre of rotationmove along the surface of concentric spheres. On the other hand, for a rigid body rotating about an axis, the points not lying on the axismove in circleswhose centres line in the axis.Kinematics of such points are those of circularmotion.Point P inthe figure, for example, has a centripetal acceleration aC and a tangential acceleration aT,where aT
aC a a P C = v2 r = .2r .....(i) aT = dv dt = ddt . r .....(ii) a = 2 2 c T a . a .....(iii) Angular displacement If a rigid bodyturns, all its points suffer the same angular displacement but different linear displacements. The number ofsuchlinear displacements is infinite.Thereforewe prefer angular displacement for describing rotation. We define infinitesinalangular displacement ( d.. ) as a vectorwhose direction
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 3 is along the thumb of right handwith fingers curled in the sense of rotation. d. d. . It isfound that suchsmallangular displacments obeycommutativelawofvector addition and therefore qualifyas vectors: 1 d.. + 2 d.. = 2 d.. + 1 d.. . d.. is a �vector�. Since there is no physical effect in the direction of d.., it is called a pseudo vector. It is defined along the axis of rotation usingright hand rule and is also called and axialvector.The cross product of realvectors is a pseudo vector.Thus r. × p. , r. × F. , etc. are psuedo vectors.You need not go into more details at this level.Youmay see other pseudo vectorswhile studyingmagnetic fields. Angular velocity The rate of angular displacement is known as angular velocity.We denote it by ... Thus, ..= ddt .. Its direction is along the rotational axis at the givenmoment and is along d.. . Unit�rad s�1 or s�1 Other units� 1 rps = 1 revolution (rotation) per second = 2. rad s�1 1 rpm = 1 revolutionperminute = 260. rad s�1 = 30 . rad s�1. Angular Acceleration The rate ofchange of angular velocityis called angular acceleration. ..= d
dt .. . . . . . . 90º = d. dt Note that ..is in the direction d.. (the change in ..).We showthe examples for (i) ..inthe direction of .., (ii) ..opposite to ..and (iii) ..perpendicular to ..in the figures. In general ..and ..can have any angle between themjust as acceleration a. and velocityV . inlinearmotion. C2: Aparticle ismoving in a circle. Its angular velocityis ..and linear velocity is v. . The radius vector fromcentre to particle is r. . Determine (a) ..× r. (b) ..× v. (c) r. × v. . Sol: The vector ..is pointing out of the place of the circle V r .
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 4 (a)Curl the fingers of right hand from ..to r. . The thumb is found to point along v. . Now we know that | ..× r. | = .r = v. Hence magnitude also equals the magnitude of v. . Two vectors are equal if they have the same direction and the samemagnitude.Hence 90º . . × r . × r (b)We put the tails of ..to v. together and curlthe fingers ofright hand from ..× v. . The thumb points towards the centre of the circle, i.e., along �r. . Themagnitude is .v.Thus ..× v. = .v ( rr. . ) 90º . . × r v . where ( rr. . ) is unit vector directed towards the centre. (c)We put the tails of r. and v. together and curl the fingers of right hand from r. to v. . The thumbpoints parallel to ... v 90º r × V v Direction of Now, |r. × v. | = rv. Hence r. × v.
= rv . .. . . . . .. . . . where . .. . . . . .. . . is unit vector along ... C3:Aparticle ismoving in a circle of radius r with speed v. Find an expression for its acceleration. Sol: Let v. be the velocity, r. be the radius vector and ..be the angular velocity of the particle. Then V . = ..× r. Differentiating both the sides relative to t, dV dt. = d dt ( ..× r. ) = ddt.. × r. + ..× dr dt . Now, ddt.. is angular acceleration, ..and dr dt . is velocity v. . Hence dv dt . = ( ..× r. ) + ( ..× v. ). . ..= ( ..× r. ) + ( ..× v
. ) Using rules of cross product, ( ..× r. ) is ofmagnitude .r and is either along + v. or along � v. . This is tangential acceleration: aT = .r
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 5 ( ..× v. ) is directed towards centre and is called centrepetal acceleration aC = .v = v 2 r = .2r. Rotation about stationary axis Let the axis of rotation be �static�in laboratory. The rigid bodyis turning at angular velocity .along this axis. Then taking Z-axis along the rotationaxis,we canwrite componentofangular velocity. The angular acceleration ..will the also be along the defined axis andwe canwrite. . . axis Z d. . dt = Z .z= z ddt . = 22 ddt. = z d d d dt . . . = z dd.. .z Uniform rotation: If the body turns by equal angles in each equal time interval, howsoever small, the rotation is said to be uniform. The angular velocityis constant.Hence ddt. = .z = constant. . .. = .z .t Thus, angle turned = angular velocity× time taken. C4:Atable fan is rotating at uniformangular speed of 200 rpm. Howmuch angle does it turn in 3s? Sol: Here .z = 200 rpm = 200 revolutiononeminute = 200 × 30 . rads�1. [. 1 rpm= 30 . rad s�1]
.t = 3s
.. = ? Using .. = .Z.t, we have .. = 200 × 30 . × 3 radian = 62.8 radian Note that radian is S.I. unit of angle.Youmay use . radian= 180º to find it in degreeswhich is not necessary unles told to do so. Rotation with uniform angular acceleration: If the angular velocity changes by equal value in each equal time interval howsoever small, the rotation is uniformly accelerated. The angular accelertion in this case is constant..Z = constant z ddt . = .Z = constant This equationis similar to x dv dt = ax = constant Hencewemayobtain kinematical equation using similar treatment.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 6 C5: Angular displacement of allpoints ofa rigid bodyis same. In case ofnon-rigid bodygreater thandistance of the point fromaxis of rotation greaterwillbe its angular displacement. 2 1 P1 P2 P2´ P1´ O 2 1 P2´ P1´ P1 P2 O Rigit body ( 1= 2 ) Non-rigit body ( 2> 1) Equation of LinearMotion and RotationalMotion LinearMotion RotationalMotion (a) If acceleration is 0, v = constant and s = vt. (a) If acceleration is 0, .= constant and . = .t. (b) If acceleration a = constant, (b) If acceleration a = constant then (i) s = (u v) 2. t (i) . = 1 2 ( ) 2 . . . t (ii) a = v u t . (ii) . = 2 1 t . . . (iii) v = u + at (iii) .2 = .1 + . t (iv) s = ut + (1/2) at2 (iv) . = .1t + 12 . t2 (v) v2 = u2 + 2as (v) .22 = .12 + 2 ... (vi) Snth = u + 12 a (2n � 1) (vi) .nth = .1 + (2n � 1) 2. (c) If acceleration is not constant, the above (c) If accelerationis not constant, the above equation equationwillnot be applicable. In this case will not be applicable. In this case (i) v = dx dt (ii) a = dv dt =
2 2 d x dt (i) . = ddt. (ii) . = ddt . = 22 ddt. (iii) vdv = ads (iii) .d. = .d . C6: Aflywheel of radius 0.5metre starts fromrest with angular acceleration of 0.34 radian/sec2.What will be its angular velocity after 100 sec ? Sol. Here .2 = .1 + .t .2 = 0 + 0.34 × 100 = 34 radian/sec. C7: Acircular disc of radius 2mis revolving at 240 revolutions per minute a torque is appliedwhich slows it at constant rate of . radian/sec2. Inwhat time the discwill completelystop ? Sol. Here .2 = .1 + .t or 0 = 2.n � .t . t 2 n 2 240 8 sec. 60 . . . . . . . . .
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 7 C8: In the above question, calculate howmany revolutions the discmakes before it comes to rest. Sol. If . is the angular displacement then, .22 = .12 � 2.. or 0 = .12 � 2.. 2 2 2 2 2 1 4 n 4 240 32 2 2 2 60 . . . . . . . . . . . . . . . . . . . . . No of revolutions = ./2. = 16 Example 1: Two gears G1 and G2 have radii and R > r. They are coupled and angular velocityofG1 varies at uniformrate of ..Determine the angle turned byG2 in time t if themotion starts fromrest. G1 G2 R T Sol: Since the contact point of gears is not sliding, the velocityis common. Hence .1 = vr (magnitude) r R .2 = vR .1 = vr . V 2 = vR (magnitude) . .2 = 1r R . (magnitude) Differentiatingthis .2 = .1 r R (magnitudes) Now we can use for gear �2 .. = .Z 0t + 12 .Zt2 Then.. = 0 + 12 r R. . . . . . . t2 = 12 r
R. . . . . . . t2 [TheGear-1 turns by 12 .t2whileGear-2 turns r R . . . . . . �times this value.] Example 2. Adisk 8 cmin radius rotates about its central axis at a constant rate of 1200 rev/min. Determine (a) its angular speed, (b) the speed at point 3 cmfromits centre, (c) the radial acceleration of a point on the rim, and (d) and total distance a point on the rimmoves in 2.00 s. Sol. (a) 1200 2 40 rad / s 60. . . . . . = 40 × 3.14 rad/s = 126 rad/s (b) v = r . = 0.03 × 126 = 3.78 m/s
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 8 (c) ar = R.2 = 0.08 × (126)2 = 1.26 km/s2 (d) . = .t = 126 × 2 = 252 radian . SR . . . S = R. = 0.08 × 252 = 10.1 m Dynamics of rotation The study ofrotationalmotionwith explanationof its cause is known as rotational dynamcis. According to rotationaldynamics, �torque� is responsible for changing angular velocity.Torque is also known asmoment of force. MOMENT OF FORCE If a body is free to rotate about a point O at rest in an inertial frame S, and a force F is applied atA, if produces rotation about O (see figure).We find by experiment that rotational effect is directly proportional to r, R and sin..We define a quantity .O r F . F . .r . = rF sin . We define torque as a vector by ..= r. × F. . Which is combined effect of force and its action point.We call it moment of force F about Oor torque about O. The torque of a force F. about a point is defined as the cross product ofthe position vector of action point of the force and the force vector. In the figure, torque about Ois ..= r. × F. .Note howthe direction is obtained using right hand rule for cross product and vector ..is shown at O. S.I. unit of torque isNmand [ML2T�2] is its dimensional formula. Moment arm : Moment arm of a torque is the perpendicular (r.) dropped on action line of force fromthe point about which torque is calculated. F r|| r. moment arm r
In the figure r.. ismoment arm.We canwrite r. as vector sumof r.. and r. ||which is parallel to F. . Then ..= ( r.. + r. ||)× F. = ( r.. × F. ) + (r. ||× F. ) = r.. × F. Giving | ..| = r. Fsin90º = r1F. C9: Aforce of 20Nacts at (2, 2 3 ) along the line y= 3 x. Calculate the torque of this force about the origin. y = 3x F (2, 2 3) Z (0, 0) Y X
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 9 Sol: The line of action ofF passes through origin. Hence r.. = 0 ..= r.. × F. = 0. C10: In the previous example, calculate the torque of the force about ponit (1, 0)m. Sol:We have 60º y = 3x (20N) F (2, 2 3) r Z O (1, 0) Y X r. ..= r. × F. = r.. × F. . | ..| = r.F Fromfigure | r.. | = (1m) sin60º = 3 2 m F = 20 N . . = 10 3 Nm(inward along - Z axis) In vector from = (10 3 Nm)(� ^k ). Example 3. Aforce of F . 2 i� . 3�j . Nis applied to an object that is pivoted about a fixed axle aligned along the z. coordinate axis. if the force is applied at the point r . 4 �i . 5�j . m, find (a) themagnitude of the net torque about the z axis and (b) the direction of the torque vector ... Sol. . . r . F . . . . . .4�i . 5�j.. .2�i . 3�j. . . . 12 k� .10 k� . 2 k� . Torque about a point due to several forces If severalforces act on a body at various points, their rotational effect about
a point is described bynet torque. Net torque or resultant torque is the vector sumof torques due to all the forces about the same point.We denote it by. . (gamma). . . = 1 ..+ 2 .. + .......... N .. where 1 .., 2 .. , .......... N ..are torques about the same point due to forces 1 F. , 2 F. , ........ N F. . Ifthese be acting at 1 r. , 2 r. , ...... N r. relative to the point about which torque is requested.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 10 1 ..= 1 r. × 1 F. 2 .. = 2 r. × 2 F. | | | N ..= N r. × N F. In the special case, when the forces are concurrent, we can obtain net torque as torque due to net force. Torque about an axis Let several forces 1 F. , 2 F. ...... be acting on a body.We want to find torque about Z-axis. Thenwe take any point Oonthis axis. The action points have positions 1 r. , 2 r. ........... relative toO.Give the totalof this torque about Ois. . : . . = ( 1 r. × 1 F. ) + ( 2 r. × 2 F. ) + ............... Its Z-component,..Z, is called torque about Z-axis. Torque about an axis is defined as the component of torque vector along the axiswhere the torque is calculated about any point on the axis. Z Z torque about Z-axis F2 F1 O Inthe figure, .z = .cos. It maybe positive, negative or zero. It is a scalar quantity. CoupleMoment Two forcesofthe samemagnitude, opposite directions and actingalong parallel lines constitute a couple. The torque due to couple of forces is called �couple moment�. In the figure, F. and � F. constitude a couple. d r.2 r.1 r1 �F r2 F Their torque aboutOis given by . . 0 = 2 r. × F.
+ 1 r. × (� F. ) = (r.2 × F. ) + r..1 × (� F. ) = d. × F. [see figure] |. . 0| = d × F (one of the forces × separation) �Couplemoment� is independent of the point about which it is calculated. It is a �Free vector�. Anticlockwise and clockwise torque The torque which tends to rotate the body clockwise is known as clockwise torque. F .F .grav mg O r C In the figure grav .. is clockwise about O.An anticlockwise torque is that whichtends to rotate the bodyanticlockwise. In the figure F .. is anticlockwise torque about O.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 11 C11: Themore is the value of r, themorewill be torque and easier to rotate the body. (i) The handle of screwdriver is taken thick. (ii) In villages the handle of flour-mill is placed near the circumference. (iii) The handle of handpump is kept-long. (iv) The rinch used for opening the tap, is kept-long. C12: Work done by torque = 2 1 d . . . . . = Torque× angular displacement. Example 4. Given that, r . 2 �i . 3�j . and F . 2i�. 6k� . . Themagnitude of torquewillbe� (A) 405 N�m (B) 410 N�m (C) 504 N�m (D) 510 N�m Sol. We knowthat, . . r . F . . . . . .2�i . 3�j.. .2i� . 6 k� . . . .12.. �j.. 6.. k� ..18�i . . . .12�j. 6 k� .18i� . Now | . | . (.12)2 . (.6)2 . (18)2 . | . | . 144 . 36 . 324 . 504 . Hence (C) is correct. Example 5. An automobile engine develops 100 kilo-watt,when rotating at a speed of 1800 rev./min.The torque developed by it will be � (A) 60 N�m (B) 531 N�m (C) 5.31 N�m (D) 6.0 N�m Sol. The power delivered by the torque . exerted on rotating body is given by P = .. or . . P. Here P = 100 kW= 100,000Watt 1800 2 60 rad / sec. 60 . . . . . . . . .. .. 105 513 N.m 60 3.14 . . . . Hence (B) is correct. Equilibrium of rigid bodies Several forces and torques acting ona rigid bodymaybe �balanced�. Then the bodyis said to be in equilibrium ormeahanical equilibrium�. It is classified into two parts: rotational equilibriumand translational equilibrium. Translational Equilibrium: Abody is said to be in translational equilibriumif the vector sumof all the forces be zero. The forces are said
to be balanced and Newton�s first law ofmotion is obeyed if . i F. = 0.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 12 Rotational Equilibrium: If the vector sumof all the torques acting on a body about a point be zero, the body is said to be in rotational equilibriumabout that point. If allthe forces acting onthe bodylie in a plane, we can classify torque into clockwise and anticlockwise. For rotational equilibrium,we get Sumofclockwise torques = Sumof anticlockwise torques Principle of moments For a body in rotational equilibriumunder coplanar forces clockwise torque is balanced by anticlockwise torque. In vector form, the vector sumof all the torques is zero. ..i = 0 Rotational equilibrium is rotational version of theNewton�s first lawofmotion inwhich forces acting on a particle are balanced producing translational equilibrium. Mechanical Equilibrium Mechanical equilibriumrefers to a systemunder rotational aswellas translational equilibrium. In this case: or, . i .. = 0 [rotationalequilibrium] or, . i F. = 0 [translationalequilbrium] Action point of resultant force. The action point of resultant force is that point about whichtorque due to the forces is zero. Centre of gravity (CG.) is the action point of resultant of gravitational forces acting on all the particles forming the body. Inthe figure, letCbe the centre of gravityofthe body.Let dmbe anelementary mass at position .Torque d.. about C is given by C.G. r C c dmg d..= r × ( dmg. ) If the total torque is evaluated,we have c .. = r dmg . . . . whole body Bydefinition, this torque is zero. If g. is uniformover thewhole body, then c .. = ( r dmg) . . . . = dm r g . . ... . . . . . = 0 [by definitionofCG] Now dmr. . is the first moment ofmass about C. IfC happens to be centre ofmass also, the firstmoment ofmass of the whole body about it is zero.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 13 whole body dmr. . = 0 [C is centre ofmass] Thus, for uniform g. , centre of gravitycoincideswith centre ofmass. CM CG If g. varies over points of the bodyCG and CMdo not coincide as indicated for a long rod shown inthe figure. C13: (i) When a rigid body is rotating about a fixed axis and a force is applied on it at some point thenwe are concernedwith the component of torque of this force about the axis of rotation not withthe net torque. (ii) The component of torque about axis of rotation is independent of the choice of the originO, so long as it is chosen on the axis of rotation, i.e., wemay choose point Oanywhere on the lineAB. (iii) Component of torque along axis of rotationABis zero if (a) F. ||AB (b) F. intersectsABat some point (iv) If F. is perpendicular toAB, but does not intersect it, thencomponent oftorque about lineAB=magnitude of force F. × perpendicular distance of F. fromthe lineAB(called the lever armormoment arm) ofthis torque. (v) If there aremore than one force 1 F. , 2 F. ........... acting on a body, the total torquewill be .. = 1 r. × 1 F. + 2 r. × 2 F. + .......... But if the forces act on the same particle, one can add the forces and than take the torque of the resultant force, or .. = r. × ( 1 F. + 2 F. + ......) (vi) In general, ifmany forces are acting on a bodythe net torque is different about different points but if the body is in translational equilibrium(i.e. 1 F. + 2 F. + ...... = 0) the net torque about different points is equal. This can be shown as follows: Let the force 1 F. + 2 F. ,.......... are acting on a rigid body at position vectors 1 r. , 2 r. ,.......... etc. The position vectors are about the originO. Then 0
. . = 1 r. × 1 F
. + 2 r. × 2 F. + ......... .....(i) Now, suppose the position vector of an another point P is p r. and nowwewant to find the torque about this point P, then, p . . = ( 1 r. � p r. ) × 1 F. + ( 2 r. � p r. ) × 2 F. + ........ = ( 1 r. × 1 F. + 2 r. × 2 F. + .....) � p r. ( 1 F. + 2 F. + ......) .....(ii) Fromequations (i) and (ii) we can see that 0 . . = p . . if 1 F. + 2 F. + ..... = 0 or the net torque of allthe forces about anypoint is same if the body is in translational equilibrium. C14: It is correct that whenever the resultant force acting ona body is zero, the bodyis in static equilibrium. Sol. No; zero resultant torque in also necessary.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 14 C15:Ameter rod is acted upon by two forces, 2N and 4N, normal to its length as inthe figure.Determine the distance of the point fromCMat which their resultant force is acting. Sol: Bydefinition, actionpoint of resultant force is that aboutwhich the net torque becomes zero. 4N 10cm 2N CM (a) 4N 10cm 2N CM (b) Case (a): In this case the two torqueswill cancel each other only if the point lies lower than the action point of 1Nforce. Let its distance fromCMbe x cm. Then 4 × (10 � x) = 2 × (50 � x) 40 � 4x = 100 � 2x x = �30 cm 4N 2N CM PA 10 � x x 50 � x 2 x(50 � x) x 4 x(10 � x) Ncm P Negative sign shows that the action point P is 30 cmbelow the centre ofmass. Case (b): In this case let the point P be at distance x-above CM. Equatingmoment about P. 4 × (x � 10) = 2 × (50 � x) x CM 10cm P 2N 4N 4x � 40 = 100 � 2x 6x = 140 x = cm= 36.7 cm The point P lies 23.3 cmabove the CM. C16:Arectangular block having uniformdensityhas centre of gravity at C. It is placed on horizontal plane. Drawthe normal contact force and its action point. C(a) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ C The plane is now tilted by an angle . (sec figure (b). Still the . block is at rest under frictionalforce.Drawthe normal force and its action point. Sol: N . , mg . and f.
and must be concurrent forces for they are balanced. Action point of mg Action point of N mg (a) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ . (b) Action point of N Action point of friction mg N Action point of mg N [The largest angle . is that forwhichmg passes through edge.Above this angle, equilibriumis disturbed]. Example 6. Ahorizontal force F =mg/3 is applied on the upper surface of a uniformcube ofmassmand side a which is resting on a rough horizontal surface having µS = 1/2.The distance between lines of action ofmg and normal reactionNis : (A) a/2 (B) a/3 (C) a/4 (D) None
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 15 Sol. In this case, torque ofmg is balanced by torque due to normal reaction and applied force F. . N a x Fa mg a 2 2 . . . . . .. .. or N a x mga mga 2 3 2 . . . . . .. .. or N a x mga mga mga 2 2 3 6 . . . . . . .. .. But N = mg . mg a mgx mga 2 6 . . or x a a 3a a a 2 6 6 3 . . . . . Hence (B) is correct. Example 7: A uniform rod of length l lies at rest. Draw forces acting on it. Considering its rotationaland translational equilibrium, determine the force of friction acting at Qas a function of angle .. Howdoes it varryby decreasing angle . ?What factor willdecideminimumvalue of .without slipping ? \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ .l Q P Smooth vertical wall Sol: Figure shows howforces can be shown to be concurrent. Forces Rough onQ are N and fwhich adjust to produce the resultant Fc that passes through intersection ofN . and mg . . Considering horizontal and vertical forces,we have f = N. .....(i) N = mg .....(ii) Fc N P mg f Q Fc N. mg N´ Considering torque about a horizontal axis passing throughQ, mg 2l cos . = N´ l sin . .N..= mg 2 cot. .....(iii)
Using these, f= mg 2 . . . . . . cot. As . is decreased, cot. increases.Hence frictional force f increases. . No Slippapge 2.s .min cot . 90º Using lawoflimiting friction f . .SN .....(iv) Thus, 12 mgcot.....smg . cot... 2.S tan... s 1 2.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 16 . .min . tan�1 s 1 2 . . . . . . . The factor 12 comes fromthe cube�s centre of gravity and .S fromnature of . . contact surfaces.These control .min. Youmay notice that a personwalking on ice slips for larger steps (small ., limiting friction crossed). Small steps (large 0) ensure a friction less thanlimiting (no slipping). Example 8. Auniformboard ofweightWandwidth 2L hangs froma light, horizontal beam, hinged at thewall and supported by a cable (fig.) Determine (a) the tension in the cable and L 2L (b) the components of the reaction force exerted by thewall on the beam, in terms of W, Land .. Sol. Taking torque about O, .2L = (T sin .) 3L .× 2 L = 3TLsin . 2.= 3 T sin . ...(1) Also, for translatoryequilibrium, Fx = T cos . ...(2) L 2L w Fx O T Fy Fy + T sin . = . ...(3) Fromequation (1)T 2 3sin . . . Fromequation (1), (2) and (3) x F 2 cot 3 . . . and y F 3. . MOMENT OF INERTIA Moment of inertia takes same role in rotatorymotion asmass in translatorymotion. Smaller themoment of inertia about anaxis, easier to produce rotation about that axis; larger themoment of inertia, harder to produce
rotation.Thismeans thatmoment ofinertia about an axis is ameasure ofrotationalinertia of the bodyabout that axis. This is its physicalmeaning. Radius of gyration: To gyratemeans to rotate. The radius of gyration about an axis is defined as that distance fromthe axiswhere totalmass can be centred so as to give the samemoment of inertia as the actualmass distribution. I = .mr i i2 CM M Fromfigure, radius of gyrationKis given byK= I M . It can be shown that it has dimensions of length and depends on geometry of the system.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 17 The radius of gyration of a homogenous sphere about its diameter, for ex,maybewritten as K = 2MR2 5 M = 25 R Theorems on Moment of Inertia We shall discuss two important theorems onMI that help it. Theoremof perpendicular axes for laminear bodies and Theorems of parallel axes. Theorem of perpendicular axes Let Zaxis be normal to a lamina. Let us drawXYaxes intersecting Z-axis at (0, 0, 0). The points of lamina have z = 0.We shall denote a point in lamina onlyby(x, y) a for simplicitywithout a loss. Let mi bemass of i-th particle at (xi, yi, 0). Its distance ri fromZ-axis is given by 2 i r = 2i x + 2i y . Itsmoment of inertia about Z-axis ismiri2. For all theN-particleswe have IZ = .mi 2 i r = .mi( 2 2 i i x . y ) = .mi 2i x + .mi 2i y = Iy + Ix Statement: If thre axes x, y and z be drosen such that two of themlie in the plane of a lamina, the sumofmoment of inertia about these two axes is equal to themoment of inertia about the third axis. This is known as theoremofperpendicular axes. C17: Auniformsquare plate hasmoment of inertia Ix and Iz about Xand Z axis shown in the figure. Showthat Iy = Ix = Iz/2. Also show that rotation of XY axes about t-axis by 45º or by . does not change the result. y z x CM Sol: Using the theoremof perpendicular axes, Iz = Ix + Iy .....(i) Bysymmetryofmass distribution Ix = Iy .....(ii) . Iz = 2Ix = 2Iy . Ix = Iy = Iz/2.
If the axes be rotated about Z-axis by ., even then Ix.. = Iy.. Ix...+ Iy......Iz . z x´´ y´´ I I I2 . .
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 18 Parallel axes theorem (Steiner�s theorem) Let a bodyhave amoment of inertia Iz about Z-axis and Icm about an axis passing throughCMof the bodyand parallel to the Z-axis. Let d be the separationbetween the two parallel axes andMbe themass of thewhole body. .i CMri.mi ri Ri Parallel axis through CM d d M Then Iz = Icm +Md2. This is known as parallel axes theorem. Themoment ofinertia of a bodyabout an axis is equalto the sumofits moment ofinertia about an axis parallel to the given axis and passing throughCM, and the product ofmass of the bodywith the square of distance between the parallel axes. Proof: Let i-th particle havemassm.Drop perpendiculars onZ-axis and CM-axis fromthis particle. TheMI about zaxis is given by (see figure). Iz = .miri .....(i) as fromZ-axis the distance ofmi is ri. The summation runs over all the particlar in the body. Now, construct vectors by arrows as in the figure. i r. = d. + i R . We have i r. . i r. = 2 i r = d2 + 2 d. . i R . + 2i R .....(ii) Using (ii) in(i) Iz = .mi(d2 + 2i R + 2 d. . i R . ) = (.mi)d2 + .mi 2i R + 2..mi i R . = Md2 + Icm + 2 d. ..mi i R . .....(iii) Let ' i r. be position vector of i-th particle fromCM. Let i . . + i R . = ' i r. . By definition ofCM, the sumof firstmoment ofmass about CMis zero. . .mi '
i r. = 0 . .mi .i + .mi i R . = 0 Now.mi i . . is a vector alongCM-axiswhile .mi i R . is normal to this axis. But their sumis zero.We knowthat a b. . . & a b. . . = 0 . a. = 0 = b. . .mi i R . = 0 .....(iv) From(iii)&(iv), we have I = Md2 + Icm This proves the theorem.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 19 Some Calculation of Moment of Inertia 1. Moment of inertia of a uniform rod about an axis passing through its CM and normal to its length: Let us choose an elementarymass dMat position x on x-axis. Its distance is also x fromaxis. Itsmoment of inertia dIz about Z-axis is given by dMdx x x =L2X x = � L2 dIz = dMx2 = Mdx L . . . . . . x2 The totalMI will be obtained by inetragion over thewhole rod (fromx= �L/2 to x= L/2). Thus Iz = ML L/ 2 2 L / 2x dx . . = ML 3 1/ 2 1/ 2 x3 . . . . . . . = M L3 L3 3L 8 8 . . .. . . . . .. . . .. = ML2 12 This is theMI of rod about CM-axis, normal to rod. About one of its ends: IfZ-axis passes normally through one end, wemay use parallel axes theorem. Here, Icm = ML2 12 CM d = L2 Z d = L2 Using parallel axes theorem
Iz = Icm + Md2 IL = M 2 L2 . . . . . . + ML2 12 = ML2 3 2. Rectangular Plate: Axis throughCM, in the plane of the plate and parallel to an edge: LetMbe themass of the plate. Thenmass per unit area is Mlb . Let us consider the plate ismade up of thin bars perpendicular to the axis one ofwhichis shown in the figure and hasmass dM. Themoment ofinertia of this bar about the given axis is given by dIx= 1 12 dM l2 CM dM b Y lX Integrating over all the bars, the totalmoment of inertia is given by I = 1 12 l2. . dM = 1 12Ml2
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 20 Similarly, themoment of inertia about anaxis (Y) throughCMin the plane ofthe plate normal to breadth b can bewritten asI = 1 12 Mb2 Moment of inertia about an axis normal to the plate & passing through CM: It can be obtained using perpendicular axis theorem. In the figure Ix = 1 12 Mb2 Iy = 1 12Ml2 . Iz = Ix + Iy = 1 12Mb2+ 1 12Ml2 = 1 12M(l2 + b2) If the plate is a square, its sides are equal then, l = b = a Iz = 16 Ma2 3. Rectangular bar:Arectangular barmaybemade up of rectangular plates of mass dM. ThenIz = dM( 2 b2 ) 12l . dM MZ b Integrating,we get h Iz = M( 2 b2 ) 12l . [Effect of h contained inM] 4. Ring orHoop:An element ofmass dMlies at a distance R fromZ-axis. . dIz = dMR2 Integrating over thewhole ring R Z I dM z = R2 . dM = MR2 5. MI orDisc about its axis: LetMbemass of uniformdisc ofradiusR. Thenmass per unit area isM/.R2.Let the disc bemade up of elementary rings of radius r andwidth dr.Themass of this ring is obtained: Area = 2.rdr dM = (mass per unit area) × area = 2 MR
. .
. . . . . × 2.rdr Z r Rdr 2.r dr Nowwemaywrite theMI of the ring about Z-axis as dIz = dM.r2
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 21 = 2 2M R r3dr Integrating over thewhole discwe get Iz = 2 2M R R 3 0 . r dr = 2 2M R R4 4 . . . . . . = 12 MR2 Diametrical axis: Let XandYbemutually perpendicular diametrical axes.Then Ix = Iy [bysymmetry] Y X Z Also, Iz = Ix + Iy [perpendicular axes theorem] Now Iz = mR2 . 12 mR2 = 2Ix = 2Iy . Ix = Iy = 14 mR2 6. Uniform solid cylinder: MI about its axis: Wemayimagine the cylinder to bemade up of large no of elementarydiscs, one of the themhasmass dM. Its MI about z axis is given by dIz= 12 dMR2 Integrating over thewhole cylinder, Iz = 12 R2 . dM = 12 MR2 R Z dM M MI about perpendicular bisector axis: It is equal to Iz = MR2 4 +
M 2 12l 7. Homogenous sphere, diametrical axis: Let us consider a homogenous sphere ofmassMand radius R. Itsmoment of inertia about diametrical axis shown in the figuremay be calculated as follows. Let us consider the sphere to bemade up of disc like slices one ofwhich is located at position z. Its thickness is dz and radius is R2 . z2 . TheMI of this elementary disc is dIz= 12 ×Mass × (radius)2 = 12 × .(R2 � z2)dz. × (R2 � z2) Z = R R2 � z2 dz z = 0 R z = �R Z where . = density = 2 M 4 R 3 .
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 22 dIz= 2 .. (R2 � z2)dz Integrating over thewhole sphere Iz = 2 .. R 4 2 2 4 R (R 2R z z ) . . . . dz = 2 .. R R R 4 2 2 4 R R R R dz 2R z dz z dz . . . . . . . . . . . . . . = 2 .. 2R5 4 R5 2 R5 3 3 . . . . .. .. = 2 .. 30 20 6 15 . . . . . . . . R5 = 815 .. R5 Putting the value of .,we get Iz = 25 mR2. Moment of Inertia for simple geometrical objects Some bodies of simple geometryare thin uniformrod, uniformrectangular plate, circular ringor hoop, uniform circular disc, uniformsolid cylinder, homogeneous sphere, hollowsphere, right circular cone etc.We shallgive you values ofmoment of inertia of these about frequentlyused axes. Table MI of some regular bodies Body Position of axis Moment of inertia Laminar body Z Y X laminar body Iz = Ix + Iy (perpendicular axis theorem) Any body d Z CM
M IZ = ICM +Md2 (ParallelAxes theorem) Body Positionof axis Moment of inertia
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 23 Uniformrod 90º CMl Total mass = m Iz = m 2 12l l Z Iz = m 2 3l Z. Iz = m 2 3l sin2. Z. CM Iz = m 2 12l sin2. CM b l Y [The axis is the plane of plate] Uniformrectangular plate CM b l X Iy = m 2 12l Z X Y [Axis Z is normal to plate] Iz = mb2 12 Iz = m12 (l2 + b2) Ring orHoop ZR [Z-axis : through CM and normal to plane of hoop] COM Iz = mR2
CM R [Axis as diameter] I = 12 mR2
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 24 UniformCircularDisc ZR [Z-axis: through CM and normal to disc] COM Iz = 12 mR2 [Axis as diameter] I = 14 mR2 Right CircularCylinder l [Axis through CM along the length] CM R I = 12 mR2 l [Axis through CM normal to length]CM R I = mR2 4 + m 2 12l Right Circular Cone CM I = 3 10 mR2 SphericalShell R CM [Diametrical Axis] I = 23 mR2 Solid homogenous sphere R [Dimetrical axis] I = 25 mR2 PartyHollowsphere R2 [Dimetrical axis] R2 I = 25 m 5 5
2 1 3 3 2 1 R R R R . . . . . . . .
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 25 C18: Theoremof parallel axes is applicable for anytype ofrigid bodywhether it is a two dimensionalor three dimensional,while the theoremof perpendicular axes is applicable for laminar type or two dimensionalbodies only. C19: In theoremof perpendicular axs, the point of intersection of three axes (x, y and z)may be any point on the plane of body (itmay evenlie outside the body).This pointmay ormay not be the centre ofmass ofthe body. C20:Moment of inertia of a part of a rigid body(symmetrically cut from thewholemass) is the same as that of thewhole body. e.g., in figure (a) moment of inertia of the section shown (a part of a circular disc) about an axis perpendicular to its plane and passing RO M (a) (b) through point Ois 12 MR2 as themoment of inertia of the complete disc is also 12 MR2. This can be shown as in figure. Suppose the given section is 1n th part of the disc, thenmass of the discwill be nM. Idisc = 12 (nM)R2 Isection = 1n Idisc = 12 MR2 C21: Calculate theM.O.I. of a circular disc about an axis passing through its centre and perpendicular to its plane; mass of the disc is 2 kg and its radius is equal to 50 cm. Sol. Here I = 12 MR2 2 I 1 2 50 0.25 kgm2 2 100 . . . . . . .. .. Example 9. In the above question calculate itsM.O.I. (1) about a tangent in its own plane; and (2) about a tangent r . to its plane. Sol. (1) Fromthe theoremof parallel axis,M.O.I. about tangent in its own plane. I =M.O.I. about a parallel diameter +MR2 I 1 MR2 MR2 4 . . 2 I 5 MR2 5 2 50 0.625 kgm2 4 4 100 . . . . . . . .. .. (2)M.O.I. about tangent r . to its plane. =M.O.I. about r . axis +MR2 1 MR2 MR2 3MR2
2 2 . . . 2 3 2 50 0.75 kgm2 2 100 . . . . . . .. ..
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 26 C22: The figure represents a thin circular disc ofmass 4kg and radius 0.60metre.Calculate itsmoment of inertia about an axisXX´, passing throughAand r . to its plane. O 0.2m A Y XY´ X´ Sol. Fromtheoremof parallel axis,M.O.I. about XX´. =M.O.I. about YY´ +Mr2 = 12 MR2 + Mr2 = 12 × 4 × (0.6)2 + 4 × (0.2)2 = 0.88 kgm2. C23: Calculate theM.O.I. of a uniformcylinder ofmass 6 kg, redius 0.4 metre and length 1metre about an axis passing through its centre and r . to its length. Sol. 2 2 I M L R 0.74 kgm2 12 4 . . . . . . . . . C24: There is a uniformcircular disc ofmass 10 kg and radius 2 metre. Calculate the radius of gyration if it is rotating about anaxis passing through its centre and r . to its plane. Sol. Here I = 12 MR2 . K = R2 = 22 = 1.414 m C25: Asolid sphere ofmass 6 kg and radius 2 metre is rotating about its diameter; then calculate the radius of gyration. Sol. Here I = 12 MR2 . K 2 R 2 2 1.26 metre 5 5 . . . . C26: For pointsmasses each ofmassmare attached to amassless string. Calculate the M.O.I. of the systemabout XX´. L L L m m m m XX´ Sol. M.O.I about XX´ = mL2 + 0 + mL2 + m(2L)2 = 6 mL2.
C27: Three point masses each ofmass mare kept as shown in the figure, the string is massless. Calculate theM.O.I. of the systemabout an axis passing through (1)AB, (2) BC and (3) about an axis passing . rlythroughA. 10m 6m 8mAB Cm mm Sol. (1) M.O.I about AB = 0 + 0 + m(6)2 = 36 m (2) 64 m (3) 164 m
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 27 Example 10. Three bodies have equalmassesm. BodyA is solid cylinder of radiusR, bodyB is a square lamina of side R, and bodyC is a solid sphere of radiusR.Which body has the smallest moment of inertia about an axis passing through their centre ofmass and perpendicular to the plane (in case of lamina) (A)A (B) B (C) C (D) D Sol. ForA 2 A MR I 2 . For B . . 2 2 2 B m MR I R R 12 6 . . . ForC 2 C I 2 mR 5 . where IA > IC > IB Hence, (B) is correct. Example 11. ApointmassmA is connected to a pointmassmB byamassless rod of length l as shown in the figure. It is observed that the ratio of the moment of inertia of the systemabout the two axiesBBandAA,whichisparallelto eachother and perpendicular mA mB A B A B l to the rod is BB AA II = 3. The distance of the centre ofmass of the systemfromthemass Ais (A) (3/4)l (B) (2/3)l (C) (1/2)l (D) (1/4)l Sol. IAA = mBl2 IBB = mAl2 BB AA I 3 I . AB m 3 m . . mA = 3mB But mAxA =mB(l � xA)
3mBxA =mB (l � xA) 3xA = l � xA 4xA = l A x 4 . l Hence (D) is correct. Example 12. Three uniformrods each ofmassmand lengthLmetre are connected to forman equilateral triangle. Calculate theM.O.I. of the systemabout an axis passing through one of the vertex and r . to plane of the triangle. Sol. M.O.I. of the rodAB, andAC about the given axis are each equal to 13 mL2. andM.O.I. of the rodBCabout the given axis (fromthe theoremof parallel axis) is 2 1 mL2 mx2 1 mL2 m 3 L 12 12 2 . . . . . . . . . .
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 28 2 1 2 3mL 10 2 mL mL 2 4 12 . . . Hence,M.O.I. of thewhole systemabout given axis I 1mL2 1 mL2 10 mL2 18 mL2 3 mL2 3 3 12 12 2 . . . . . A B C Dm, L x L, m m, L C28: The figure belowrepresents two spheres ofmassMand radiusR. Theyare kept in contact as shown. CalculateM.O.I. about axisYY´. YSol. Y´ I 7 mR2 7 mR2 5 5 . . I 14 mR2 5 . Example 13. Four spheres each ofmassMand radius r are kept with their centres on the corners of a squre of side L. CalculateM.O.I. of the systemabout an axis passing through one side of the square. A BD C M M ML M L L L Sol. M.O.I. aboutAB 2 Mr2 2Mr2 2Mr2 ML2 2Mr2 ML2 5 5 5 5 . . . . . . . . . . .. .. .. .. 8Mr2 2ML2 2M.4r2 5L2 . 5 5 . . . . C29: Two point masses are kept at distance of 9metre on a thin uniformrod of negligiblemass.The rod remains horizontal if it is pivoted atA. CalculateM.O.I. about a vertical lineXX´ through the pivotA. Sol. Since the rod remains horizontal aboutA, hence 1 × g(9 � y) = 2g × y . y = 3 metre . I = 2(3)2 + 1(6)2 1kg 2kg XX´ A 9 � y y
I = 54 kgm2 Example 14. TheM.O.I. of a thin uniformrod ofmassMand lengthL about an axis passing through its centre is I1. This rod is bent in formof a ring, and if I2 is theM.O.I. of the ring formed about its centre thencompare I1 and I2. Sol. For thin rod, 2 1 I 1 ML 12 . For the ring, 2.r = L . r L 2 . . 2 2 2 I Mr M L 2 . . . .. . .. . 2 2 1 2 2 2 I 1 4 ML I 12 ML 3 . . . . .
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 29 Example 15: Three rods each ofmassmand length l are joined together to form an equilateral triangle as shown in figure. Find the moment of inertia of the systemabout an axis passing through its centre ofmass and perpendicular to the plane of the triangle. A B C CM Sol:Moment of inertia of rod BC about an axis perpendicular to plane of triangle ABC and passing through themid-point of rod BC (i.e.,D) is I1 = m 2 12l Fromtheoremof parallel axes,moment of inertia of this rod about the asked axis is I2 = I1 + mr2 A B C CM 30º Dr m 2 2 m 2 m 12 2 3 6 . . . . . . . . l l l .Moment of inertia of all the three rods is I = 3I2 = 3 m 2 6 . . . . . . l = m 2 2l Example 16. Three identical thin rods each ofmassm&length l are placed along x, y&z-axis respectively they are placed such that, one end ofeach rod is at origin �O�. Thenmoment ofinertia of this systemabout z-axis is (A) m 2 3l (B) 2m 2 3l (C) ml2 (D) m 2 4l Sol. Since the one end of each rod is at origin �O�
. 2 x y m I I 3 . . l According to perpendicular theorm. Iz = Ix + Iy 2 2 2 z m m 2m I 3 3 3 . . . l l l Hence (B) is correct. Example 17: Find themoment of inertia of a solid sphere ofmassMand radius R about an axisXXshown in figure. xx Sol: Fromtheoremof parallel axis, IXX = ICM + Mr2 = 25 MR2 + MR2 = 75 MR2 x x r = R CM
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 30 Example 18:Consider a uniformrod ofmassmand length 2lwithtwo particles of massmeachat its ends. LetABbe a line perpendicular to the lengthof the rod and passing through its centre. Find themoment of inertia ofthe systemabout AB. l l m m B A Sol: IAB = Irod + Iboth particles = m(2 )2 12l + 2(ml2) = 73 ml2 Example 19: Find themoment of inertia of the rodAB about an axisYYas shown in figure.Mass of the rod ismand length is l. Ay y B Sol:Mass per unit length of the rod = ml Mass of an element PQ of the rod, dm= . m . . . . l . dx Perpendicular distance of this elementalmass about yy is r = x sin.. .Moment of inertia of this small element of the rod (can be assumed as a point mass) about yy is, dI = (dm)r2 = . m dx . . . . l . (x sin.)2 = ml sin2. x2 dx . Moment of inertia of the complete rod, I = xx 0 dI . . .l = ml sin2. 1 2 0 . x dx Ay y B Q
P x r = m 2 3l sin2. Note: (i) I = 0 if . = 0 (ii) I = m 2 3l if . = 2. or 90º Example 20. Two rods ofequalmassmand length l lie along the x-axis and y-axiswith their centres origin.What is themoment of inertia of bothabout the line x= y : (A) m 2 3l (B) m 2 4l (C) m 2 12l (D) m 2 6l Sol. 2 2 x y m I I sin 12 . . . l 45º x y m 2 m 2 I 24 24 . . l l , 2m 2 I 24 . l m 2 I 12 . l Hence (C) is correct.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 31 Example 21. Arigid bodycan be hinged about any point on the x-axis.When it is hinged such that the hinge is at x, themoment of inertia is given by I = 2x2 � 12x + 27 the x-coordinate of centre ofmass is (A) x = 2 (B) x = 0 (C) x = 1 (D) x = 3 Sol. I = 2x2 � 12x + 27 or dI 4x 12 dx . . Forminimummoment of inertia dI 0 dx . . 4x � 12 = 0 . x = 3 ButM.I. about an axis passing throughcentre isminimum. Hence (D) is correct. Example 22. The figure shows a uniformrod lying along the x-axis. The locus of all the points lying on the xy-plane, about which themoment of inertia of the rod is same as that about Ois y O x (A) an ellipse (B) a circle (C) a parabola (D) a straight line Sol. 2 0 m I 3 . l Here 2 r x y2 2 . . . . . .. .. l . 2 2 P m I mr 12 . . l r y P (x, y) l/2 x � l/2 But I0 = IP or 2 2 2 m m m x y2 3 12 2 ... . .. . . .. . . . . ... . .. l l l
or 2 2 2 m m m x y2 3 12 2 ... . .. . . .. . . . . ... . .. l l l or 2 2 3m m x y2 12 2 ... . .. . .. . . . . ... . .. l l or 2 2 x2 x y2 4 4 . . . . l l l or x2 + y2 � xl = 0 or 2 2 x2 y2 x 0 4 4 . . . . . l l l or 2 2 x y2 2 4 . . . . . .. .. l l This is an equation of circle. Hence (B) is correct.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 32 Question No. 23 to 26 The figure shows an isosceles triangular plate ofmassMand baseL. The angle at the apex is 90º. The apex lies at the origin and the base is parallel to X-axis. l X M Y Example 23. Themoment of inertia of the plate about the z-axis is (A) ML2 12 (B) ML2 24 (C) ML2 6 (D) None of these Sol. tan 45º h/ 2 . l . h 2 . l In .ACD, tan 45º yx 2h . . l A B C 45º 45º y dx D h l/2 l/2 . y x 2 / 2 . .ll . y = x . dA= 2y dx = 2xdx . dm= 6 dA = 2 x .dx . . .2 2 2y dI dm dm(x) 12 . .2 dI 2x dx 4x 2x dx x2 12 . . . . . . . . . .
3 2x dx 3 dI 2x dx 3. . . . dI 8 x3 dx 3 . . . /2 4 3 0 8 x dx 8 x 3 3 4 . . . . . . . . . .l 8 4 4 3 64 24 . . . . l l Here M area of plate . . 2 M 4M 12 2 . . . . l . l l . 4 2 2 4M M I 24 6 . . . l l l ML2 I 6 . (. l = L ) Hence (C) is correct.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 33 Example 24. Themoment of inertia of the plate about the x-axis is (A) ML2 8 (B) ML2 32 (C) ML2 24 (D) ML2 6 Sol. dI = (dm) x2 dI = (2x . dx) x2 dI = 2. x3 dx . /2 3 0 I . 2.. x dx l x4 4 I 2 2 4 64 . . . . . . . . . . . . . . . . l 4 4 4 2 4M I 2 64 32 32 . . . . . . . l l l l M 2 I 8 . l Hence (A) is correct. Example 25. Themoment of inertia of the plate about its base parallel to the x-axis is (A) ML2 8 (B) ML2 36 (C) ML2 24 (D) None of these Sol. 2 dI dm x 2 . . . . .. .. l
. 2 / 2 x 0 I dm x . 2 . . . . . .. .. l l ML2 I 24 . Hence (C) is correct. Example 26. Themoment of inertia of the plate about the y-axis is (A) ML2 6 (B) ML2 8 (C) ML2 24 (D) None of these Sol. . . 2 dm 2 4dmx dI 2y 12 12 . . . /2 2 0 dmx I 3 . .l M 2 I 8 . l Hence (B) is correct.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 34 TORQUE AND NEWTON�S SECOND LAW In section 2.4,we have discussed about the turning effect of a force, torque.We knowthat for a bodyto be in totationalequilibrium, the sumof allthe torque acting on itmust be zero.Nowwhat happens ifnet torque is not zero. The case is similar if net forces acting on a body is not zero, the body wil accelerate according to Newton�s second law. In rotationalmotion also the lawholds good, but require somemodificationaswhen net torque ona bodyabout a given axis of rotationis not zero, bodywillhave angular acceleration.Themagnitude of angular acceleration can be obtained byNewton�s Second lawin rotationalmotion. In translationalmotionwe use F = ma ...(1) In rotationalmotionwe use . = I . ...(2) Let hand side is the net torque acting on the body and on right hand side I is themoment of inertia of the body about the given axis and ..is the angular acceleration of the body. This corresponds toNewton�s second lawfor translationalmotion, a ..F,where torque has taken the place of force and correspondinglythe angular acceleration ..takes the place of the linear acceleration a. In the linear case, the acceleration is not onlyproportional to the net force but is also inverselyproportional to the inertia of the body, whichwe call itsmassm, thuswe canwrite a = F/m. In case of rotationalmotionmoment of inertia plays the role ofmass. Aswe have discussed that the rotationalinertia of an object depends not onlyonitsmass, but also on howthat mass is distributed. For example, a large diameter cylinderwillhave greater rotational inertia thanone of equal mass but smaller diameter (longer than previous).The formerwillbe harder to start rotating, and harder to stop as itsmoment of inertia is larger.When themass is concentrated farther fromthe axis of rotation, the rotational inertia is greater. This is the reason why in rotationalmotion the mass of a body can not be considered as concentrated at its centre ofmass. For understanding the application ofNewton�s second lawin rotationalmotion, consider the rod shown in figure, pivoted at an end about which it can rotate. If two forces F1 and F2 are applied on it as shown fromopposite directions, tend to rotate the rod. The respective torque of these forces are given as 90º r2 r1 F1 Clockwise torque due to force F1 is .1 = F1 sin . . r1 Anticlockwise torque due to force F2 is .2 = F2 . r2 Let .1 > .2, the rod rotates in clockwise directionwith an angular acceleration ., which can be shown from equation (2), as net torque is in clockwise direction, we have 2 1 1 2 2 Fr sin F r M3 . . . . . . . . . .
l ...(3) Asmoment of inertia of the rod ofmassMand pivoted at one ofits end is givenas M 2 3l .Above equation (3) willgiveninitial angular acceleration ofthe rod.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 35 APPLICATION OF . = I. For problems based on pure rotation : Solution technique : Step (i) Drawthe free body diagram(if required) Step (ii) Identify, the axis of rotation and angular accelerationw.r.t. axis of rotation. Step (iii) Calculate net torquew.r.t. axis of rotation and then us . = I.with+ ve/ � ve sign convention. Example 27. Arod of lengthL andmassMis free to rotate about a vertical axis passing through its one end.Aconstant force F starts acting in horizontal plane. Initially the direction ofF . is perpendicular to length of the rod. Calculate angular velocityof rod as a function of angular displacement . at any time t. L cos L t = t F Sol. Let ..= angular displacement of rod at time t F ..= angular velocityat this time. ..= angular acceleration. . = FL cos .. ML2 3 = . . = 3Fcos ML . . d 3Fcos d ML . . . . . d 3F cos d ML .. . . . . . . 2 3Fsin C 2 ML . . . . When ..= 0 ; ..= 0 ; C = 0 . 6Fsin ML . . . KINETIC ENERGY OF ROTATING BODY Arigid bodyismade up oflarge number of particles.The kinetic energyof a rotating bodyis the sumof kinetic energyof its particles. Let .be the angular velocity. Then kinetic energy of i-th particle at distance ri fromrotation-axis is
2 i i K 1 m v 2 . where .ri = vi. The sumof such terms over thewhole body gives K = . 12 mi(.ri)2 = . 12 mi.2 2 i r . mi ri = 12 (.mi 2 i r ).2
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 36 The quantity.mi 2 i r is themoment of inertia of the bodyabout rotation axis, denoted byI. Then K = 12 I.2 Multiplying and dividing byI, K = 12 I2 2 I. . K = L2 2I Here Lis angularmomentumabout the axis. C30:Whether a particle is in translationalmotion, rotationalmotion or in both it merely depends on the reference point with respect towhich themotion of the particelis desribed. For example: Suppose a particle P ofmassmismoving in a straight line as shown in figure (a), (b) and (c). P v A r P v B90º . v cos. v sin.P v r C (a) (b) (c) Refer figure (a):With respect to pointA, the particle is in pure translationmotion. Hence, Kinetic energy of the particle can bewritten as K.E. = 12 mv2 Refer figure (b):With respect to point B, the particle is in pure rotationalmotion. Hence, the kinetic energy of the particle can bewritten as K.E. = 12 l.2 = 12 (mr2) 2 vr . . . . . . = 12 mv2 Refer figure (c):With respect to point C, the particle can be assumed to be in r
otational aswell as translational motion. Hence, the kinetic energyofthe particle can bewritten as K.E. = 12 m(v cos.)2 + 12 I.2 = 12 m(v cos.)2 + 12 (mr2) 2 vsin r . . . . . . . = 12 mv2 Thus, in all the three cases, the kinetic energy of the particle comes out to be the same. C31: If aparticle ismoving ina circle it is in pure rotationalmotionabout the centre ofthe circle,while for amoment it maybe in pure translationalmotion about some other point.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 37 C32: If a particleP ismoving in a circle, its angular velocity about centre ofthe circle (.c) is two times the angular velocity about anypoint on the circumference of the circle (.0) or .c = 2.0 This is because .P.CP = 2.P.OP (by property of a circle) P.P C O .c = pp P CP t . . . , .0 = pp P OP t . . . Fromthese relations we can see that .c = 2.0. C33: If a rigid body is rotating about a fixed axiswith angular speed ., all the particles in rigid bodyrotate same angle in same interval of time, i.e., their angular speed is same (.). They rotate in different circles of different radii. The planes of these circles are different. Linear speed of a particle situated at a distance r fromthe rotational axis is r . v = r. or v . r C34:Angular velocity of a rigid body (.) is ddt. .Here . is the angle between the line joining any two points (sayAandB) on the rigid body and anyreference line (dotted) as shown in figure. . B A . For exampleABis a rod oflength 4m. EndAis resting against a verticalwall OYand Bismoving towards right with constant speed vB = 10m/s. To find the angular speed of rod at . = 30º, we can proceed as under. OB = x =AB cos. . x = 4 cos. . Bv = 10 m/s B AOY or X dx dt = �4 sin. d
dt. . . . . . . ddt. . . . . . . = � (dx / dt) 4sin . [ dx dt = vB = 10 m/s] or . = � 10 4sin 30º = �5 rad/s Here negative sign implies that . decreases as t increases d 0 dt. . . . . . . . . C35: Acircular disc ofmass 2 kg and radius 1 metre is to be rotated about its centre in a horizontal plane with angular acceleration of 4 rad/sec2. Calculate the torque required. Sol. . . I. 1 MR2 2 . . . . a = 12 × 2 × (1)2 × 4 = 4 kgm2/sec2. C36: Aconstant torque of 40 Nm is applied to a wheel pivoted on a fixed axis. At what rate power is being furnished to thewheelwhen it is rotating at 120 revolution perminute. Sol. P = . . . = t . (2.n) = 40 × 2. × 120 60 = 502.6Watt.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 38 C37: Aflywheel ofmoment of inertia 7.5 kgm2 is rotating at 240 revolution perminute; calculate itsK.E. Sol. . = 2.n = 2. × 240 60 = 8 . . K.E. = 12 I.2 = 12 [7.5] [8.]2 = 12 × 7.5 × 64 × p2 = 2366 joule. Example 28.Auniformrod ofmassmand length l hinged at its end is released fromrest when it is in the horizontal position. The normal reaction at the hingewhen the rod becomes vertical is : (A) Mg 2 (B) 3Mg 2 (C) 5Mg 2 (D) 2Mg Sol. Loss in P.E. = gain inK.E. or mg 1 I 2 2 2 l . . or 2 m 2 mg 3 . . l l or 3g . .2 l F mg F mg m 2 2 . . l . or F mg m 2 2 . . l .F mg m 3g 5 mg 2 2 . . l . l Hence (C) is correct. Problems with pure rotation and pure translational of two or more connected objects. Solution techinque : Step (i) Drawthe free body diagram. Step (ii) Assume linear acceleration of translatory object&angular acceleration ofrotatoryobject. Step (iii) Use F . ma . . for translationmotion and . . I. .
for rotationalmotion. Step (iv) Relate linear and angular acceleration and solve the obtained equation and get the answer. Example 29. Consider a pulley fixed at its centre of mass by a clamp. Alight rope is wound over it and the free end is tied to a block ofmassm. Find the linear acceleration of block. Sol. For rotationalmotionof pulley: TR = MR2 . 2 . ...(1)
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 39 For translationalmotion of object : Mg � T = ma ...(2) and a = Ra ...(3) . MR2 a MaR TR . 2 R 2 . . . . .. .. ...(4) Solving (2) and (4) mg Ma ma 2 . . a T Mg . . a mg m M/ 2 . . Example 30. a uniformrod of length l, hinged at the lower end is free to rotate in the verticalplane. If the rodis held verticallyinthe beginning and thenreleased, the angular acceleration of the rodwhenit makes an angle of 45º with the horizontal(I =ml2/3) (A) 3g 2 2 l (B) 6g 2 l (C) 2 g l (D) 2 g l Sol. Loss in P.E. = gain inK.E. or . . 2 mg 1 cos 1 I 2 2 l . . . . Differenting both sides,mg sin d I d 2 dt dt . . l . . . or mg sin I 2 l . . . . 2 mg sin 45º 2m3 . . l l . 3g 2 2 . .
l Hence (A) is correct. Example 31. Aperson pulls along a rope wound up around a pulleywith a constant force F for a time entervalof t seconds. If a and b are the radiiof the inner and the outer circumference (a < b), then find the ratio ofwork done by the person in the two cases shown in the figure isW1/W2. F F Sol. W Case I Case II 1 = F a . Fa = I .1 . 1 Fa I . . . 2 1 1 t 2 . . .
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 40 . 2 1 W Fa 1 Fa t 2 I . . . .. .. 2 2 2 1 F a W t 2I . Similarly, 2 2 2 2 F a W t 2I . . 2 1 2 2 W a W b . Example 32. Auniformcylinder ofmassmcan rotate freely about its own axiswhich is horizontal.Aparticle of massm0 hengs fromthe end of a light stringwould round the cylinder which does not slip over it.When the systemis allowed to move, the acceleration of the descendingmasswill be (A) 0 0 2m g m. 2m (B) 0 0 m g m.m (C) 0 0 2m g m.m (D) 0 0 m g 2m. m Sol. m0g � T = m0a Tr = I . or mr2 Tr 2 . . or T mr ma 2 2 .
. . m0 or 0 0 m g ma m a 2 . . or 0 0 m g m m a 2 . . . . .. .. . 0 0 0 0a m g 2m g m m m 2m 2 . . . . Hence (A) is correct. Example 33. For thepivoted slender rod oflength l as shown in figure, the angular velocity as the bar reaches the verticalposition after being released in the horizontalposition is l/4 l l/4 (A) gl (B) 24g 19l (C) 24g 7l (D) 4g l Sol. Loss in P.E. = gain inK.E. or mg 1 I 2 4 2 l . . or 2 2 mg 1 m m 2 4 2 12 4 .. . . .. . . . . . . . .. . . .. l l l or 2 2 mg 1 m m 2 4 2 12 16 . . . . . . . . . l l l . 24g 7 . . l Hence (C) is correct.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 41 Problem based on situation in which same object is performing translation and rotational motion. This type of motion of an object can be considered as superposition of two types of motion taking place simultaneously. (A) translatorymotion ofcentre ofmass (B) rotationalmotionw.r.t. axis passing throughcentre ofmass Solution technique : Step (i) Drawthe free body diagramof object. a mg T mg T Step (ii) Assume linear accelerationofcentre ofmass and angular accelerationw.r.t. anaxis passing through the centre ofmass. Step (iii) Use F =ma for translatingmotion of centre ofmass and . = I. for rotationalmotionw.r.t an axis passing throughcentre ofmass. Step (iv) Relate a and . and solve the equations. Example 34.Acylinder ofmassMis suspended through a two stringswrapped around it an shown infigure. find the linear acceleration of centre ofmass of the cylinder. Mg Sol. a mg T mg T a mg For translatorymotion of centre ofmass Mg � T =Ma ...(1) For Rotationalmotion of cylinderw.r.t. an axis passing through the centre ofmass MR2 TR 2 . . ...(2) Relation between a&. a = R. ...(3) solving (1), (2) and (3) we get Mg MR . a Ma 2 R . . . . .. .. Mg = 3Ma/2 a = 2g/3 Example 35. Two discsAand B touch eachother as in figure.Arope lightlywound onAis pulled down at 2m/s2. Find the friction force betweenAandBif slipping is absent. R 2R 1 kg 2 kg 2 m/s2
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 42 Sol. mR2 TR fR 2 . . . .T f . mR a 2 R . . .T f . ma 1 2 1 2 2 . . . . . f. T � f = 1 ...(1) f Also, m´.2R.2 2 4 R2 a f 2R ´ 2 2 2R . . . . . F = a = 2 N LECTURE � 7 ROLLING MOTION (a) Rollingmotionof an object canbe consider as superposition of twomotions taking place simultaneously. (i) Translatorymotion of the centre ofmass (ii) Rotationalmotionof bodyw.r.t. anaxis passing through the centre ofmass. (b) In pure translationalmotion, all pointsmove with the same linear velocityVcm. In rotationalmotion about centre ofmass all pointsmovewiththe same angular velocityabout the central axis. v v + R v � R v2 + 2R2 v v v v v RR R Perfect Rolling : (a) If an object is in perfect rolling on a surface thanpoint of contact ofbodywith surface must be relativelyat rest w.r.t. surface. A B v surface for perfect rolling vAB = 0 net velocityofA= net velocityofB. (b) Perfect rolling on ground : (i) The relative instantaneous speed of the point of contact during rolling is zero. (ii) For perfect rollingmotion,work done against friction is zero. (iii) In case of perfect rolling the rotating object remain�s relatively at rest w
.r.t. surface on which it is rolling therefore the force of frictionwill be static in nature. (iv) Due to static nature of friction inperfect rolling its value lives in betweenOand µN.Where µ is coefficient of friction between rolling object and surface. (v) For perfectly rolling v � .R = 0where v is velocity of centre ofmass and ..is angular velocity. v v v � R VG = 0
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 43 C38: If sphere is in perfect rolling on platformor on plank v = .R = v1 sphere v v1 Solution technique : (i) Drawthe free bodydiagramof different components of system. (ii) Assume linear and angular acceleration of different components of system. (iii) User F . ma . . and . . I. . for translational and rotationalmotion respectively. (iv) Relate linear and angular acceleration. (v) Finally solve the equations. Note : Friction is not necessary for perfect rolling. Example 36. The spool shownin figure is placed onrough horizontal surface and has inner radius r and outer radiusR. The angle . betweenthe applied force and the horizontal can be varied. The critical angle (.) for which the spool does not roll and remains stationary is given by R r F (A) cos 1 r R . . . .. .. . . (B) cos 1 2r R . . . .. .. . . (C) cos 1 r R . . . (D) sin 1 r R . . . .. .. . . Sol. For equilibrium F cos . = f ...(1) and (F)r = fR or f r F R . ...(2) Fromequation (1) and (2) cos f r F R . . . . cos 1 r R . . . .. .. . . Hence (A) is correct. Example 37. Around object ofmassMand radiusRrollswithout slipping down an inclined plane set onangle . to horizontal. Find � (a) Linear acceleration ofcentre ofmass. (b) Minimumcoefficient of friction required for perfect rolling. assumingM.I.w.r.t. an axis passing through centre ofmass isMK2. Sol. For translatorymotion mg sin . � fr = ma ...(1)
For rotatorymotion fr R = MK2 (.) ...(2) and a � R. = 0 ...(3) R a mg mg cos mg sin fr solve equation (1), (2) and (3) 2 2 a g sin 1 K / R . . . and r 22 f mg sin 1 RK . . . . . . . . .
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 44 . fr . µmg cos . (for perfect rolling) Where 22 R 5 K 2 . . . . . . . . µ 2 tan 7 . . Imperfect Rolling (i) When point of contact of Rolling object is not relatively at rest. w.r.t. surface onwhich it in rolling then it is called imperfectRolling. (ii) Nature of frictionin this case is kinetic&it should opposing themotion of point of contact (strictly). (iii) Direction of linear acceleration and angular acceleration should be taken according to direction of friction (strictly). (iv) Nature of friction is kinetic therefore its value is µKN.Where µK is kinetic friction coefficient between surface and object. Transformation of Imperfect Rolling into perfect Rolling 0 N v0 imperfect Rolling v0> 0R v = R perfect Rolling µN 0 N v0 imperfect Rolling v0< 0R v = R perfect Rolling v µN 0 N v0 imperfect Rolling v0< 0R µN v v If becomes zero before v. If v becomes zero before . v = 0 & = 0 if v and both becomes zero simultaneously. Perfect rolling Z
Solution technique : Step (i) Drawthe free body diagram. Step (ii) Indicate the direction of friction incorrect sense.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 45 Step (iii) Assume linear and angular acceleration. Step (iv) Use F = ma and T = I.. Calculate a and .. Step (v) then use the concepts of kinematics to solve the problems. Kinetic energy of a rolling body: Let the centre ofmass bemovingwith velocity C V . . Relative to the centre ofmass, a particle at position i r. will have a velocity ..× i r. . Relative to ground the particle bemovingwith velocity i V . = C V . + ..× i r. The speed square is given by mi ri . VC 2 i V. = i V . . i V . = ( C V . + ..× i r. ).( C V . + ..× i r. ) = 2 C V. + | ..× i r. |2 + 2 C V . .( ..× i r. ) The kinetic energyof thewhole bodyis given by 2 i i K 1 m v 2 . . = 2 i C 1 m V 2 . + . 12 mi | ..× i r. |2+ 2.mi C V. .( ..× i r. ) .....(i) Now, .mi = totalmass =M . 2 i i 1 m V 2 . = 2 C
1MV 2 .....(ii) Let i r. be resolved along ..and normal to ... Let i r. . be normal to ..and i|| r. be along such that i r. = i r. . + i|| r. Then ..× i|| r. = 0. | ..× i r. | = ..× i r. . + ..× i r. . = | ..× i r. . | = .ri. . 12 .mi | ..× i r. |2 = 12 .mi .2 2 i r . = 12 Icm.2 .....(iii) The third termcan bewritten as 2.mi C V . .( ..× i r. ) = 2 C V . .( ..× .mi i r. ) = 0 .....(iv) In equation(iv) the summation is the first moment ofmass about centre ofmass, which is zero bydefinition of CM. Hence thewhole termbecomes zero. Using these,K = 12 2 C MV + 12 Icm.2
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 46 The quantity 12 2 C MV is called translationalkinetic energyand 12 Icm.2is called rotational kinetic energyof the rolling body. Example 38. Auniformdisc ofmassMand radiusRrolls on a smooth horizontal surfacewithout slippingwith a linear velocity v. Calculate itsK.E. Sol. K.E. of a rolling body(without slipping) 2 2 2 2 2 2 1 1 1 1 v Mv I Mv MK 2 2 2 2 R . . . . . . 2 2 2 1Mv 1 K 2 R . . . . . . . . for disc, since K = R2 Hence K.E. of a disc = 34 Mv2. C39: Asolid sphere ofmassMand radiusRis rotating aswell asmovingwith a linear velocity v; then calculate its K.E. Sol. K.E. = 12 Mv2 22 1 KR . . . . . . . ; for sphere, K = 25 R . K.E. for a spherewhich is rolling (without slipping) 7 10 . Mv2. Example 39. Asolid cylinder ofmassMand radiusR has a lengthL. If it is rolling aswell asmovingwith linear velocityv, thencalculate itsK.e. (assuming no slipping). Sol. We have, K.E. = 1
2 Mv2 22 1 KR . . . . . . . for a solid cylinder, K= R2 . K.E. = 34 Mv2 (same as that for a circular disc ofmassMandmovingwith same v). Example 40:Adisc ofmassmrollswithout slidingwith speed v0. (a)write its kinetic energy. (b)Write the ratio of translationalto rotationalkinetic energies. Sol: (a) The kinetic energyof a rolling bodyis given by K = 2 CM 1MV 2 + 12 Icm.2 Here, Vcm =V0, .= CM Vr (no slipping)
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 47 = 0 Vr Icm = 12 mr2 (for disc) . V0 . K = 201 mv 2 + 12 12 mr2.2 = 201 mv 2 + 201 mv 4 = 203 mv 4 (b) TranslationKE = 201 mv 2 Rotational KE = 12 12 mr2.2= 201 mv 4 Required ratio = Translation KE Rotational KE = 2020 1 mv 21 mv 4 = 2 The ratio is, thus, 2 : 1. Example 41.Asolid disc is rollingwithout slipping aswell asmovingwithlinear velocityv.Whatwillbe the ratio of translatoryK.E. and totalK.E. Sol. 2 2 2 2 Translatory K.E. 1/ 2M Total K.E.[Trans.K.E. Rot.K.E.] 1/ 2M [1 K /R . . .v v
2 2 2 2 2 2 R R 2 R K R R / 2 3 . . . . . C40:Acircular disc ofmass 2 kg and radius 0.2metre rolls down an inclined plane fromrest.What time it will take to cover a distance of 1metre along the plane, if angle of inclination of the plane is 30º. Sol. Acceleration, 2 2 2 2 g 1 a gsin 2 g 1 K / R 1 R 2R 3 . . . . . . . . s 1 at2 2 . 1 1 g t2 2 3 . . t 6 0.78 9.8 . . s
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 48 Mechanical Energy The sumof kinetic energy (K) and potential energy (U) ismechanical energy. If a rigid body has its CMat height hcm above gravitational zero levelofpotential energy, U=mg hcm, Themechanicalenergy (E) of rolling bodywill be E = 2 CM 1MV 2 + 12 Icm.2 +Mghcm In other fields like spring force, other termsmayappera in the expression for U. If there be no sliding, thework done byfrictionforce is zero. Themechanical energyremains conserved in that situation. Example 42. Athin rod of lengthLis placed at angle . to verticalon a frictionless horizontal floor and released. If the center ofmass has acceleration =A, and the rod an angular acceleration = . at initialmoment, then (A)A= (L.). sin . (B)A/2 = (L.).sin. (C) 2A= (L.).sin . (D)A= L. Sol. . cm a A L sin 2 . . . . . A L sin 2 . . . L/2 L/2 sin 90º� mg acm N . 2A= L ..sin . Hence (C) is correct. Example 43.Adisc of radiusRis rolling purelyon a flat horizontal surface,witha constant angular velocity.The angle between the velocityand accelerationvectors of point Pis P C (A) zero (B) 45º (C) 135º (D) tan�1(1/2) Sol. ar = R.2 (radial acceleration is directed towards centre at point P. Here cm cm P cm v v 1 cos v 2 v 2 . . . . . = 45º ar vcm R vcm vcm R vP= vc2m+ (R )2 = 2 v Hence (B) is correct. Example 44. Awheel, of radius 1 m, is rolling purelly on a flat, horizontal surface. It�s centre ismovingwith a
constant horizontal acceleration= 3m/s2.At amoment when the centre of thewheelhas a velocity3m/s, then find the acceleration of a point 1/3 vertically above the centre of thewheel. Sol. acm = r. v = r. . vr . . . . .2 2 t cm r a . a . a . a ar at +acm Here t a 13 . .
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 49 2 r a 13 . . acm = 3 m/s2 v = 3 m/s Putting the value, a = 5 m/s2. Example 45. Asolid sphere is spin of about its on axiswith angular velocity.. in clockwise direction and is gently placed on a rough horizontal surface. Coefficient of friction betweensphere and the surface is µ. Calculate. (a) Time afterwhichsphere starts perfect rolling. (b) Final linear and angular velocity. (c) Energyloss against friction. Sol. For translatorymotion µN=ma and N = ma . µmg=ma . a = µg ...(1) For rotationalmotion N a mg µN N m µ R 0 µNR 2 mR2 5 . . 5µg 2R . . ...(2) Let after time �t� sphere starts perfect rolling. For translatorymotion . v = 0 + µgt ...(3) For rotationalmotion a = µg u = 0 v t = 0 t = t 0 5µg . t 2R . . . . ...(4) at t = t for perfectRolling v � .R = 0 v = .R ...(5) From(3), (4) and (5) 0 µgt R 5µg t 2 . . . (a) Therefore time after which sphere starts perfect rolling t = 0 t = t 0 0t Rµg 5µg 2
.
.
.0 2 R t 7µg . . . . . . . . (b) Linear velocityat the time ofperfect rolling 0 0 2 R 2 v .µg R 7µg 7 . . . . (c) Work done byfriction or energy loss 2 2 2 0 1 I 1 mv 1 I 2 2 2 . . . . . . . .. ..
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 50 Example 46. Asolid cylinder ofmassmismovingwith velocity v0 in horizontaldirection. It is gently placed on a rough horizontal surface. If coefficient of friction between sphere and surface is µ. Calculate time interval afterwhich it starts perfect rolling. R v0 Sol. For translationalmotion : µ µmg=ma a = µg ...(1) For rotationalmotion µNR = I . mR2 µmgR 2 . . N a mg µN v0 2µg R . . ...(2) Let v is the linear speed and . is angular speed at the time of perfect rolling. . v = v0 � at = v0 � µgt and 0 t 0 2µg . t R . . . . . . . . . .. .. ...(3) . v � R.= 0 (for perfect rolling) vR . . ...(4) From(3) and (4) v 2µg . t R R . 0 v µgt 2µg . t R R . .0 3µg . t v R R . , 0 v t 3µg . . . . . . . Example 47. Aspherical shell ofmassmand radius r ismovingwith linear velocity v0 in forward direction and rotatingwith .0 inanti clockwise direction.Nowit is gentlyplaced on a rough horizontalsurface. If coefficient of friction between shell and surface is µ. Find the relation between .0 and v0 for . to be zero before v0. Sol. For translatorymotion µmg=ma
a = µg ...(1) For rotationalmotion µmgr 2 mr2 . 3 . . . = 3 µg / 2r ...(2) at time t : v = v0 � µgt ...(3) and 0 3µg t 2r . . . . ...(4) when . = 0 N a mg µN v0 0 3µg t 0 2r . . . 0 2 r t 3µg . . and 0 0 2 r v v µg 0 3µg . . . . . . . . . . 0 0 v 2 r 3 . .
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 51 Example 48. Acylinder ofmassMand radius R is resting on a horizontal platform(which is parallel to the x�y plane)with its axis fixed along the y axis and free to rotate about its axis. The plateformis given amotionin the x-direction given byx=Acos (.t).There is no slipping betweenthe cylinder and plateform. Find themaximum torque acting on the cylinder during itsmotion. Sol. x v dx A sin t dt . . . . . . ax = � A.2 cos .t . amax = .2A Since, no sliping takes place. . acm = R. . 2A R . . . . MR2 I 2 . . . . . 2 2 MR A 1 2 mRA 2 R 2 . . . . . . Example 49. Abit ofmud stuck to a bicycle�s front wheel of radius r detaches and is flung horizontally forward when it is at the top of thewheel.The bicycle ismoving forward at a speed v and it is rollingwithout slipping. Find the horizontaldistance travelled bythemud after detaching fromthewheel. Sol. x = 2vt 2r 1 gt2 2 . . t 4r g . 2v x . x 4r 2v g . . r 16rv2 x 4v g g . . Example 50. Find the relationship between v0 and .0 so that sphere of radiusRandmass mrolls up the inclined plane before it starts falling down the plane finallyin given fig.
v 0 Sol. For translationalmotion v0 µN�mg sin .=ma µmg cos . �mg sin .=ma a = (µ cos . � sin .)g ...(1) for rotationalmotionµmg cos ..R= 25 mR2 . 5µg cos 2R . . . ...(2)
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 52 assume t is the timewhen v = 0 . 0 = v0 � (µ cos . � sin .) gt . . 0 t v µcos sin g . . . . at this time .> 0 0 a mg cos mg cos µN N 0 5µg cos . t 0 2R . . . . . . . . . 0 0 5µg cos v 2R µcos sin g . . . . . . . . . 0 0 5µv cos 2R µcos sin . . . . . . Example 51. Two spheres are rolling with same velocity (for their C.M.) their ratio of kinetic energy is 2 : 1 & radius ratio is 2 : 1, theirmass ratio willbe : (A) 2 : 1 (B) 4 : 1 (C) 8 : 1 (D) 2 2 : 1 Sol. T 1 mv2 1 I 2 2 2 . . . T 1 mv2 1 2 mR2 2 2 2 5 . . . T 1 mv2 1 2 mv2 2 2 5 . . . . .. .. T 1 7 mv2 2 5 . . . .. .. . 1 1 2 2 T m T m . Hence (A) is correct. Example 52. Two thin circular discs ofmass 2 kg and radius 10 cmeach are joined
by a rigid massless rod of length 20 cm.The axis of the rod is along the perpendicular to the planes of the disc throughtheir centres. This object is kept on a truck insuch awaythat the axis of the object is horizontal and perpendicular to the direction ofmotion ofthe truck. Its frictionwith the floor ofthe truck is large enough so that the object canroll on the truckwithout slipping.TakeX-axis as the direction of motion of the truck and Z-axis as the vertically upwards direction. If the truck has an acceleration 9m/s2, Calculate : (i) the force of friction oneach disc. (ii) Themagnitude and direction of the firctional torque acting on each disc about the centre ofmassOof the object. Express the torque in the vector formin terms of unit vectors�i, �j and �kinX,Yand Z directions. Sol. Givenmass of disc,m= 2 kg and radius, R= 0.1 m (i) FBDof anyone disc is shown in figure : Truck a = 9m/s2 ZY X
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 53 Frictional force on the disc should be in forward direction. Let a0 be the linear acceleration ofCOMof disc and . the angular acceleration about its COM. Then : 0 a f f m 2 . . ...(i) 2 f .R 2f 2f 10f I 1/ 2mR mR 2 0.1 . . . . . . . . ...(2) Since there is no slipping between disc and truck, therefore, a = 9m/s2 a0 f Pf Q a0 + R. = a or f (0.1) 10f a 2 . . . . .. .. or 3 f a 2 . Þ f 2a 2 9.0 N 3 3 . . . . f = 6 N Since this force is acting inpositive x-direction, f . .6�i.N . (ii) . . r . f . . . Here f . .6�i.N . (for both the discs) P 1 r . r . .0.1 �j. 0.1 k� . . z x y P Q 2 20cm = 0.2m O f f and Q 2 r . r . 0.1 �j. 0.1 k� . . Therefore, frictional forque on disc 1 about point O(centre ofmass) : . . . . . . 1 1 . . r . f . .0.1�j. 0.1k� . 6i� N.m . 0.6k� . 0.6�j . . . or . . 1 . . 0.6 k� . �j N.m .
and 2 2 | .1 | . (0.6) . (0.6) . 0.85 N . m . Similarly, . . . . 2 2 . . r . f . 0.1�j. 0.1k� . 6i� N.m . . . . . 2 . . 0.6 .�j. k� N.m . and | .2 | . | .1 | . 0.85 N .m . . INSTANTANEOUS AXIS OF ROTATION v . . The combined effects of translation of the centre ofmass and rotation anaxis through the centre ofmass are equivalent to a pure rotationwitht he same angular speedabout anaxis passing through a point of zero velocity. Such an axis is called the instantaneous axis of rotation. (IAOR).This axis is always perpendicular to the plane used to represent themotion and the intersectionof the
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 54 axiswith this plane defines the location of instantansus centre ofzero velocity(IC). v . . r Vp = r.. IC r.Vp For example consider awheelwhich rollswithout slipping. In V this case the point of contactwiththe ground has zero velocity. Hence, this point represents the IC for the wheel. If it is imagined that thewheel ismomentarilypinned at his point, the velocity of anypoint on thewheelcan be found using v = r.. Here r is the distance ofthe point fromIC.Similarly, the kinetic energyofthe bodycanbe assumed to be pure rotational about IAOR or, Rotation+Translation . Pure rotation about IAORpassing through IC] K.E. = 2 cm 1 mv 2 + 12 Icm.2 . K.E. = 2 IAOR 1 I 2 . C41: (i) Although the ICmay be conveniently used to determine the velocity of any point in a body, it generally does have zero acceleration and therefore, it should not be used for finding the acceleration of anypoint in the body. (ii) When a body is subjected to general planemotion, the point determined as the instantaneous centre of zero velocityfor the bodycan onlybe used for aninstant of time. Since the bodychanges its position fromone instant to the next, then for each position of the body a unique instantaneous centremust be determined. The locus of points which defines the IC during the body�smotion is called a centrode. Thus, each point on the centrode acts as the IC for the body only for an instant of time. Locaion of the IC If the location of the IC is unknown, it maybe determined by using the fact that the relative position vector extending fromthe ICto a point is always perpendicular to the velocityof the point. Following three pssibilities exist. Give the velocity of a point (normally the centre of mass) on the body and the angular velocity of the body. If v and .are known, the ICis located along the line drawnperpendicular to v. at P, such that the distance fromP to ICis, r = v. . Note that IC lie on that side of P which causes rotation about the IC, which is consistent with the direction ofmotion caused by ..
and v. . . P v r IC Example 53:Arotating disc moves in the positive direction of the x-axis. Find the equation y(x) describing the position of the instantaneous axis of rotationifat the initialmoment of the centre c of the discwas located at the point Oafterwhich it movedwith constant velocityvwhile the disc started rotating counterclockwisewith a constant angular acceleration .. The initial angular velocity is equalto zero. O c v y x IC y O c v y x x Sol: t = xv and . = .t = x v . The positionof IAORwill be at distance
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 55 y = v. or y = vx v . or y = v2.x or xy = v2 . = constant. This is the desired x-y equation. This equation represents a rectangular hyperbola. C42: The equation text = l. does not hold good ina non-inertial frame.However, there exists a veryspecial case when text = l. does hold even if the angular acceleration a ismeasured froma non-inertial frame. That special case is, when the axis of rotation passes through the centre ofmass and otherwise the pseudo forces produce a pseudo torque about the axis. This is the reason, in above exampleswe calculated the angular acceleration a about an axis passing through the centre ofmass and perpendicular to the plane ofmotion of the particles. Because about that axiswe can apply, . = ext .l . C43: Work done byfriction in pure rolling on a stationaryground in zero as the point of applicationof the force is at rest. Therefore,mechancal energycan be conserved ifall other dissipative forces are ignored. C44: In accelerated pure rolling the velocity of the bottommost point is zero but despite the relatin a = R., the accelertion of the bottommost point is not zero. Because acceleration of any ponit P can be given as P a. = C a. + PC a. Here, PC a. has two components: Tangential acceleration at = r. (which is perpendicular to CP) and radial or normal acceleration an = r.2 (which is along PC) Thus P a. = C a. + ( PC a. )t + ( PC a. )n For the botommost ponit, C a. + ( PC a. )t = 0 R C r P .... v, a as C a. = a [in forward direction] and ( PC a. )t = R. [in backward direction] and since, a = R. therefore C a. + ( PC a. )t = 0. But ( PC a. )n . 0. It is R.2 towards centre. Thus, acceleration of bottommost ponit is R.2 towards centre. Similarlyintheproblems like showninfigure, it iswrong tosaythat acceleration of point Pis equal to acceleration of blockA.Althoughwe canwrite, A v, a P
aA = a + r. In case of pure rolling, problems can also be solved byusing the energy conservation principle (provided no other dissipative forces are present). So, in this casewewillwrite the energyequation as, Decrease in gravitationalpotential energyofA= increase inkinetic energyofA(onlytranslational) + increase in
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 56 kinetic energy ofspoot (both rotational aswell as translational). C45: Incaseswherepulleyis having somemass and friction is sufficient enoughto prevent slipping, the tensionon two sides of the pulleywill be different and rotatinoalmotion ofthe pulleyis also to be considered. C46: At a given instant the value of .for a rigid bodywill be same for every point. C47: The torque equation (. = l.) can be applied only about two points. These are, (i) centre ofmass (ii) point about which body is in pure rotation. Example 54. There is rod of length l . The velocities of its two ends are v1 and v2 in opposite directions normal to the rod. The distance of the instantaneous axis ofrotation fromv1 is (A) zero (B) 2 1 2 v v . v l (C) 1 1 2 v v . vl (D) l / 2 Sol. . 1 2 v v x x . . . l . or v1(l � x) = v2x or v1 l � v1 x = v2 x xl � x O l v1 . v2 1 1 2 v x v v . .l Hence, (C) is correct. Example 55. Aladder of lengthL is slippingwith its ends against a verticalwall and a horizontal floor.At a certain moment, the speed of the end in contact with the horizontal floor is v and the laddermakes and angle . = 30º with the horizontal. Then the speed of the ladder�s centermust be (A) 2v/ 3 (B) v/2 (C) v (D) None Sol. v/ 2 . . l. . 2v l
Also, vc = r. = 2 2 3 4 4 . . . . . . . . . . . . . . l l l /2v2 l /2 30º 60º 60º 30º 30º 3 l/4 l /4 O v l /2 C r intantaneous axis of rotation = 1 3 1 4 2 l . . . l . 2v v 2 . l . . l Hence (C) is correct. Example 56. A rectangular rigid fixed block has a long horizontal edge. A solid homogeneous cylinder of radius R is placed horizontally at rest with its length parallel to the edge such that the axis of the cylinder and the edge ofthe block are in the same vertical plane as shownin figure. there is sufficient friction present at the edge so that a verysmalldisplacement causes the cylinder to roll ofthe edge without slipping.Determine : R (a) the angle .c throughwhich the cylinder rotates before it leaves contact with the edge.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 57 (b) the speed of the centre ofmass of the cylinder before leaving contact with the edge, and (c) the ratio of the translational to rotationalkinetic energies ofthe cylinderwhen its centre ofmass is in horizontal linewiththe edge. Sol. (a)The cylinder rotates about the point ofcontact.Hence themechanicalenergyofthe cylinderwillbe conserved i.e., R v R Rcos v´ (1) (2) (3) . (P.E. + K.E.)1 = (P.E. + K.E.)2 . mgR + 0 = mgR cos . + 1 I 2 1 m 2 2 2 . . v But .= v/R (No slipping at point of contact) and I = 1 mR2 2 Therefore, mgR = mgR cos . + 2 . 2 2 . 2 1 1 mR v / R 1 mv 2 2 2 . . . .. .. or 34 v2 = gR (1 � cos .) or v2 4 R 3 . g (1 � cos .) ...(1) At the time of leaving contact, normal reactionN= 0 and . = .c.Hence mg cos ..= mv2 R ........ v mg mg sin mg cos N = 0 or v2 R = g cos . ...(2) From(1) and (2) 43 g(1 � cos .c) = g cos .c or 74 cos .c = 1 or cos .c = 4/7 or .c = cos�1(4/7) (b) . . v 4 gR 1 cos 3 . . . (Fromequation 1) At the time it leaves the contact cos ..= cos .c = 4/7 . . . v 4 gR 1 4 / 7
3 . . or v 4 gR 7 .
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 58 Therefore, speed ofCOMof cylinder just before it leaves the contact is 4 gR 7 . (c) At themoment, when cylinder leaves the contact v 4 gR 7 . Therefore, rotational kinetic energy, KR = 12 I.2 or 2 . 2 2 . 2 R K 1 1 mR v / R 1 mv 1 m 4 gR 2 2 4 4 7 . . . . . . . .. .. .. .. or KR = mgR 7 ...(3) Nowonce the cylinder loses its contact,N= 0, i.e. the frictional force,which is responsible for its rotation, also vanishes.Hence its rotationalkinetic energynowbecomes constant,while its translationalkinetic energyincreases. Applying conservation of energyat (1) and (3) : Decrease ingravitationalP.E. =Gainin rotationalK.E. +translationalK.E. . TranslationalK.E. (KT) =Decrease in gravitational P.E. �KR or KT = (mgR) � mgR 7 = 67 mgR ...(4) From(3) and (4) we have TR K 6 / 7mgR K mgR / 7 . or TR K 6 K . ANGULAR MOMENTUM Let a particle ofmassmhave a velocityv . in a frame S. Let O be a point at rest in S. The particle has a position vetor r. relative to O. v m rThen the cross prodcut r O . × mv . is called angularmomentumabout O. The angularmomentumof a particle about Ois equal to the cross product of its position vector relative to O and themomentumvector. O . l = r. × p. . Its S.I. unit is kgm2s�1 and dimensional formula [ML2T�1].
Moment arm: In the figure r. is themoment armfor .l.We have r. = moment arm O . l = ( r.. + || r. ) × p. rs r. r Op = r.. × p. [. || r. ×p. = 0]
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 59 For a particle to have angular momentum, it must posses a linear momentum. However, a particle having a linearmomentummaynot have anangularmomentum. The concept ofangularmomentumis introduced formotion of particles in space here.However, youwill find in your studies ofmodern physics or atomic physics another angularmomentumwhich is not related tomotion in space- that is spin angular momentumof particles like electron. It is instrinsic to a particle, say, electron. There are severalparticleswhich have spinangularmomenta.We shallnot discuss then here.We shall consider motion-related angularmomenta only. C48:Aparticle ismovingwithuniformvelocity.Does its angularmomentumabout anypoint change as itsmoves ? Sol: The angularmomentumabout O(see previous figure)maybewritten as .l= r.. × p. = r.p inward, which does not change. Example 57. Aparticle ofmass 0.5 kg is rotating in a circular path of radius 2mand centrepetal force on it is 9 Newtons. Its angularmomentum(in J.sec) is : (A) 1.5 (B) 3 (C) 6 (D) 18 Sol. L=mvr But mv2 F r . or v Fr m . L m Fr r m . L 0.5 9 2 2 0.5 . . . L . 36 . 6 J s Hence (C) is correct. Example 58.Aparticle ofmassmis rotating in a plane is a circular path of radius r, its angularmomentumisL.The centripital force acting on the particle is : (A) L2 mr (B) L2m r (C) 22 L mr (D) 2
3 L mr Sol. L=mvr . v L mr . . 2 2 m L F mv mr r r . . .. .. . . . 23 L F mr . Hence (D) is correct. Example 59. Aparticle ofmass 2 kg located at the position .�i . �j.mhas a velocity 2 .�i . �j . k� . m/s. Its angular momentumabout z�axis in kg-m2/s is : (A) zero (B) 8 (C) 12 (D) �8
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 60 Sol. The angularmomentumabout origin L . r .mv .. . . L . .�i . �j.. 4.�i . �j. k� . .. L . 4..k� .�j.k� .�i. .. L . 4�i . 4�j. 8 k� .. . Lz = �8 kg m2/s Hence (D) is correct. Example 60. The position vector of a particle ofmass 2.00 kg is given as a function of the by r . 6 �i . 5t �j . m. Determine the angularmomentumof the particle about the origin, a function of time. Sol. v dr 5�j dt . . . . . L . r .mv .. . . L . .6i� . 5t �j.. 10 �j .. L . 60 k� .. Example 61. Aparticle ofmassmis projectedwith a velocity u at an angle of . with horizontal. Find the intial angularmomentumof the particle about the highest point of its trajectory. Sol. u . u cos.�i . u sin .�j . . | L | . | r . mu | .. . . Here u2 sin2 h 2g . . O h x y R/2 u u2 sin2 R g . . r h �j R i� 2 . . . . mu3 sin2 cos L 2g. . .
Angular momentum of many particles If a systemhas severalparticles, having angularmomenta 1 .l, 2 .l, ........ n .l about a point Oat rest in frame S, the totalangularmomentumof the systemis defined as their vector sum: L. = 1 .l+ 2 .l+ ........ + n .l Angular momentum about centre of mass For calculatingangularmomentumabout centre ofmass of a system,wemust take a frame S. inwhichCMis at rest. Fromthis framewe observemomenta
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 61 '1p. , '2 p. , ........ 'N p. of the particles located at '1r. , '2r. ,.... 'Nr. relative to centre ofmass. Then the particle has angularmomentum ' i r. × 'i p. about centre of Z. Y. X.Vi.pi. Frame fixed to CM but not to the body mass. Total angularmomentumis the vector sumof such angularmomenta. We denote it by cm L . : cm L . = . i' r. × 'ip. . Relationship to angular momentum about a Laboratory point: Let Obe a point in laboratory andO. be the centre ofmass of the system. Let i-th particle, havingmassmi be movingwith velocity 'iv. in the frame of laboratory.Then angularmomentumof this particle about Owillbe i r. ×mi 'iv. . Thewhole bodywill have an angularmomentum O L. = . i r. × mi 'iv. . Relative to the frame inwhichO. is at rest (CM- frame) the i-th particle has a different velocity i v. , it is relative velocity i v. � CM V. , where CM V. is the velocity of centre ofmass. Simmilarllyits position vector is i' r. = i r. � cm r. The figure shows i V. = ' i V. + CM V. i r. = i' r. + cm r. VCM vi. VCM O. rcm ri L ri. vi. Using thesewe have L .
= . i r. × mi i v. = ..( cm r. + i' r. ) × mi i v. = . cm r. × mi i v. + . i' r. × mi i v. = cm r. × .mi i v. + . i' r. × mi i v. = cm r. × M CM V. + . i' r. × mi ( 'iv. + CM V. ) [. M CM V. = .mi i v. ] = cm r. × M CM V. + . i' r. × mi 'iv. + mi CM V. = cm r. × M CM V. + cm L. + .mi i' r. × CM V. = ( cm r. × M CM V. ) + cm L. [. .mi i' r. = 0] The first termgives angularmomentumassociatedwithmotion of centreofmass,while the secondwithmotion about CM. Example 62:Arod of length 3mand negligiblemass connects two balls of masses 1 kg and 3 kg. The systemis thrown inair negligible (friction) as in the figure.At a certainmoment, the centre ofmass is located at
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 62 (2m, 4m) andmovingwith 10ms�1 at 60º with horizontal. The rod is turning at .= 2 rad s�1 about the centre ofmass. 1kg 10 m/s CM 2kg 4m Yj ^ k ^O X . = 2 rads�1 (a) Determine the angularmomentumof systemabout centre ofmass and express it using ^i , ^j and ^k . (b) Determine the angularmomentumof the systemabout origin. Sol: (a) See figure for quantities needed. 1m 2m 6 ms�1 2 ms�1 1kg 2kg j ^ k^ i^ We have cm L. = . i r. × mi 'iv. . . = {2m(� ^i ) × 1 kg (6 ^j ms�1)} + {(1m) ^i × (2 kg) 2m�1(� ^j )} = �16 ^k (kgm2s�1) (b) O L. = ( cm r. × cm MV . ) + cm L. Here, cm r. = (4 ^i + 4 ^j )m, = �16 ^k (kg m2s�1) CM V. = 10 ms�1(cos 60º ^i + sin60º ^j ) M = 3 kg Using thereO L. = 27.92 ^k (kg m2s�1) Angular momentum about an axis The component of angularmomentumabout a point on the given axis is known as angularmomentumalong that axis. If angularmomentumabout a point Oon Z-axis isL . . It ismaking an angle .with Z-axis, then the angularmomentumabout Z-axis (LZ) is given by LZ = L cos. It is a scalar quantity. Angular momentum of a body rotating about an axis at rest: Let Z-axis beoriented along the stationaryaxis of rotating of thebody.An i-th particlewill have angularmomentum i .labout a point Oin the Z-axis. Then total angularmomentumL . is given by ItsZ-component is called angularmomentumabout Z-axis, LZ.Now LZ = .liZ , where liz is the Z-component of angularmomentum i .l. To get liZ,we figure, mi Vi
ri. R. i O li .z Z Here li = Rimivi . liZ = Rimivi cos. = Rimi.Zri cos. Now, Ricos. = ri . liZ = 2 i i z m r . . LZ = 2 i i z (.m r ). LZ = IZ.Z where IZ = 2 i i .m r is calledmoment of inertia about Z-axis. Torque and angular momentum Nowwe shallconsider fundamental lawof rotationaldynamics-also calledNewton�s second lawof rotational
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 63 dynamics. It relates the rate ofcharge of angularmomentumto torque. Single Particle System The figure shows a particle havingmomentum p. . There is a point Ofixed inXYZ-frame. Let angularmomentumabout Obe denoted by o .l. Then, 0 L .. = r. × p. . Differentiatingwe get ddtl = d r dt. × p. + r. × d p dt. = v. × p. + r. × F. [. d p dt. = F. ] = 0 + .. = .. Thus . ddt .l = .. The rate of change of angularmomentumabout a point is equalto unbalanced torque about the same point. The case of many particle system Let there bemanyparticles havingmomenta i p . . Then i dp dt . = i F. [Newton�s second lawofmotion] .....(i) where i F. is unbalanced force. Nowit arises due to two types of sources-particleswithin the systemboundary and the particles outside the boundary of the system. Let be the force exerted by internal particles.We call it internal force. Let iext F. be the force exerted by externalparticle;we call it external force. Then i F. = iint F. + iext F. .....(ii)
Nowwewrite angularmomentumLin the form: L. = . i .l= . i r . × i p. .....(iii) Differentiallyrelative to time t, dL dt . = . i i i i dr dp p r dt dt . . . . . . . . . . . .. .. = . i v. × i p. + . i r . × ( iint F. + iext F. ) [Using (i) and (ii)] = 0 + . i r . × iint F. + . i r . × iext F. = in t ..+ ext ..
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 64 Itis experimentallyfound that internal torque can�t change angularmomentumof the system. . int F. = 0 [postulate ofrotationaldynamics] Thus, dL dt . = ext .. .....(iv) The rate of change of angularmomentumof a systemof particles about a point in anintertialreference frame is equal to net external torque about the same point. Taking Z-component of the above relationship (equation -iv) we get .ext z = z dL dt Now, LZ is angularmomentumabout Z-axis LZ = IZ.Z where IZ ismoment of inertia about Z-axis. Hence, .Z= d dt (IZ.Z) = z dI dt wz + Iz z ddt . . .z = z dI dt .z + Iz.z [non-rigid system] Ifmoment of inertia. IZ constant then, .ext Z = IZ.Z [rigid bodies] The external torque acting on a rigidbodyabout an axis is equal to theproduct ofmoment ofinertia and angular acceleration about the same axis. Example 63. The angularmomentumof a flywheel having amoment of inertia of 0.4 kgm2 decreases from30 to 20 kgm2/s in a period of 2 second. The average torque acting on the flywheel during this period is : (A) 10 N . m (B) 2.5 N . m (C) 5 N . m (D) 1.5 N . m Sol. Lt . . . . 10 5 Nm 2 . . . Hence (C) is correct. Angular Impulse Theorem As ext . . is equal to rate of change of angularmomentum L. ,we have ext. . dt . . dL..
= f L. � i L. The LHS is called angular impulsewhich equals the change in angularmomentum. Some dynamical Relationships
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 65 Quantities Single Particle System of Particles Angularmomentum � about a ponit .l= r. × p. L. = . i r. × i p. L. = cm L. + cm r. × M cm V . � about z-axis lZ = lcos. LZ = L cos. = Iz.z Torque� about a point .. = r. × F. . . = . i r. × i F. ; . . = . . ext � about z-axis ..Z = .cos. .Z = ..cos. Equation ofRotationalDynamics - about a point .. = d.l dt .. = dL. dt - about z-axis .2 = z d.l dt .2 = z dL dt .z = Iz.z .z = Iz.z + z dI dt . . . . . . .z Angular Impulse � about a point dt .. . = f i . . l . l . .ext dt . Lf . Li . .. .. - about z-axis z. . dt = lzf � lzi .extzdt = Lzf � Lzi Conservation of angularmomentum � about a point .. = 0 . .l= constant ext . . = 0 .... L. = constant � about z-axis .2 = 0 . lZ = constant .ext z = 0 . LZ = constant Example 64. The moment of inertia of semicircular plate of radiusR andmassMabout
axisAA´ in its plane passing throughits centre is A(A) A´ MR2 2 (B) 2 MR 2 cos 4 . (C) 2 MR 2 sin 2 . (D) MR2 4 Sol. Here 2 x mR I 4 . and 2 y mR I 4 . .x = . cos . A A´ x x´ y .y.....sin.. . x x y y L . I . �i . I . �j ..
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 66 or �e = unit vector along the axis. e� . e cos.i� . esin .�j e� . cos.�i . sin .�j . 2 2 2 2 axis � mR mR L I L . e cos sin 4 4 . . . . . . . . . .. . 2 2 mR 2 mR 2 I cos sin 4 4 . . . . mR2 I 4 . Hence (D) is correct option. Question No. 65 & 66 Auniformrod is fixed to a rotating turntable so that its lower end is on the axis of the turntable and it makes an angle of 20º to the vertical. (The rod is thus rotatingwith uniformangular velocityabout a vertical axis passing throughone end). If the turntable is rotating clockwise as seen fromabove. 20º Example 65.What is the direction of the rod�s angularmomentumvector (calculated about its lower end) ? (A) verticallydownwards (B) down at 20º to the horizontal (C) up at 20º to the horizontal (D) verticallyupwards. Sol. . . . ..cos.�i ..sin .�j .. Ix = 0 2 y m I 3 . l x x´ y´ x . Lx = Ix .x =0 and 2 y y y m L I sin 3 . . . . . . l . m 2 � L sin i
3 . . . . .. lm 2 � L sin 20 j 3 . . . .. l Hence (B) is correct. Example 66. Is there a torque acting on it, and if so inwhat direction ? (A) yes, vertically (B) yes, horizontally (C) yes at 20º to the horizontal (D) no Sol. mg cos mg mg sin O . Torque ofmg about Ois
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 67 (mg sin.) 2l Hence (B) is correct. CONSERVATION OF ANGULAR MOMENTUM The angularmomentumof a systemabout a ponit canbe changed onlybyan external torque about that point. This is suggested by dL dt . = ext . . . If the external torque about a point be zero (either the vector sumis zero, or the torque is absent) then dL dt . = 0 The angularmomentumwillnot changewithtime.This is knownas lawof conservationof angularmomentum. If the net torque or external torque about a point be zero, the total angularmomentumabout that point is conserved. If .z = 0, Lz = constant. If torque about an axis be zero, the angularmomentumabout the axis conserved: Iz.z = constant. C49: Is the angularmomentumof the Earthconserved ?Explain. Ans: Yes. The line of action of gravitational force due to the Sun on the earth passes through the Sun. . ext . . = r. × F. = 0 F r VCM S L . s L. = constant. Thus, torque on the Earth about theSun is zero and hence angularmomentumabou the Sun is conserved. C50: ClassicalAngularmomentumof the electronabout nucleus inH-atomis conserved. Explainwhy. Ans: Nucleus exerts electrostatic pull on the electron. Its line of action is passing through the nucleus. Hence its torque is zero. ext . . = r. × F. = 0 Fr When external torque is zero, angularmomentumremains constant. This is whyclassicalangularmomentumofthe electron about the nucles is conserved. C51:Aperson is rotatingwith turn tablewithout considerable friction. His hands are stretched. Nowhe folds his hands.His angular speed increases. Explain.
Ans: Since friction is neglected, there is no torque about vertical axis of the system.Hence I.1 = I2.2 As theman folds hands, hismoment of inertia decreases. The angular speed increases. Example 67. Aparticle ismoving in a circular orbit of radius r1 with an angular velocity .1. It jumps to another
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 68 circular orbit of radius r2 and attains an angular velocity.2. If r2 =0.5 r1 and assuming that no external torque is applied to the system, then the angular velocity .2, is given by : (A) .2 = 4.1 (B) .2 = 3.1 (C) .2 = 2.1 (D) .2 = .1 Sol. Applying conservation principle of angularmomentum, mr12.1 = mr22 .2 or r12 .1 = r22 .2 or r12 .1 = (0.5r1)2 .2 or .1 = 0.25 .2 or .2 = 4 .1 Hence (A) is correct. Example 68.Aman, sitting firmlyover a rotating stoolhas his arms streched. If he folds his arms, thework done by theman is (A) zero (B) positive (C) negative (D)may be positive or negative. Sol. The kinetic energyisT 1 I 2 2 . . But L = I. . LI . . . 1 L 2 L2 T I 2 I 2I . . . . .. .. But in this case, If < Ii . Tf > Ti (Because angularmomentumremains conserved) But w = .T = Tf � Ti > 0 Hence,workdone is positive. Hence (B) is correct. Example 69.Asmallobject is attached to a light stringwhichpasses through a hollowtube.The tube is held byone hand and the string by the other. The object is stet into rotation in a circle of radius r1.The string is then pulled down, shortening the radius of the circle to r2. The ratio of the newkinetic energy to originalkinetic energy is (A) 12 rr (B) 1 (C) 2 12 rr . . .. .. (D) 2 21
rr . . .. .. Sol. Byconservation principle of angularmomentum, mv1r1 = mv2r2 . v1r1 = v2r2 ...(1) But ratio = 2 2 2 2 2 1 2 1 2 1 1 mv v r 21mv v r 2 . . . . . . . . . . . . . . Hence (C) is correct. Example 70. Asmall bead ofmass mmoving with velocity v gets threaded on a stationary
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 69 semicircular ring ofmassmand radius R kept on a horizontal table. The ring can freely rotate about its centre. The bead comes to rest relative to the ring.What willbe the final angular velocity ofthe system? O R v m (A) v/R (B) 2v/R (C) v/2R (D) 3v/R Sol. Applying conservation principle of angularmomentum, mvR = (mR2 +mR2). . 2 vR v 2R 2R . . . Hence, (C) is correct. Example 71. Achildwithmassmis standing at the edge of a disc withmoment of inertial I, radiusR, and initial angular velocity .. See figure given below.The child jumps off the edge of the discwith tangentialvelocityvwith respect to the ground. Then newangular velocity of the disc is v (A) I 2 mv2 I . . (B) (I mR2 ) 2 mv2 I . . . (C) I mvR I . . (D) .I mR2 . mvR I . . . Sol. Applying conservation principle of angularmomentum, (I + mR2).= I.´ + mv1R ...(1) But v1 � R.´ = v (Byrelative concept) ...(2) Fromequation (1) and (2) (I mR2 ) mvR ´ I . . . . . Hence (D) is correct. Example 72. Two men, each ofmass 75 kg, stand on the rimof a horizontal large disc, diametrically opposite to each other.The disc has amass 450 kg and is free to rotate about its axis. Eachmansimultaneouslystart along the rimclockwisewith the same speed and reaches their original starting points on the disc. Find the angle turned through bythe discwith respect to the ground. Sol. vrel = v1 + R. ad vrel = v2 + R. . v1 = vrel � R.
and v2 = vrel � R. . Li = Lf . O = mv1R + mv2R � MR2 2 . v1 vrel vrel v2 or MR2 2 . = mR (vrel � R. + vrel � R.) = 2mR (vrel � R.) ...(1) . 2. = (v1 + R.) t . 1 rel rel 2 2 2 v R v R R v . . . . . . . . . . .
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 70 . rel t 2v. . . . . . . ...(2) Fromequation (1) rel MR 2mR 2mR v 2 . . . . or rel M 2m 2m v 2 . . . . . .. .. . rel 2mv M/ 2 2m . . . . rel rel rel 2 2 2mv v v M/ 2 2m . . . . . . . . . . 4 m 4 75 4 75 M 2m 225 150 375 2 . . . . . . . . .. . . . . . . 4 radian 5. . Example 73. Auniformdisc ofmassmand radiusRrotates about a fixed vertical axis passing through its centre with angular velocity..Aparticle of samemassmand having velocity2.Rtowards centre of the disc collides with the discmoving horizontallyand sticks to its rim. Find (a) the angular velocityof the disc. (b) the impulse on the particle due to disc. (c) the impulse on the disc due to hinge. Sol. (a)Applying conservationprinciple of angularmomentumabout centre of disc. 2 2 mR mR mR2 ´ 2 2 . . . . . . . . . . O . 2 2 mR ´ 3 3 2 mR 2 . . . . .
. (b) Fydt = Jy = 2 m.R x x F dt J mR ´ mR3 . . . . . . 2 2 2 x y J J J mR 2 19 . . . . . mR 37 3 . . = 37 m R 3 .
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 71 Example 74. Ablock ofmassmis attached to a pulley disc of equalmassm, radius r by means of a slack string as shown. The pulley is hinged about its centre on a horizontal table and the block is projectedwith an initial velocityof 5m/s. Its velocitywhen the string becomes taut will be (A) 3 m/s (B) 2.5m/s (C) 5/3m/s (D) 10/3m/s Sol. For block, �Tdt =mv �mv0 For disc, rTdt = I. or 2 Tdt I Iv r r . . . or 0 mv mv mv 2 . . or 0 v v v 3 v 2 2 . . . . 0 v 2 v 2 5 10 m/ s 3 3 3 . . . . Hence (D) is correct. Example 75. Auniformrod of length l is given an impulse at right angles to its length as shown. Find the distance of instantaneous centre ofrotation fromthe centreof the rod. cmimpulse x Sol. Let instantaneous axis of rotation is passing through point O. vcm � y. = 0 . vcm = y. \ cm y . v. y x Fdt O vcm C Here Fdt = mvcm xFdt = I..= m 2 12 . l or m 2 Fdt 12 x. . l or 2 cm mv m
12x. . l 2 cm y v 12 x . . . l Example 76. Asolid sphere ofmassmand radiusR is placed on a smooth horizontal surface.Asudden belowis given horizontallyto the sphere at a height h = 4R/5 above the centre line. If I is the impulse of the blowthen find (a) theminimumtime afterwhich the highest point Bwilltouch the ground. (b) the displacement ofthe centre ofmass during this internal.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 72 Sol. (a) Here Fdt = mvcm and hFdt = Icm . or h I 2 mR2 5 . . Fdt or 4R I 2 mR2 4 5 . . . 2 I mR . . . . = .t or . = .t . t mR 2I . . . . . (b) Fdt = mvcm or I = mvcm . cm v I m . . cm S v t I mR R m 2I 2 . . . . . . Angular momentum conservation in collision Ifwe consider a ball collidingwith a bat, and neglected the force exerted byhand during collision, there is no external torque and bodies are freely colliding. Ifwepivot a bodyand then consider another bodycollidingwith it, the pivot exerts an impulsive external force, in general. The external force acting onthe systemduring collisionis not negligible. Case (a): Free collision of free extended bodies:We use two conservation laws in this case. (i) Totalmomentumis conserved. (ii)Angularmomentumabout any point is cosnerved. Case (b) Pivoted bodies: In this case, external force ( ext F. ) on the system arises frompivot (O) and is considerable. O r. N (i) Totalmomentumis not conserved. N (ii) Angularmomentumis conserved about the pivot. ROTATIONAL COLLISIONAND ANGUALR MOMENTUM We have discussed head on and oblique collisions. In this sectionwewilldiscuss the different cases of collision of two bodies inwhich during or after collision rotationalmotion ofthe body is a
lso taken into account.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 73 m u AL B A L v Bw We first discuss teh simples case of a collsionwhich is shown in figure.ArodAB ofmassMand length Lis hinged at the pointA, is hangingvertically.Asmallballofmassmmovingwitha speed uhorizontallystrikes the endBof the rod elastically.Let us consider that after collision the rodwill start rotatingwithangular speed . and the ball continues to move forward with a less speed v. Sot the ballmay rebound but we need not to consider this case, as ifballwill rebound, the result willgive thevelocityvnegative. The values ofv andwin this problemcan be obtained in twoways using energy and angular momentumconservation or using impulse equations. Note about hinge at pointA: Students should note that when ballwill strike the rod, anexternal impulsewill be developed at the hingewhichwillprevent and rod to move forward, as rod as onlyrotate, can not translate. Due to this external linear momentumof the systemcan not be conserved but angular momentumcan be conserved about the hinge. As stated in above paragraph, herewe cannot use linearmomentumconservation but as so external torque is present we use angularmomentumconservation about the hinge as muL=mvL+ ML2 3 . . . . . . . .....(i) Heremulis the angularmomentumof the ballbefore striking the rod and it is the onlyangularmomentumbefore collision as rodis at rest.After collisionaas ballwillmovewitha speedvin same direction, its angularmomentum ismul and that of rod is I.as it starts rotationwith initial angular velocity.. As collision is elasticwe use kinetic energyconservation before and after collison is 12 mu2 = 12 mv2 + 12 ML2 3 . . . . . . .....(ii)
Nowusing the above two equations, we get the values of v and .. The other wayof solving this problems is by breaking equation (i). It can be broken in two parts if required in some problems using impulse equations. Again consider the initial case when ball strikes the rod, a rnormal force is developed between the surface of ball and that of rod due to the push of ball against rod, the situtation is shown in figure. This interaction for F acts for a short duration (say dt) when the ball and rod are in contact. It retards the ball and its torque accelerates the rod in anticlockwise direction.We can use the impuse equatons for ball (linear) and rod (angular). m u AL Fdt B Formotion ofball, we have the linear impulse equation is Fdt mu = Fdt = mv .....(iii) Formotion ofrod, we have the angular impulse equation as 0 + Fdt = I. or FLdt = ML2 3 . . . . . . . .....(iv)
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 74 Ifwe use the above two equations alongwith equation(ii),we can solve the problem.Here ifeqauation(iii) and (iv) aremerged, it results equation (i). This type ofworking might be ofmore utility in solving the problem instead of directlyusing energyconservations. The previous casemight also be ofinelastic or partical elastic collision. If in previous problemwith same initial conditions the collisionis partical elastic and the coefficient of restitution is given as e, the angularmomentum conservationequation(i) remains same as no externaltorque is actingbut nowwe cannot use enrgyconservation as collision is not perfectlyelastic.Herewe use the definition of coefficient of restitution that it is the ratio of velocityof separtion after collision on the velocityof approach before collision. In this case it is used as e = L v u . . or L.= v eu .....(v) Solving the equations (i) and (v), we get the results v and .. If this collisionwere perfectlyinelastic, we use e = 0,which comes fromequation (v), v = L., as no separation occurs inperfectlyinelastic collision.Here it is important to be noted that the ball and rodwillbe separate even if inelastic collision takes place because ball is in translationalmotion and rod is in rotationalmotion. Figure explains the situation. m u AL B A L v B. e = 0 v = L. m u AL B A B . Now we consider onemore when ball sticks to the end of the rod and will rotate alongwith the rod. It is a specific type of inelastic collision as ballsticks to rod.Here againwe can neither use energy conservation nor the coefficient ofrestitutionequation.Onlyangularmomentumconservationequationis sufficient for solving the solving the problemas one of previous varialbes reduces, the linear velocityof the ball. Consider figure. Ball and rodwillstart together in rotationwith an initial angular velocityw.Thuswewrite the angularmomentum conservation equation as muL= I. [H = I =M.I. of rod plus ball] muL =
2 ML mL2 3 . . . . . . . . .....(vi) Heremoment of inertia of the systemtakencombined that of rod pulsballas both are inrotationalmotionwith angular velocitywafter collision. Example (vi)willgive us the angular velocity.ofthe system. Now we discuss another case when rod is not hinged at end end. Consider the collision shown in figure. Here rodAB is placed on a smooth surfacewhich is free to move on it and ballmoving shown in figure. m u A LB A L B v1 . d v2 As no externalforce is acting onthe system, herewe canalso conserve linearmomentumalongwith angularmomentum.Whenthe ballstrikes the rod, the instantaneous axis of rotationis taken at the centre of the rod hence the equation of angularmomentumconservation iswritten abut the centre of the rod. If after collision, ballmoveswith speed v1 and centre of rodmoveswith some angular velocity, let it be ., then according to angularmomentumconsevationwe have.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 75 mud = mv1d l. or mud = mv1d + ML2 12 . . . . . . . ....(vii) Fromlinearmomentumconservation, we have mu = mv1 +Mv2 .....(viii) Equations (vi), (vii) and (viii) gives the unknown parameters after collisions v1, v2 and . ifthis collision is not elastic,we use coefficient of restitution instead of energyconvervationequation (iv), as e = 2 1 (v d ) v u . . . .....(ix) Here v2 + d. is the linear speed of point P after collision as rod moves translationallywith velocity v2 and rotateswith angular velocity.and v1 is the finalvelocityof the ball after collision. For inelastic collisions, e can be taken as zero. Example 77. Arod of lengthRandmassMis free to rotation about ta horizontal axis passing through hinge P as in figure. First it is taken aside such that it becomes horizontaland thenreleased.At the lowest point the rod hits the blockB ofmassm and stops. Find the ratio ofmasses such that the block B completes the circle. Neglect anyfriction. M P R m B Sol. Minimumvelocityrequired byblock �m� to complete themotionin 5gR conservingmechanical energy 1 I 2 Mg R 2 2 . . . MgR I . . Cons. angularmomentumwrt P before&after collision. I. . m.R 5gR I . MgR mR 5gR I . R.M m P 5gR MgRI = m2R2 5 gR puting ML2 MR2 I 3 3 . . (since L= R) M 15 m .
Example 78:Arod ofmassMis lying on a smooth horizontal table.Asmall disc sliding at speedV0 hits the rod at one end normal to its length. The disc comes to a stop just after the collision. (a) Write the velocityof the centre ofmass of the rod just after the collision. (b) Write the angular velocity ofthe rod about its CMjust after collision. (c) Howmuch distance does the rodmove in one rotation about CM? Sol: Since there isno external force onthe rod + disc system,momentumis conserved.Let Vbe thevelocityofCM of the rod.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 76 (a) Conservingmomentum, mv0 + M× 0 = m × 0 + MV V = 0 mv M .....(i) . M m v (b) Angularmomentumconservation about a point of space initiallycoincidingwith the centre of the rod: mv0 L2 = I. ML2 12 . . ...... = 0 6mv ML .....(ii) (c) As there is no friction, the rod slideswith constant velocity after the collision. Let itmove by s in time t. In one rotation angle turned is 2..Hence 2. = .t .....(iii) . s = v2. . = 0 mv M × 0 2 6mv ML. . . . . . . = L3 Example 79. ArodACoflengthLandmassmis kept on a horizontal smooth plane. It is free to rotate andmove. Aparticle of same massmmovingwith velocity v strikes rod at point B which is at a distance L/4 frommid pointmaking angle 37º with the rod.The collision is elastic.After collision find (a) the angular velocityof the rod. l/4 37º A B C (b) the distancewhich centre of the rodwill travel in the time inwhich it makes halfrotation. (c) the impulse ofthe impact force. Sol. (a) The ballhas v´, component ofits velocityperpendicular to the length of rod immediatelyafter the collision u is velocity ofCOMof the rod and .is angular velocityof the rod, just after collision. The ball strikes the rod with speed v cos 53º inperpendicular direction and its component along the lengthof the rod after the collision is unchanged. Using for the point of collision.
Velocity of separation=Velocity of approach V´ D u 3v u v´ 5 4 . . . . . . . . . . l ...(1) Conserving linearmomentum(of rod + particle), in the direction . to the rod. mv . 3 mu mv´ 5 . . ...(2) Conserving angularmoment about point �D� as shownin the figure
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 77 m 2 0 0 mu 4 12 . . . . . . .. . . l l u 3 . . l ...(3) Bysolving u 24 v , W 72 v 55 55 . . l (b) Time taken to rotate by . angle t = .. l/4 Ndt In the same time, distance travelled = u2 . t = 3 . l Using angular impulse-angularmomentumequation. m 2 72 v N . dt . . 4 4 55 . . l l l N . dt 24mv 55 . . or using impulse �momentumequation onRod Ndt mu 24m 55 . . . Example 80: In the previous example for what ratio mM is the collision elastic ? Sol: For elastic collisioninitialkinetic energyshould be equal to theKE just after the collision. That is, Ki = Kf 201 mv 2 = 0 + 12 MV2 + 12 I.2 = 12 MV2+ 12 ML2 12 .2 = 12 M.
2 0 mv M . . .. .. + 12 ML2 12 2 0 6mv ML . . . . . . . m = m2 M + 3 m2 M 1 = 4 mM . mM = 14
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 78 Example 81:Asmall disc and a thin uniformrod of length L, whosemass is . times greater than themas of disc, lie on a smooth horizontalplane.The disc is set inmotion, in horizontaldirection and perpendicular to the rod, with velocityv, after which it elastically collideswith the end of the rod. Find the velocity of the disc angular velocity of the rod after the collision.At what value of hwill the velocityof the disc after the collision be equal to zero ?Reverse the direction ? Sol: The situation is shown in figure. Ifmass of disc ism, thenmass of rod hm. if v1 and v2 be the velocities of the disc and rod after collision, using linearmomentumconservation,we have mv =mv1 + hmv2 or v = v1 + .v2 ....(x) If after collision, rod starts rotatingwith angular speedw, using consrvation of angularmomentumwe have mv L2 = mL2 12 . . . . . . . . + mv1 L2 or v = 16 .L.+ v1 or . = 1 6(v v ) L . . .....(xi) As collision is elastic, using kinetic energyconservation,we get 12 mv2= 21 1 mv 2 + 221 v 2 . + 1 I 2 2 . or v2 = 21 v + 3. (v - v1)2 + . 2 1 2 (v . v ) . Solving for v1,we get v1 =
44 . . . . v Herewe can see that v1 is zero if h = 4 and v1will be negative when . > 4. Example 82. On a smooth table two particles of mass m each, travelling with a velocity v0 in opposite directions, strike the ends of a rigidmassless rod of length l, kept perpendicular to their velocity. The particles stick to the rod after the collision. Find the tension in rod during subsequent motion. v0 m l v0 m Sol. 0 0 mv mv I 2 2 l . l . . . 2 0 m mv 2 . . l l or 2 0 m mv 2 . . l l . 0 2v . . l . 2 2 0 2v T m m 2 2 . . . . . . . . . .. .. .. .. .. .. l l l . 20 2mv T . l
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 79 Example 83. Asolid uniformsphere of radius RandmassMrollswithout slipping with angular velocity w0 when it encounters a step of height 0.4 R. Find the angular velocityimmediatelyafter inelastic impact withthe rough step. 0 h Sol. Applying conservatio principle of angularmomentumabout point of impact, mvcm (R � h) + Icm .0 = I.´ . . 2 2 0 0 mR R h 2 mR 7 mR ´ 5 5 . . . . . . or 2 2 0 0 7 mR mRh 7 mR ´ 5 5 . . . . . or 0 0 7 R h 7 R ´ 5 5 . . . . . . 0 ´ 5 7 5h 7R 5 . . . . . . . . . . 0 ´ 5 7R 5h 7R 5. . . . . . . . . . ´ 0 .7R 2R. 7R . . . . 0 ´ 57. . . CONCEPT OF TOPPLING Rolling is generallya difficult concept to grasp. The primaryquestionthat rises in the students�mind is howcan a point in the body have zero velocityand yet the Fbodymoves forward. Let us consider a simple example as shown in figure. Fig. : Force acting on a flat body Assume that a force F acts on a flat body as shown above. How will the body move ? Obviouslythe bodywillmove forward translationally.The situation is shown infigure. v Fig. : Motion of the body v All points in the body willmove with the same velocity and acceleration. Therefore, the bodywill move translationally. Nowassume that the bodybecomes lesswide. What will the situation be ? This is shown in figure. Fig. : Force acting on a body with less width F
In this case as well, the body is likely to move translationally. However as the body becomes less and lesswide, something else happens. Let us nowconsider the situation infigure. In this case as the force Fincreases, the frictional force also increases.At some point
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 80 the bodystarts toppling.At this instance the bodytopples about point A. themotion in this situationis shown in figure. Fig. : Force on a very thin body F As the force increases, the bodystarts toppling about point A.The bodynowhas fallen down.Let us step back and consider the situationin greater depth. Fig. : Body topples F A Obviously this kind ofmotion is not translational. The bodyhasmoved forward as awhole.We can say that though because the centre ofmass of the body has moved, yet themotion has not been translational.What canwe sayabout the motion ?We can observe the following points : � Different points in the bodymovewith different velocities and accelerations.Whereas the topmost point has the greatest velocity, the point Ahas zero velocity. � The point of contact Ais at rest, evenwhile the bodymoves. � The bodyhasmoved forward evenwhen one point has zero velocity. We cantherefore conclude that the bodycanmove forward even if a single point in the bodyhas zero velocity. each one of these points is shown below: Different points in the bodymovewith different velocities and accelerations.Whereas the topmost point has the greatest velocity, point Ahas zero velocity.This is shown in figure. Fig. : Different points have different velocities v v1 v2 The point of contact Ais at rest, evenwhile the bodymoves.This is shownin figure. Fig. : Point A has zero velocity. Velocity profile of other points 0 A The body hasmoved forward evenwhenone point has zero velocity and zero acceleration. This is shown in figure. Fig. : Point has moved F B C F The centre ofmasswas initially at B. It has thenmoved forward to point C. Therefore the bodyhasmoved forward. We call this kind ofmotion toppling.Nowlet us look at when exactly a bodywill topple : � Abodywill topple when the base is smaller. � Abodywilltopplewhen the force exerted is higher fromthe ground level. � Abodyresting on a surface,whichhas friction, ismore likely to topple. Look at the situation below. Which body is more likely to topple ? F B
F C Fig. :Which body will topple ? Which body will topple ?
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 81 Obviously body B, because bodyC is wider then bodyB. This illustrates the first point that for a body to topple, the base has to be smaller. Consider the next situation now:F B F C Fig. :Which body will topple ? In this case, thewidth is the same. Which body will topple ? In the second case, the height at which the force is applied is lower.Therefore the bodyas shownfirst is likely to topple. If the surface has no friction, the bodywill bemore likely to move by pure translation than by toppling. Look at the situation below. C F F C Fig. :Which body will topple ? Which body will topple ? Obviously the the square, because it has a lesswide base.Why? Explanations follow. Similarly, we can see that a pentagon ismore likelyto topple than a square. Similarly an octagonwill bemore likely tomove by toppling than a pentagon. Therefore, we have seen that as the body becomes less and less wide, it tends to move more and more bytoppling. F Fig. : An octagon is more likely to move by toppling The question, however, still remains unanswered. How and when will a body move by translation and when will it move by pure translation ? Let us consider the situation shownin figure F B b a A Fig. : When will the body topple ? Assume that thewidth is a and height is b. Let us drawthe force diagram, just as the bodyis about to topple.Assume that (µ) is the coefficient of friction.We know fromthe earlier discussions that the bodywill topplewhen the normal force shifts to the rightmost point. For the bodyto topple about A, takingmoments about A, (whydowe take point A? Becausemaximumnumber of force i.e. friction and normalforce pass throughA. It is also correct to take torque about centre ofmass, though thiswillbe lengthier). F B b a A Fig. : Force diagram on the body mg N Total torque = Fb �
mga 2 If torque > 0 then the body topples.
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 82 There F > mga 2b What thismeans is that � If F < mga 2b then bodywillmove in pure translation andwillnot topple. � If F > mga 2b , the bodywill topple. Let us studythe concept to toppling in a littlemoremathematical detail.Whilewe outline every scenari, tryto visualize physicallythemotion of the body. We have alreadyestablished that for force F > mga 2b , the bodywill start moving by topping. Let us nowbring into this picture the friction,width and height of the force and study toppling. Consider first a square body ofwidth a.Assume that the force is applied at the top of the square.Assume also that the coefficient of friction is µ. The situation is shown in figure. F mg f Coefficient of friction = µ Fig. : Condition for toppling a square N Using the equations described above, the bodywilltopple for force F > mga 2b (since a = b). The bodywill topplewhen F > mg 2 or F > 0.5 mg If the coefficient of friction = 0.2 what will happen to the square ? Will it topple or translate ? For the body to just topple or translate, the force equations are same. The force equation for just toppling or just translation is F � f = 0 . F = f ...(1) N � mg = 0 . N = mg; f.µmg If µ =0.2,Maximumfrictional force = 0.2mg Therefore using equation (1) the force required to move the body by translation is F = 0.2 mg. The Force required to topple the body is F = 0.5mg. Therefore transalation happens first. The bodywill translate and not topple. What happens if the coefficient of friction is 0.6 ? In such a case, the force required to topple is F = 0.5 mg. This remains unchanged. However the force required to move the body by translation is 0.6mg. In this case topplingwill take place first. We can nowsummarize the scenario below: For µ > 0.5, the bodywillmove by toppling. For µ < 0.5, the bodywillmove by translation.
ROTATIONAL MECHANICS ROTATIONAL MECHANICS This off course assumes that the force is exerted at the top.
What happens if the force is exerted at any other point ? Insuchacase, thecoefficientoffrictionwillhavetobehighertomovethebodybytoppling.Visualizethis. It isverydifficult for asquaretotopple.it requiresaveryroughsurface(µ>0.5).
What is the value of frictional forces when µ < 0.5 and µ > 0.5 ?
The figure explains it all.
F=µmg F<µmg
N Coefficient of friction = µ < 0.5 Moves by translation Friction = µmg Coefficient of friction = µ > 0.5 Moves by translation Friction < µmg mg mg f = µmg f < µmg A A N Note the interesting fact in the second case. The body topples and moves. but during this time f < µmg. We allknow that when a body translates frictional force = µmg. The interesting thing to note is that a body a shown above can move in other methods translation. It can topple. � not necessarilyby just And what happens when the body topples ? The body moves, but yet frictional force < µmg. This is a very important point. Students need to visualize this scenario very clearly in their minds. When the body topples, frictional force is less than µmg.Also it requires less effort to move a body. This is a classic property of toppling. Let us summarize the characteristics of toppling that we have alreadydiscussed earlier : � All points have different velocities and accelerations. � The lowermost point will have zero velocity and acceleration when the body just starts toppling. � Frictional force is less than the maximumvalue of friction. If this is the case, why doesn�t a square body not topple so easily ? Thereasonisverysimple. Thecoefficient offrictionsneedsto behigherthan0.5for asquarebodyto topple (assuming that the force acts at the topmost point). This does not happen in reality because if µ = 0.5, the surface must reallybe very rough. Most surfaces are smoother than this. Hence a square bodyprefers to move bytranslationandnotbytoppling. � Itrequireslesseffortorforcetomovethebodybytoppling. mg NAF
But now let us consider a hexagon. a hexagon. Again assume that the coefficient of friction is µ and that the forceis being exerted at the top.Assume the widthof the
sides to be a.
Fig. : Force required to topple a hexagon
Again we can take torque about A. The side of the hexagonis a. Therefore the height
www.physicsashok.in 83 will be a tan 60. You can easily see that this is the case, by looking at figure. Fig. : Height is a tan 60 a a tan 60 60º
ROTATIONAL MECHANICS ROTATIONAL MECHANICS mga
The torque about A is, Fatan60� =0
2 mga
Therefore the body will topple for F >
2 tan 60 Now tan 60 ~ 1.732. Therefore the body will topple for F > 0.21 mg
You can now see the huge difference. for µ < 0.21, the body will translate and for µ > 0.21, the body will topple. The situation is again reproduced below Coefficient of friction = µ < 0.21 Moves by translation Friction = mµg mg N F A f = µmg Coefficient of friction = µ > 0.21 Moves by translation Friction < mµg mg N F A f = µmg Fig. : Motion for toppling a hexagon body Again the same conditions that were discussed before, regarding toppling hold true. When a body just topples, the coefficient of friction is less than µmg. The point of constact is a rest (i.e. point A). Different points have different accelerations. These are typical characteristics of toppling and you should now be clear about these. Note that we can make an interesting observation. The value of µ required to make a square topple is 0.5, whereas the value of µ required to make a hexagon topple is 0.21. this obviously bears out the fact that it is easier to topple a hexagon than a square. Similarly you will find that the minimum value of µ required to make a body topple will decrease if the body is an octagon. It will reduce stillfurther for a decagon. This means that a decagon willmove by toppling more easily than bypure translation. The force required to move it by toppling is less. What can we say about an infinite-sided polygon ? Extending the same logic, you will find that for values ofµ < 0, the body will topple. What does this mean ? The body always topples. Even if a small force is applied, the body will topple. Therefore note the interesting observation. Circular bodiesmove predominantlybytoppling.Againlet us reproduce theproperties oftoppling here.
� All points have different velocities and accelerations.
� Thelowermostpointwillhavezerovelocityandaccelerationwhenthebodyjuststartstoppling. � Frictionalforceislessthanthemaximumvalueoffriction. � Itrequiresverylittleefforttomoveabodybytoppling. Thesearepreciselythepropertiesoftopplingacircularbodyaswell. Theonlydifferenceisthat sincetheside of the circle is 0, virtuallyno forceis requiredto topplethebody.Thebodymovesbytopplingand F<µN.
Now let us introduce a very interesting property. Toppling is Rolling. Theyare one and the same. Visualize toppling motion and you canvisualize rolling motion. If you look at the properties oftoppling described above, these are exactlythe properties of rolling.
www.physicsashok.in 84
ROTATIONAL MECHANICS ROTATIONAL MECHANICS Toppling is thereforethe same as rolling. To visualize rolling motion always think oftoppling motion. It is easier to visualize toppling.
How does a body move when it is toppled ?
This is shown below
F F FC B A C A B C Fig. : Visualization of toppling To start with, A is at the edge. During the first round of toppling, A is at rest and B falls down. Now the body topples about point B. Point C touches the gound. In the next round, the body topples about point C. An interesting point to observe here is that the body topples about different points. First it is point A, which is at rest, and the body topples about A. Then the body topples about point B and B is at rest. Therefore different points have zero velocities and accelerations. This is exactly the situation with rolling as well. In rolling the point of contact has zero motion. This point however keeps on changing. It is veryimportant for the students to be able to visualize toppling and understand rolling in this contaxt. To summarize � Toppling is the same as rolling. � During toppling or rolling F < µmg. � It requires negligible force to topple or roll a disc. � Point of contact has zero motion with respect to the surface. � Different points have zero motion at different points of time. Another questionthat the student mayask is what happensin the case ofa hexagon if the value of the force F > 0.21 mg. In such a case, the bodywilldifinitelytopple, but the point of contact will also move. This will not have zero acceleration. This is shown in figure. Coefficient of friction = 0.21 A Acceleration f = µmg F > 0.21 mg Fig. : Value of force is higher than the force required to topple. The point of contact will move. The body will topple, but in a different way. the point of contact moves. We call this kind of motion sliding or skidding. Notice that when a body slides or skids, Frictionalforce � µmg and not less than µmg.
What is the minimum force required for the body to topple ?
Thebodywilltopplewhenthenormalforcepassesthroughtheedge. Theforce diagramat this stage is shownin figure.
taking torque about point A we have
body is just going to topple.
mg (1/2)= 4F F = (1/8)Mg = 0.125 mg Theforcerequiredforthebodytomovebytranslationis0.2mg, butthebodytoppleswhentheforceis0.125
Fig. : Force diagram when mg A N Mass=1 kg F f www.physicsashok.in 85
PHYSICS ROTATIONAL MOTION ROTATIONALMECHASNHIECEST www.physicsashok.in 86 mg. Therefore, it is obvious that the bodywill topple before it translates. Example 84. Auniformcube of side �b�and massMrest on a rough horizontal table.A horizontal force Fis applied normal to oneof the face at a point, at a height 3b/4 above the base.What should be the coefficient of friction(µ) between cube and table so that iswill tip about an edge before it starts slipping ? F b 3b/4 (A) µ > 2/3 (B) µ > 1/3 (C) µ > 3/2 (D) None Sol. mg b F 3b 2 4 . ...(1) F = f < µ mg ...(2) Fromequation (1) and (2) F < µmg 2mg µg 3 . or µ 23 . Hence (A) is correct. Example 85. Auniformcylinder rests on a cart as shown.The coefficient of static friction between the cylinder and the cart is 0.5. If the cylinder is 4 cmin diameter and 10 cm in height, which of the following is theminimumacceleration of the cart needed to cause the cylinder to tip over ? (A) 2m/s2 (B) 4 m/s2 (C) 5 m/s2 (D) the cylinderwould slide before it begins to tip over. Sol. Taking torque about O, ma0 × 5 > mg × 2 or 0 a 2g 4 . ma0 a0 mg O . 2 0min a 2 10 4 m/ s 5 . . . Hence (B) is correct.
ROTATIONALMECHANICS www.physicsashok.in 87 THINKING PROBLEMS 1. Can the mass of a body be taken to be concentrated at its centre of mass for the purpose of calculating its rotationalinertia? 2. If two circular discs of the sameweight and thickness aremade frommetals ofdifferent densities,whichdiscwill have the largermoment of inertia about its central axis? 3. About what axiswould a uniformcube have itsminimumrotational inertia? 4. Consider four bodies : a ring, a cube, a disc and a sphere.All the bodies have the samemass. The ring, disc and sphere have the same diameter, equalto the length ofthe cube on eachedge.All rotate about their axes through their respective centres ofmass.Which one has the largestmoment of inertia? Which is the smallest? 5. Aperson can distinguish between a rawegg and a hard boiled one byspinning each one on a table. Explain. 6. Awooden sphere rolls down two different inclined planes of the same height but different inclines.Will it reach the bottomwith the same speed in each case ?Will it take longer to roll down one incline than the other? If so, which one and why ? 7. Aman, withtwo equalmasses held at arm�s length, stands on a rotating table. If hemoves themasses and puts themover his shoulders, does this change his speed of rotation? Explain. 8. Astudent stands on a platformthat can rotate onlyabout a vertical axis. In his hand he holds the axle of a rimloaded bicyclewheelwith its axis vertical. Thewheel is spinning about the verticalaxiswithan angular speed .0, but the student and the platformare at rest. The student turns the wheel through 180º, that is, the axle of the wheel is held downward by the student.What happens ? 9. Adiver can turn several somersaults before strikingwater. Explain. 10.Showthat in the course of diving, the rotational kinetic energy of a diver, turning somersaults before striking water, increases.What is the source of this increased energy? 11. If the polar ice capsmelt and spread uniformly, howwill the length of the daybe affected ? 12.Acircular turntable rotates at constant angular velocity about a vertical axis. There is no friction and no driving torque.Anicepan containing ice also rotates alongwith it.The icemelts but none of thewater escapes fromthe pan. Is the velocity nowgreater, the same as, or less than the originalvelocity?Give reasons for your answer. 13.In order to get a billiard ball to rollwithout sliding fromthe start, the cuemust hit the ball not at the centre but exactly at a height of 2/5 Rabove the centre, where Ris the radius of the sphere. Explain. 14.Astudent standson a turntablewitha rim-loaded bicyclewheel. he holds the shaft of thewheelverticaland inthe beginning, there is no motion of the turntable. Now the student is asked to rotate the wheel in a clockwise direction relative to him.What happens to themotion ofturntable + student? 15.The melting of the polar ice caps is supposed to be a possible cause of the variation of the earth�s time of rotation.Explain.
ROTATIONALMECHANICS www.physicsashok.in 88 SOLUTION OF THINKING PROBLEMS 1. No. 2. 2 1 1 12 I = mr and 2 2 2 12 I = mr 2 1 12 2 2 I r I r = Now 2 2 1 1 2 2 m . . r t. . . r t. . 2 1 2 2 2 1 rr .. . . 1 2 2 1 II .. . or I 1. . Thus, the one with greater density will have less rotational inertial. 3. About a diagonal, because the mass is more concentrated about a diagonal. 4. I (ring) = (1/4)ma2 where a is the length of each edge of the cube, I (cube) = (1/6)ma2, I (disc) = (1/8)ma2, I (sphere) = (1/10) ma2. I1 : I2 : I3 : I4 = 1/4 : 1/6 : 1/8 : 1/10 = 30 : 20 : 15 : 15. Thus, the ring has the greatest moment of inertia and the sphere the smallest. 5. The hard boiled egg will continue to sping if they are momentarily stopped and then let go. 6. The acceleration down an inclined plane is given by a = g sin ../(1 + k2/r2) where k is the radius of gyration. h (height of the plane) = s sin . where s is the length of the plane. Now 2 2 ( 2 2 ) 1 1 sin / 1 / 2 2 s = at = g a t + k r . t2 sin2 a = constant . t .1/ sin. Thus, the greater the time of descent the less the inclination to the horizontal
. Now 2 2 2 2 2 sin 1 / sin v as g h k r . . . . . . v = a constant Thus, the sphere will reach the bottom with the same speed, whatever be the inclination. 7. The only external force acting on the system (man + table) is gravity and reaction of the ground, and those exert no torque about the axis of rotation. Hence, the angular momentum of the system is conserved about this axis. When the man puts the masses on his shoulders, the moment of inertia of the system about the axis decreases and so the angular speed of the system increases. 8. The platfor starts rotating with almost double the initial angular momentum of the wheel. To justify this, let us consider �wheel + student + platform� asour system. The initial angular momentum of the system is I0.0, where I0 is the moment of inertia of the wheel about the vertical axis. Since there is no external torque on the system about the vertical axis, the angular momentum about the vertical axis must be conserved about this axis. Let I be the moment of inertia of �student + platform� about the vertical axis. The angular momentum of the wheel about the vertical axis is now �I0.0. Let . be the angular velocity of the platform. Then by the principle of conservation of angular momentum I0.0 = �I0.0 + I. or I. = 2I0.0 . 0 . . 2. on account of negligible m.i. of the platform.
ROTATIONALMECHANICS www.physicsashok.in 89 9. Let us consider the diver as a system and his limbs as its components. When he leaves the diving board, he has a certain angular speed .0 about the horizontal axis through his centre of mass. Now there are no external forces acting on him except gravity, and gravity exerts no torque about his centre of mass. His angular momentum remains constant, and I0.0 = I.. When he pulls in his limbs, his moment of inertia decreases and consequently, his angular velocity increases. 10. Since the diver is free from external torque about the horizontal axis through his centre of mass, his angular momentum remains constant. Therefore. I0.0 = I. or 2 2 2 2 I0w0 = I w ....(i) When he pulls in his limbs, his moment of inertia about the horizontal axis decreases. . I < I0 or 0 1 1 I I > or 2 2 2 2 0 0 0 1 I 1 . I I I . . . . or 2 2 0 0 1 1 2 2 I. . I . The source of this increased kinetic energy is the diver himself, who does work when he pulls the parts of his body together. His body energy is converted into his kinetic energy. 11. If the polar ice caps were to melt, the water formed would spread on the surface of the earth and so its moment of inertia would increase. By the law of conservation of angular momentum, the speed of rotation would be reduced, that is, the length of the day would increase. 12. Due to the accumulation of mass near the edge, the moment of inertia of the system increases and so the angular velocity of the system decreases. 13. Let the ball be struck by a horizontal force F at a height d above the centre. Then the torque exerted by it on the ball about its centre of F.d. By the formula . = I., we have F.d = 25 MR2 × . Let a be the acceleration of the centre of mass produced by this force. Then F = Ma. For rolling without sliding, v = .R and a = ...R; . F = M ..R . 2 2 5 M. R . d . MR .. . 25 d = R
14. The system (turntable + student) will turn in the opposite direction. Since the entire system is free from external torques, the angular momentum of the system (wheel + turntable + student) remain constant at zero. If I = moment of inertia of turntable + student and I0 = moment of inertia about the vertical axis of rotation of the table and .0 = angular velocity of rotation (relative to the earth), then I0.0 + I..= 0 or 0 0 I I. . . . 15. A liquid of given mass has less inertia for rotational motion than the same mass of solid because a liquid cannot sustain shearing force. It always tends to move translation-wise. Hence, the melting of the ice caps would result in the reduction of the rotational inertia of the earth and hence, the angular speed of rotation of the earth would increase or time of rotation decrease. If water were to spread uniformly, the result would be as in Example 11.
ROTATIONALMECHANICS www.physicsashok.in 90 ASSERTION & REASON Astatement of Statement-1 is given and a Corresponding statement of Statement-2 is given just belowit of the statements,mark the correct answer as � (A) If both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation ofStatement-1. (B) If both Statement-1 and Statement-2 are true and Statement-2 isNOT correct explanation ofStatement-1. (C) If Statement-1 is true but Statement-2 is false. (D) If both Statement-1 and Statement-2 are false. (E) IfStatement-1 is false but Statement-2 is true. 1. Statement-1 : In pure rollingmotion, net work done by friction is zero. Statement-2 : Sumof translationalwork done by friction and rotationalwork done by friction is zero. 2. Statement-1 : Auniformsolid cylinder rolling with angular velocity . along a plane surface strikes a vertical rigid wall.Angular velocity of cylinder when it begins to roll up a wall is less than the initialAngular velocity (.). Statement-2 : After striking the verticalwall angular velocity 3. Statement-1 : If bodies slide down an inclined plane without rolling then all bodies reach the bottom simultaneously. Statement-2 : Acceleration of all bodies are equal and independent of the shape. 4. Statement-1 : The force of friction in the case of a disc rollingwithout slipping down an inclined plane is zero. Statement-2 : When the disc rolls without slipping, friction is required because for rolling condition velocity of point of contact is zero. 5. Statement-1 : As the radius of earth reduces by 50% without any change in mass, length of a day reduces. Statement-2 : Angular momentumconservation provides drop in time as . decreases to 25%of the original. 6. Statement-1 : If there is no external torque on a body about its centre ofmass, then the velocity of the centre ofmass remains constant. Statement-2 : The linearmomentumof an isolated systemremains constant. [JEE, 07] MATCH THE COLUMN 1. A circular body of mass M and radius R, initially spinning about its centre of mass with 0 . is gently placed on a rough horizontal surface. The moment of inertia of body about its C.M. is 2 .CM . MK . If the coefficient of friction between the body and the surface is . then: Column I Column II (A) Translational work done by the friction (P) �ve (B) Rotational work done by the friction (Q) +ve (C) Larger the moment of inertia of body, the time (R) smaller required for rolling motion (D) Larger the moment of inertia of the body, (S) longer work done by the friction (T) greater
ROTATIONALMECHANICS www.physicsashok.in 91 2. In the following column - I mass of each object is m and circular of radius R. Column - II represents moment of inertia. Column I Column II (A) Full ring (P) mR2 (B) Half ring (Q) 2 2 mR (C) Quarter ring (R) 2 4 mR (D) Arc making an angle . at the centre (S) 2 2 mR .. 3. Suppose a force Of is applied at the top most point of a rigid body of radius R and mass M. Column I Column II (A) Force of friction will be zero for (P) Solid sphere (B) Force of friction will be forward for (Q) Zero (C) Force of friction will be backward for (R) Ring (D) If a . R. then force of friction (S) No body 4. A rigid body of mass M and radius R rolling without slipping on the inclined plane, then the magnitude of force of friction. Column I Column II (A) For ring (P) Mg Sin. 2.5 (B) For solid sphere (Q) Mg Sin. 3 (C) For solid cylinder (R) Mg Sin. 3.5 (D) For hollow sphere (S) Mg Sin. 2 5. A rigid body is rolling without slipping on the horizontal surface Column I Column II (A) Velocity at point A i.e. A V (P) V 2 (B) Velocity at point B i.e. B V (Q) Zero (C) Velocity at point C i.e. C V (R) V (D) Velocity at point D is i.e. D V (S) 2 V LEVEL � 1 1. A ring of mass 0.3 kg and radius 0.1 m and a solid cylinder of mass 0.4 kg and of the same radius are given, kinetic energy and released simultaneously on a flat horizontal surface such that they begin to roll with the same KE as soon as released towards a wall which is at the same distance from the ring and the cylinder. Then: (A) The cylinder will reach the wall first. (B) The ring will reach the wall first. (C) Both will reach the wall simultaneously. (D) None of the above. 2. A solid homogeneous sphere is moving on a rough horizontal surface, partially rolling and partly sliding. During this kind of motion of the sphere: (A) total KE is conserved. (B) angular momentum of the sphere about the point of contact is conserved. (C) only the rotational KE about centre of mass is conserved. (D) angular momentum about the centre of mass is conserved.
ROTATIONALMECHANICS www.physicsashok.in 92 3. The angular momentum of a projectile (in X-Y plane) about origin (point of projection) at time t has: (A) only Z component. (B) only X and Y components. (C) All the three X, Y and Z components. (D) only Y-component. 4. A solid sphere of radius R and mass M is pulled by a force F acting at the top of the sphere as shown in figure. Friction coefficient is sufficient F Rough enough to provide rolling without slipping. Work done by force F when the centre of mass moves a distance S is : (A) FS (B) 2FS (C) Zero (D) 3 2FS 5. A body of mass M slides down an inclined plane and reaches the bottom with velocity v. If a ring of same mass rolls down the same inclined plane, what will be its velocity on reaching the bottom? (A) v (B) v 2 (C) 2 v (D) v 2 6. A sphere cannot roll on: (A) a smooth horizontal surface. (B) a smooth inclined surface. (C) a rough horizontal surface. (D) a rough inclined surface. 7. A thin spherical shell of radius R lying on a rough horizontal surface is hit sharply and horizontally by a cue. At what height from the ground should it be hit so that the shell does not slip on the surface? (A) 2 3.R (B) 5 4.R (C) 5 3.R (D) 3 2.R 8. What is the moment of inertia of a triangular plate ABC of mass M and side BC = a about an axis passing through A, A B C 45° a 45° shown in figure and perpendicular to the plane of the plate? (A) 2 6 Ma (B) 3 2 4 Ma (C) 2 24 Ma (D) 2 12 Ma 9. A thin circular ring of mass �M and radius r is rotating about its axis with a constant angular velocity w, Two objects, each of mass m, are attached gently to the opposite ends of a diameter of the ring. The wheel now rotates with an angular velocity. (1983) (A) . . wM M .m (B)
. .
. . 2 2 w M m M m . . (C) . 2 . wM M . m (D) w.M 2m. M. 10. Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. the rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of (1995) (A) 0.42 m from mass 0.3 kg (B) 0.70 m from mass 0.7 kg (C) 0.98 m from mass 0.3 kg (D) 0.98 m from mass 0.7 kg 11. A cubical block of side a is moving with velocity V on a horizontal smooth plane as shown in Figure. It hits a ridge at point O. The M V O a angular speed of the block after it hits O is(1999) (A) 3V .4a. (B) 3V .2a. (C) 3V . 2a . (D) Zero 12. An equilateral triangle ABC formed from a uniform wire has two small identical beads initially located at A. The triangle is set rotating about the vertical axis AO. Then the beads are released from rest simultaneously and allowed to slide down, one along AB and the other along AC as shown. Neglecting frictional effects, the quantities
ROTATIONALMECHANICS www.physicsashok.in 93 that are conserved as the beads slide down, are (2000) (A) angular velocity and total energy (kinetic and potential) (B) Total angular momentum and total energy (C) angular velocity and moment of inertia about the axis of rotation. (D) total angular momentum and moment of inertia about the axis of rotation. 13. One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M. It is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is (2001) (A) 1 2 2 MR (B) 1 2 4 MR (C) 1 2 8MR (D) 2MR2 14. A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The direction of the frictional force acting on the cylinder are (2002) (A) up the incline while ascending and down the incline descending. (B) up the incline while ascending as well as descending. (C) down the incline while ascending and up the incline while descending (D) down the incline while ascending as well as descending. 15. A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. A tortoise is sitting at the edge of the platform. Now, the platform is given an angular velocity 0 . . When the tortoise move along a chord of the platform with a constant velocity (with respect to the platform), the angular velocity of the platform . .t . will vary with time t as (2002) (A) .0 .(t) (B) .0 .(t) (C) .0 .(t) (D) .0 .(t) 16. Consider a body, shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulse M MJ = MV L J = MV is imparted to the body at one of its ends, what would be its angular velocity? (2003) (A) V L (B) 2V L (C) V 3L (D) V 4L 17. A particle undergoes uniform circular motion. About which point on the plane of the circle, will the angular momentum of the particle remain conserved? (2003) (A) centre of the circle (B) on the circumference of the circle. (C) inside the circle (D) outside the circle. 18. A horizontal circular plate is rotating about a vertical axis passing through its centre with an angular velocity 0 . . A man sitting at the centre having two blocks in his hands stretches out his hands so that the moment of inertia of the system doubles. If the kinetic energy of the system is K initially, its final kinetic energy will be
(2004) (A) 2 K (B) K 2 (C) K (D) K 4 19. A disc is rolling without slipping with angular velocity . . P and Q are two points equidistant from the centre C. The C P Q order of magnitude of velocity is (2004) (A) Q C P V . V .V (B) P C Q V . V . V (C) P C V .V , 2 Q C V .V (D) P C Q V .V . V
ROTATIONALMECHANICS www.physicsashok.in 94 20. A block of mass m is at rest under the action of force F against a wall as shown in figure. Which of the following statement is incorrect? (2005) (A) . = mg [where . is the friction force] (B) F = N [where N is the normal force] (C) F will not produce torque (D) N will not produce torque 21. A particle is confined to rotate in a circular path decreasing linear speed, then which of the following is correct? (2005) (A) L. (angular momentum) is conserved about the centre (B) only direction of angular momentum L. is conserved (C) It spirals towards the centre (D) Its acceleration is towards the centre. 22. Select the correct statement(s): (A) A particle performs uniform circular motion with an angular momentum L. If the frequency of particle�s motion is doubled and its KE is halved, the angular momentum becomes one-fourth. (B) If a particle moves in X-Y plane, the resultant angular momentum about origin has only Z-component. (C) A particle is moving along a straight line parallel to X-axis with constant velocity. Its angular momentum about the origin decreases with time. (D) Action of angular impulse is to change the angular momentum. 23. Aring, a cube, a cylinder, a prism and a sphere all having equalmasses, equal heights and equalmaximum width: (A) The cube has the largest moment of inertia about an axis perpendicular to cross-section and passing through the centre of mass. (B) The ring has the largest moment of inertia about the axis mentioned above. (C) The cylinder has the smallest moment of inertia. (D) The sphere has the smallest moment of inertia. 24. Select the correct alternative(s): (A) The mass of a body can be taken to be concentrated at its centre of mass for the purpose of calculating its rotational inertia. (B) Two circular discs of the same mass and thickness are made from metals of different densities, the one with greater density will have less rotational inertia about its central axis. (C) About an axis passing through diagonally opposite ends of a uniform cube, have its minimum rotational inertia. (D) None of these. 25. Select the correct statement(s): (A) A very small particle rests on the top of a hemisphere of radius 20 cm. The smallest horizontal velocity to be given to it, if it is to leave the hemisphere without sliding down its surface, is 3.2 m/s. (B) A solid cylinder at rest at the top of an inclined plane of height 2.7m rolls down without slipping. If the same cylinder has to slide down a frictionless inclined plane and acquires the same velocity as that acquired by the centre of mass of the rolling cylinder at the bottom of the inclined plane, the height of the inclined plane should be 1.8m.
(C) Three thin rods each of mass M and length L are welded so as to form an equilateral triangle. The moment of inertia of the triangle about an axis perpendicular to the plane of triangle and passing through of its vertices is 3 2 2 ML . (D) None of these. 26. Two boys of equal masses are sliding freely on smooth horizontal surface with the same speed v on parallel straight paths in opposite directions. The paths are separated by a perpendicular distance d. One of the boys carries a light pole of length d, held firmly at one end. The other boy grasps the other end of the pole just when they are crossing each other. Which of the following statements is/are true for the subsequent motion?
ROTATIONALMECHANICS www.physicsashok.in 95 (A) Each boy moves in a circular path of radius d 2 with a constant speed v. (B) The tension in the pole remains constant. (C) They come to rest when the pole has rotated through 90° to lie along the direction of original motion. (D) The tension in the pole varies with time. 27. A uniform rod AB of length . is free to rotate about a horizontal axis passing through A as shown in figure. The rod is released from rest from horizontal position. If the rod gets broken at mid-point C when it becomes AB C vertical, just after breaking of the rod: (A) Angular velocity of upper part starts to decrease while that of lower part remains constant. (B) Angular velocity of upper part starts to decrease while that of lower part starts to increase. (C) Angular velocity of both the parts is identical. (D) Angular velocity of lower part becomes equal to zero. (E) lower part falls vertically. 28. If 1 . is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and 2 . is the moment of inertia of the ring formed by bending the rod, then: (A) 2 2 1 . . . 4. (B) 2 2 1 . . . . (C) 2 1 . . . 0.3 (D) 2 1 2 . . .. 3 29. If the polar ice caps melt and spread uniformly: (A) The moment of inertial of the earth about centroidal axis would increase. (B) The moment of inertia of the earth about centroidal axis would decrease. (C) Length of the day would increase. (D) The speed of rotation would be reduced. 30. A sphere is rolling down a plane of inclination . to the horizontal. The acceleration of its centre, down the plane is: (A) g sin . (B) less than g sin . (C) greater than g sin . (D) greater than .g sin. . 2 31. The moment of inertia of the pulley system shown in figure is 4 kg m2. The radii of bigger and smaller pulleys are 2m and 1m, respectively. The 1m 2m 4 kg 5 kg angular acceleration of the pulley system is: (A) 2.1 rad/s2 (B) 4.2 rad/s2 (C) 1.2 rad/s2 (D) 0.6 rad/s2. 32. Two masses 1 kg and 2 kg are connected by an inextensible light thread passing over a pulley of mass 2 kg and radius 20 cm. There is no slipping anywhere: (A) Tensions at the two sides of the pulley will be different. (B) Accelerations of the two blocks will be g 4 . (C) Ratio of tensions on the two sides will be 5 6 (D) Ratio of tensions on the two sides will be 1 2 . 33. A tube of length L is filled completely with an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of tis ends with a uniform angular velocity .. The force
exerted by the liquid at the other end is : (A) M.2 L 2 (B) M.2L (C) M.2 L 4 (D) M.2 L2 2 34. Let . be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle . with AB. The moment of inertia of the plate about the axis CD is then equal to: (A) . (B) . sin2. (C) . cos2. (D) . cos2 .. 2.
ROTATIONALMECHANICS www.physicsashok.in 96 35. A rod of mass M kg and length L metre is bent in the form of an equilateral triangle as shown in the figure. The moment of inertia of O triangle about a vertical axis to perpendicular to the plane of triangle and passing through the centre (in units of kg m2) is (A) 2 12 ML (B) 2 54 ML (C) 2 162 ML (D) 2 108 ML 36. A cord is wound round the circumference of a solid cylinder of radius R and mass M. The axis of the cylinder is horizontal. A weight mg is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the cylinder will be (A) 2 2 mg M . m (B) 2gh R (C) . . 2 4 2 mgh M . m R (D) 37. A disc of moment of inertia . is rotating about its axis, which is initially along the vertical direction, making �n� revolution per minute. The axis gradually tilts becomes horizontal and the disc rotates about the horizontal axis with the same revolution per minute. The total time, taken in doing so is t seconds then the torque acting on the body is (A) Zero (B) 260nt . . (C) 2 2 60 n t . . (D) None of the above. 38. From the circular disc of radius R and mass M, a concentric circular disc of
small radius r is cut and removed, the mass of which is m. the moment of inertia of the angular disc remaining, about its axis perpendicular to the plane and passing through the centre of mass will be (A) . 2 2 . 12 M R . r (B) 1 . .. 2 2 . 2 M .m R . r (C) . .. 2 2 . 12 M .m R . r (D) 1 . 2 2 . 2 MR .mr 39. Three discs each of mass M and radius R are placed in contact with each other as shown in the figure here. Then the M. of the system about an XX� is: (A) 2 4 MR (B) 11 2 4 MR (C) 2 2 3 MR (D) 7MR2 40. In which case, it is easier to rotate? (A) Case I (B) Case II (C) same (D) data is insufficient 41. A small mass m is attached to the rim of a circular disc of mass M and radius R. The disc rotates with angular velocity . about an axis passing through the centre O of the disc and perpendicular to its plane. Then the angular momentum of the system will be (A) 1 2 2 MR . (B) MR2. (C) 2 2 2 M m R . . . . . . . . (D) .M m.R. .
ROTATIONALMECHANICS www.physicsashok.in 97 42. When a force F acts on a side of the hexagonal body will (A) topple (B) translate (C) both (D) None 43. A particle of mass m is moving along a straight line y = x + 4 with constant velocity v, then angular momentum of the particle about the origin is (A) Zero (B) constant (C) decreasing continuously (D) increasing continuously 44. A cylinder can maintain its rolling motion on a (A) rough horizontal surface (B) smooth horizontal surface (C) rough inclined surface (D) smooth inclined surface 45. Choose the correct statement(s) for a particle moving along a circular path of constant radius (A) centripetal acceleration is always perpendicular to velocity vector. (B) centripetal acceleration is always perpendicular to angular velocity vector. (C) angular acceleration vector is always perpendicular to angular velocity vector. (D) angular velocity vector is always perpendicular to the linear velocity vector. 46. A solid sphere is executing rolling with slipping motion on a rough horizontal surface (A) The frictional force will always perform a negative work on the sphere. (B) The work done by friction cannot be zero. (C) The angular momentum of the sphere is conserved about its centre of mass. (D) The angular momentum of the sphere is conserved about any point on the horizontal surface. 47. A solid sphere of mass M and radius R is pulled horizontally on a rough surface as shown in the figure. Choose the incorrect alternatives (A) The acceleration of the centre of mass is F M . (B) The acceleration of the centre of mass is 2 3 FM . CM F (C) The frictional force on the sphere acts forward. Rough Surface (D) The magnitude of the frictional force is F 3 . 48. A constant force F is applied at the top of ring as shown in figure. Mass and the radius of the ring are M and R respectively. Angular F M R momentum about the point of contact at time �t� (A) is constant (B) increases linearly with time (C) is 2. FRt (D) 2FRt 49. A disc of radius R rolls on a horizontal surface with linear velocity V and angular velocity .. There is a point P on circumference of disc at angle . , which has a vertical velocity. Here . is equal to (A) sin 1 V R . . . . (B) sin 1 2 V R
.
.
. . (C) cos 1 V R . . . . (D) cos 1 V R . . . . 50. Which of the following statement(s) is/are correct for a spherical body rolling without slipping on a rough horizontal surface at rest (A) the acceleration of the point of contact with the ground is zero. (B) the speed of some of the point(s) is/are zero. (C) frictional force may or may not be zero (D) work done by friction may or may not be zero.
ROTATIONALMECHANICS www.physicsashok.in 98 51. A 2kg mass attached to a string of length 1m moves in a horizontal circle as a conical pendulum. The string makes an angle . = 30° with the vertical. Select the correct alternative(s) (g = 10 m/s2). (A) The horizontal component of angular momentum of mass about the point of support P is approximately 2.9 kg m2/s. (B) The vertical component of angular momentum of mass about the point of support P is approximately 1.7 kg m2/s. (C) The magnitude of dL dt ( L. angular momentum of mass about point of support P) is approximately 2 2 10 kgm s . (D) dL dt . . . will not hold good in this case. 52. A particle of mass m is travelling with a constant velocity v .V0 . along the line y = b, z = 0. Let dA be the area swept out by the position vector of (from origin) particle in time dt and L is themagnitude of angular momentum of particle about origin at any time t. Then (A) L = constant (B) L . constant (C) dA 2L dt m . (D) 2 dA L dt m . 53. A uniform rod kept on the ground falls from its vertical position. Its foot does not slip on the ground. (A) No part of the rod can have acceleration greater than g in any position (B) At any one position of rod, different points on it have different acceleration. (C) Any one particular point on the rod has different acceleration at different positions of the rod. (D) The maximum acceleration of any point on the rod, at any position is 1.5 g. 54. A thin uniform rod of mass m and length . is free to rotate about its upper end. when it is at rest, it receives an impulse J at its lowest point, normal to its length. Immediately, after impact (A) the angular momentum of rod is jl . (B) the angular velocity of rod is 3J ml . (C) the kinetic energy of rod is 3 2 2Jm . (D) the linear velocity of mid point of rod is 32
Jm . 55. The torque . . on a body about a given point is found to be equal to A. L . . where A . is a constant vector and L. is the angular momentum of the body about this point. It follows that: (A) dL dt is perpendicular to L. at all instant of time. (B) the component of L. in the direction of A . does not change with time. (C) the magnitude of L. does not change with time. (D) L. does not change with time. 56. Two particles A and B initially at rest, move towards each other under mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, the speed of the centre of mass of the system is (A) 3 V (B) V (C) 1.5 V (D) Zero (JEE, 1982) 57. When a bicycle is in motion, the force of friction exerted by the ground on the two wheels is such that it acts (A) in the backward direction on the front wheel and in the forward direction on the rear wheel. (B) in the forward direction on the front wheel and in the backward direction on the rear wheel. (C) in the backward direction on both the front and the rear wheels. (D) in the forward direction on both the front the rear wheels. (JEE, 1990)
ROTATIONALMECHANICS www.physicsashok.in 99 58. Two particles, each of mass m and charge q, are attached to the two ends of a light rigid rod of length 2R. The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. the ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is (JEE, 1998) (A) q/2m (B) q/m (C) 2q/m (D) q/.m 59. Let . be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle . with AB. The moment of inertia of the plate about the axis CD is then equal to (A) . (B) . sin2. (C) . cos2. (D) . cos2 .. 2. (JEE, 1998) 60. Let I be themoment of inertia of a uniformsquare plate about an axisABthat passes through its centre and is parallel to two of its sides. CDis a line in the plane of the plate that passes through the centre of the plate and makes as angle .withAB.Themoment of inertia of the plate about the axis CDis then equal to [JEE, 98] (A) I (B) I sin2 . (C) I cos2 . (D) I cos2 (./2) 61. The torque . . on a bodyabout a given point is found to be equal to A. L .. .. whereA.. is a constant vector andL.. is the angularmomentumof the body about that point. Fromthis it follows that [JEE, 98] (A) dL/ dt .. is perpendicular to L.. at all instants of time (B) the components ofL.. in the direction ofA.. does not changewith time (C) themagnitude ofL.. does not changewith time (D) L.. does not changewith time 62. Asmooth sphereAismoving on a frictionless horizontalplanewithangular speed .and centre ofmass velocity v. It collides elastically and head onwith an identical sphere B at rest. Neglect friction everywhere.After the collision, their angularspeeds are .A and .B, respectively.Then [JEE, 99] (A) .A < .B (B) .A = .B (C) .A = . (D) .B = . 63. A disc of massM and radius R is rolling with angular speed . on a horizontal as shown. Themagnitude of angularmomentumofthe disc about the originOis : [JEE, 99] y O x M (A) (1/2)MR2. (B)MR2. (C) (3/2)MR2. (D) 2MR2. 64. Acubicalblock of sideLrest ona rough horizontal surfacewithcoefficient of friction µ.Ahorizontalforce F is applied on the block as shown. If the coefficient of friction is sufficientlyhigh so that the block does not slide before toppling, theminimumforce F L required to topple the block is : [JEE, (Scr) 2000] (A) infinitesimal (B)mg/4 (C)mg/2 (D)mg(1 � µ) 65.Athinwire oflengthLand uniformlinearmass density. isbent into a circular loop
with centre at Oas shown. Themoment of inertia of the loop about the axisXX´ is : [JEE, (Scr) 2000] 60º O x x´ B (A) .L3/8.2 (B) .L3/16.2 (C) 5.L3/16.2 (D) 3.L3/8.2 66. Two particles eachofmassMare connected by amassless rod of length l. The rod is lying on the smooth surface. If one of the particle is given an impulseMVas shown in the figure thenangular velocityof the rodwould be : [JEE, (Scr) 03] (A) v/l (B) 2v/l (C) v/2l (D) None Mv M
ROTATIONALMECHANICS www.physicsashok.in 100 67. Adisc hasmass 9m.Ahole of radius R/3 is cut fromit as shown in the figure. The moment ofinertia ofremaining part about an axis passing through the centre �O� ofthe disc and perpendicular to the plane of the disc is : [JEE, (Scr) 05] 2R/3 R QR/3 (A) 8 mR2 (B) 4mR2 (C) 40 9 mR2 (D) 37 9 mR2 68. Aparticlemoves in circular pathwith decreasing speed.Which of the following is correct (A) L.. is constant (B) only direction of L.. is constant [JEE, (Scr) 05] (C) acceleration a. is towards the centre (D) itwillmove in a spiral and finally reach the centre 69. Asolid sphere ofmassM, radiusR and havingmoment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickkness t, whosemoment of inertia about an axis passing through its edge and perpendicular to its plane remains I.Then, radius of the discwill be [JEE, 06] (A) 2R / 15 (B) R 2 /15 (C) 4R / 15 (D) R/4 70. Asolid cylinder ofmassmand radius r is rolling on a rough inclined plane of inclination .. The coefficient of friction between the cylinder and incline is µ. Then [JEE, 06] (A) frictionalforce is always µmg cos . (B) friction is a dissipative force (C) bydecreasing ., frictional force decreases (D) friction opposes translation and supports rotation 71. Aballmoves over a fixed track as shown in the figure. FromAto Bthe ball rolls without slipping. SurfaceBC is frictionless. KA, KB andKC are kinetic energies of the ball atA, B and C, respectively. Then hA hC C A B (A) hA > hC ; KB > KC (B) hA > hC ; KC > KA (C) hA = hC ; KB = KC (D) hA < hC ; KB > KC [JEE, 06] 72. Asmallobject of uniformdensityrolls up a curved surfacewith aninitialvelocityv. It reaches up to amaximumheight of 3v2/(4g) with respect to the initialposition. The object is [JEE, 07] v (A) ring (B) solid sphere (C) hollowsphere (D) disc FILL IN THE BLANKS 1. A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point that is directly above the centre of the face, at a height 3a 4 above the base. The minimum value of F for which the cube begins to tip about the edge is ........... (Assume that the cube does not slide). (1984) 2. According to Kepler�s second law, the radius vector to a planet from the sun sweeps out equal area in equal
intervals of time. This law is a consequence of the conservation of ................ (1985) 3. A smooth uniform rod of length L and mass M has two identical beads of negligible size, each of mass m, which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with an angular velocity 0 . about an axis perpendicular to the rod and passing through the midpoint of the rod (see figure). There are no external forces.
ROTATIONALMECHANICS www.physicsashok.in 101 When the beads reach the ends of the rod, the angular velocity of the system is .............. (1988) 4. A cylinder of mass M and radius R is resting on a horizontal platform (which is parallel to the x-y plane) with its axis fixed along the y-axis and free to rotate about its axis. The platform is given a motion in the x-direction given by x = A cos ( . t). There is no slipping between the cylinder and platform. The maximum torque acting on the cylinder during its motion is ................ (1988) 5. A stone of mass m, tied to the end of a string, is whirled around in a horizontal circle. (Neglect the force due to gravity). The length of the string is reduced gradually keeping the angular momentum of the stone about the centre of the circle constant. Then, the tension in the string is given by T = Arn where A is a constant, r is the instantaneous radius of the circle and n = ....................... (1993) 6. The ratio of Earth�s orbital angular momentum (about the Sun) to its mass is 4.4 x 1015 m2/s. The area enclosed by Earth�s orbit approximately ........... m2. (1997) 7. A uniform disc of mass m and radius R is rolling up a rough inclined plane which makes an angle of 30° with the horizontal. If the coefficients of static and kinetic friction are each equal to . and the only forces acting are gravitational and frictional, then the magnitude of the frictional force acting on the disc is ........... and its direction is .............. (wire up or down) the inclined plane. (1997) 8. A rod of weight w is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. the knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is ........... and on B is ................. (1997) 9. A symmetric lamina of mass M consists of a square shape with a semicircular section over of the edge of the square as shown in Figure. The side of the square is 2a. The moment of inertia of the lamina about an axis through its centre of mass and perpendicular O2a A B to the plane is 1.6 Ma2. The moment of inertia of the lamina about the tangent AB in the plane of the lamina is .................... (1997) TRUE & FALSE 10. A triangular plate of uniform thickness and density is made to rotate about an axis perpendicular to the plane of the paper and (a) passing through A, (b) passing B, by the application of the same force, F, at C A C B F (midpoint of AB) as shown in the figure. The angular acceleration in both the cases will be the same. (1985) 11. A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity .. Another disc of the same dimensions but of mass M 4 is placed gently on the first disc coaxially. The angular velocity of the system now is 2. 5 . (1986)
ROTATIONALMECHANICS www.physicsashok.in 102 12. A ring of mass 0.3 kg and radius 0.1 m and a solid cylinder of mass 0.4 kg and of the same radius are given the same kinetic energy and released simultaneously on a flat horizontal surface such that they begin to roll as soon as released towards a wall which is at the same distance from the ring and the cylinder. The rolling friction in both cases is negligible. The cylinder will reach the wall first. (1989) PASSAGE PASSAGE � 1 Moment of inertia is a physical term which oppose the change in rotational motion. Moment of inertia depends on distribution of mass, shape of the body as well as distance from the rotational axis. Moment of linearmomentum is called angular momentum. If no external torque act on the system then angular momentum of the system remains conserved. Geometrical meaning of angular momentum relates to the areal velocity. 1. Mass M is distributed over the rod of length L. If linear mass density ( . ) linearly increases with length as . = Kx. The M. . . of the rod about one end perpendicular to rod i.e. (YY�) (A) 2 3 ML (B) 2 12 ML (C) 2 2 3 ML (D) 4 4 KL 2. Four holes of radius R are cut from a thin square plate of side 4R and X Y mass M. The moment of inertia of the remaining portion about z-axis. (A) 2 12 MR . (B) 4 2 3 4 MR . . . . . . . . (C) 4 2 3 6 MR . . . . . . . . (D) 8 10 2 3 6 MR . . . . . . . . 3. A particle of mass m is moving along the line y = 3x + 5 with speed V. The ma
gnitude of angular momentum about origin is (A) 5 12mV (B) 52 mV (C) 12 mV (D) 13 mV 4. A hollow sphere is rolling without slipping total energy during rolling is 60 J, then (A) translational K.E. = 36 J (B) rotational K.E. = 36 J (C) translational K.E. = 24 J (D) rotational K.E. = 24 J 5. Acceleration of block of mass m1 is (given moment of inertia of pulley is . and string does not slip over the pulley). (A) 1 2 1 2 m m g m m . . . . . . . . (B) . . 1 2 1 2 2 m m g m m R . . . . . . . . . . (C) 1 2 2 1 2 m m g R m m . . . . . . . . . . (D) 1 2 2 1 2 2 m m g R m m R. . . . . . . . . . . . . . . . . .
ROTATIONALMECHANICS www.physicsashok.in 103 PASSAGE � 2 Angular velocity is defined as the rate of change of angular displacement. The angular velocity is a measure of the degree of rotation of body. For a rigid body . is constant. All angular variable (such as angular displacement, angular velocity and angular acceleration) are directed along the axis of rotation and perpendicular to plane. Rotating rigid object has two acceleration, one centripetal and other tangential both accelerations being normal to each other. 6. A disc of radius R rolls on a horizontal ground with linear acceleration �a� and angular acceleration . as shown in figure. The magnitude of acceleration of point P at an instant, when its linear velocity is V and angular velocity is . , will be: (A) .a . ra.2 . .r.2 .2 (B) ar R (C) r2a2 . r2.4 (D) r. 7. A disc is rotating clockwise at 20 radian/sec. Its centre has velocity 30 m/s in the forward direction. It has radius of 20 m. Then (A) velocity of topmost point is 430 m/s in forward direction. (B) velocity of lowermost point is 370 m/s in backward direction. (C) velocity of topmost point is 400 m/s (forward). (D) velocity of lower most point is 50 m/s (forward). 8. The topmost and bottom most points have velocities V1 and V2 in the same direction. The radius of sphere is R. Then the correct option are (A) Angular velocity of sphere about cm is 1 2 2 V V clockwise R . . . (B) The Linear velocity at point P is 1 2 3 P 4 V V V . . . (C) The Linear velocity at point Q is 1 2 3 1 3 2 2 V V . . . . . . . . . (D) Velocity of centre of mass is 1 2 2 .V .V . . . . . . 9. Three identical rods, each of mass m and length . are joined to form a rigid equilateral triangle. Its radius of gyration about an axis passing . . . 60° through a corner and perpendicular to the plane of the triangle is
(A) 2l (B) 32 l (C) 2 l (D) l 3 10. A Disc of mass M and radius R is rolling with angular speed . on Y O X M . V a horizontal plane as shown in figure, The magnitude of angular momentum of the disc about O is (A) 2 2 MR . (B) 2 MR . (C) 3 2 2 MR . (D) 2MR2. PASSAGE � 3 In figure 1, the winch is mounted n an axle, and the 6-sided nut is welded ot the winch. By turning the nut with a wrench, a person can rotate the winch. For instance, turning the nut clockwise lifts the block off the ground, because more and more rope gets wrapped around the winch. Three students agree that using a longer wrench makes it easier to turn the winch. But they disagree about why. All three students are talking about the case where the winch is used, over a 10-second time interval, to lift the block one meter off the ground.
ROTATIONAL MECHANICS ROTATIONAL MECHANICS Student : 1 By using a longer wrench, the person decreases the average force he must exert on the wrench, in order to lift the block one meter in ten seconds.
nut winch block Figure : 1 person grips wrench here Figure : 2 Student : 2 Using a longer wrench reduces the work done by the person as he uses the winch to lift the block one meter in ten second. Student : 3 Using a longer wrench reduces the power that the person must exert to lift the block one meter in ten seconds. 11. Student 1 is: (A) Correct, because the torque that the wrench must exert to lift the block doesn�t depend on the wrench�s length (B) Correct, because using a longer wrench decreases the torque it must exert on the winch. (C) Incorrect, because the torque that the wrench must exert to lift the block doesn�t depend on the wrench�s length. (D) Incorrect, because using a longer wrench decreases the torque it must exert on the wrench. 12. Which of the following is true about student 2 and 3? (A) Students 2 and 3 are both correct. (B) Student 2 is correct, but student 3 is incorrect. (C) Student 3 is correct, but student 2 is incorrect. (D) Students 2 and 3 are both incorrect. 13. If several wrenches all apply the same torque to a nut, which graph best expresses the relationship between the force the person must apply to the wrench, and the length of the wrench? (A) length force (B) length force (C) length force (D&&) length force PASSAGE � 4 When a force Of is applied on a block of mass m resting on a horizontal surface then there are two possibilities, either the block moves by translation or it moves by toppling. If the surface is smooth then the block always translates but on a rough surface it topples only when the torque of the applied force Of is greater than the torque of mg about a point in contact with the ground.
When the force Of is applied the body may topple about A or it may translate. F m a A h Answer the following question.
www.physicsashok.in 104
ROTATIONALMECHANICS www.physicsashok.in 105 14. When the block topples about A, the normal force � (A) passes through centre of mass (B) is / zero (C) shifts to the right and passes through right most edge containing A. (D) is zero if the surface is smooth. 15. The block will move in pure translation if � (A) F mga h . (B) 2 F mga h . (C) 2 F mga h . (D) None 16. The block will topple about A if � (A) F mga h . (B) 2 F mga h . (C) 2 F mga h . (D) None 17. If the block be a cube of edge a and . = 0.2 then � (A) The body will translate (B) The body will topple (C) The body may translate or topple (D) None 18. If the block is a cube of edge a and . = 0.6 then� (A) The body will translate (B) The body will topple (C) The body first translates and then topples (D) None PASSAGE � 5 Two discs Aand B are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I respectivelyabout the common axis.DiscAis imparated an initialangular velocity2.usingthe entire potential energy of a spring compressed bya distance x1.DiscBis imparted anangular velocity.bya spring having the same spring constant and compressed bya distance x2. Both the discs rotate in the clockwise direction. 19. The ratio x1/x2 is [JEE, 07] (A) 2 (B) 1/2 (C) 2 (D) 1/ 2 20. When discBis brought in contact with discA, theyacquire a common angular velocityin time t.The average frictional torque on one disc bythe other during this period is [JEE, 07] (A) 2I./(3t) (B) 9I./(2t) (C) 9I./(4t) (D) 3I./(2t) 21. The loss of kinetic energy during the above process is [JEE, 07] (A) I.2/2 (B) I.2/3 (C) I.2/4 (D) I.2/6 LEVEL � 2 1. A uniform solid sphere of mass 1 kg and radius 10 cm is kept stationary on a rough inclined plane by fixing a highly dense particle at B. Inclination of plane is 37° with horizontal and AB is the diameter of the sphere which is parallel to the plane, as shown in figure. Calculate:
ROTATIONALMECHANICS www.physicsashok.in 106 (i) mass of the particle fixed at B, and (ii) minimum required coefficient of friction between sphere and plane to keep sphere in equilibrium. 2. A ball of radius R = 20 cm hasmassm = 0.75 kg andmoment of inertia (about its diameter) . . 0.0125 kg m2 . The ball rolls without sliding over a rough horizontal floor with velocity v0 = 10 ms�1 towards a smooth vertical wall. If coefficient of restitution between the wall and the ball is e = 0.7, calculate velocity v of the ball long after the collision. (g = 10 ms�2). 3. AB is a horizontal diameter of a ball of mass m = 0.4 kg and radius R = 0.10 m. At time t = 0, a sharp impulse is applied at B at angle of 45° with the horizontal, as shown in Figure so that the ball immediately starts to moves with velocity v0 = 10 ms�1. (i) Calculate the impulse If coefficient of kinetic friction between the floor and the ball is . = 0.1, calculate (ii) velocity of ball when it stops sliding. (iii) time t at that instant, (iv) horizontal distance ravelled by the ball upto that instant. (v) angular displacement of the ball about horizontal diameter perpendicular to AB, upto that instant, and (vi) energy lost due to friction. 4. A solid ball of diameter d = 11 cm is rotating about its one of the horizontal diameters with angular velocity 0 . .120 rad sec . It is released from a height so that it falls h = 1.8 m freely and then collides with the horizontal floor. Co-efficient restitution is e . 5 6 and co-efficient of friction between the ball and the ground is . . 0.2 . Calculate fraction of energy lost during collision and the distance between the points where the ball strikes the floor for the first and second time. 5. A heavy plank of mass 102.5 kg is placed over two cylindrical rollers of radii R = 10 cm and r = 5 cm. Mass of rollers is 40 kg and 20 kg respectively. Plank is pulled towards right by applying a horizontal force Of = 25 N as shown in Figure. During first second of motion the plank gets displaced by 10 cm. If plank remains horizontal and slipping doesnot take place, calculate magnitude and direction of force of friction acting between (i) plank and bigger roller. (ii) plank and smaller roller. (iii) bigger roller and floor, and (iv) smaller roller and floor. 6. A wheel of radius R = 10 cm and moment of inertia I = 0.05 kg-m2 is rotating about a fixed horizontal axis O with angular velocity 0 . = 10 rad/sec. A uniform rigid rod of mass m = 3 kg and length l = 50 cm is hinged at one end A such that it can rotate about end A in a vertical plane. End B of the rod is tied with a thread as shown in figure such that the rod is horizontal and is just in contact with the surface of rotating wheel. Horizontal distance between axis of rotation O if cylinder and A is equal to a = 30 cm. If the wheel stops rotating after one second after the thread has burnt, calculate co-efficient of friction
. between the rod and the surface of the wheel. (g = 10 ms�2)
ROTATIONALMECHANICS www.physicsashok.in 107 7. Two heavy and light cylindrical rollers of diameters D and d respectively rest on a horizontal plane as shown in figure. The larger roller has a string wound round it to which a horizontal force P can be D d P applied as shown. Assuming that the coefficient of friction . ahs the same value for all surfaces of contact, determine the limits of . so that the larger roller can be pulled over the smaller one. 8. A block of mass m height 2h and width 2b rests on a flat car which moves horizontally with constant acceleration �a� as shown in figure. Determine: (a) the value of the acceleration at which slipping of the block on the car starts, if the coefficient of friction is µ. (b) the value of the acceleration at which block topples about A, assuming sufficient friction to prevent slipping and (c) the shortest distance in which it can be stopped from a speed of 72 km/hr with constant deceleration so that the block is not disturbed. The following data are given: b = 0.6 m, h = 0.9 m; µ = 0.5 and g = 9.8 m/s2. 9. A block of and M = 4 kg of height �h� and breath �b� is placed on a rough plank of same mass M. A light inextensible string is connected to the upper end of the block and passed through a light smooth pulley as shown in the figure. A mass m = 1 kg is hung to the other end of the string. (a) What should be the minimum value of coefficient of friction between the block and the plank so hat, there is no slipping between the block and the wedge? (b) Find the minimum value of b/h so that the block does not topple over the plank, friction is absent between the plank and the ground. 10. A homogeneous cylinder and a homogeneous sphere of equal massm = 20 kg and equal radii R are connected together by a light frame and are free to roll without slipping down the plane inclined at 30° with the horizontal. Determine the force in the frame. Assume that the bearings are frictionless. Take g = 10 m/s2)
ROTATIONALMECHANICS www.physicsashok.in 108 11. A uniform ring of mass m, radius a and centre C lies at rest on a smooth horizontal table. The plane of the ring is horizontal. A point P on the circumference is struck horizontally and it begins to move in a direction at 60° to PC. If the magnitude of impulse is mv 7 , find the initial speed of point P. 12. A solid cube of wood of side 2a and mass M is resting on a horizontal surface. The cube is constrained to rotate about an axis passing through D and perpendicular to face ABCD. A bullet of mass m and speed v is shot at a height of 43a as shown in figure. The bullet becomes embedded in the cube. Find the minimum value of v r equired to topple the cube. Assume m << M. 13. A semicircular ring of mass m and radius r is released from rest in the position shown with its lower edge resting on a horizontal surface. Find the minimum coefficient of static friction s .. Which is necessary to prevent any initial slipping of the ring. [Hint: centre of mass of a semicircular ring lies at a distance of 2 r . from centre] 14. A spherical ball of radius r and mass collides with a plank of mass M kept on a smooth horizontal surface. Before impact, the centre of the ball has a velocity v0 and angular velocity 0 . as shown. the normal velocity is reversed with same magnitude and the ball stops rotating after the impact. v0 .0 Find the distance on the plank between first two impacts of the ball. The coefficient of friction between the ball and the plank is . . Assume that plank is large enough. 15. A disc of mass M is lying on a frictionless horizontal surface. A small particle of mass m strikes the edge of the disc with a velocity v0 in a tangential direction of disc and gets embedded in it, as shown in the figure. M mv0 Find the angular velocity of the disc just after the collision? 16. A solid hemisphere of mass M = 8 kg and radius R = 5m is supported at the ends of the diameter of the circular cross-section. the supports are frictionless. A particle of mass m = 1 kg is dropped freely from a height h and strikes the end of the diameter of the circular section perpendicular to the diameter passing thought he supports and sticks, as shown in the figure. m M The hemisphere just reaches a position where the flat face of it is vertical. Find h. Also calculate the reactive impulse �J� of the support during collision.
ROTATIONALMECHANICS www.physicsashok.in 109 LEVEL � 3 1. A block of mass M with a semicircular of radius R, rests on a horizontal frictionless surface. A uniform cylinder of radius r and mass m is released from rest at the top point A (see Figure). the cylinder slips on the semicircular frictionless track. How far has the block moved when the cylinder reaches the bottom (point B) of the track? How fast is the block moving when the cylinder reaches the bottom of the track? (JEE, 1983) 2. A homogeneous rod AB of length L = 1.8 m and mass M is pivoted at the centre O in such a way that it can rotate freely in the vertical plane (Figure). The rod is initially in the horizontal position. An insect S of the same mass M falls vertically with speed V on the point C, midway between the points O and B. Immediately after falling, the insect moves towards the end B such that the rod rotates with a constant angular velocity .. (A) Determine the angular velocity . in terms of V and L. (JEE, 1992) (B) If the insect reaches the end B when the rod has turned through an angle of 90°, determine v. 3. A uniform thin rod of mass M and length L is standing vertically along the y-axis on a smooth horizontal surface, with its lower end at the origin (0, 0). A slight disturbance at t = 0 causes the lower end to slip on the smooth surface along the positive x-axis, and the rod starts falling. (JEE, 1993) (A) What is the path followed by the centre of mass of the rod during its fall? (B) Find the equation to the trajectory of a point on the rod located at a distance r from the lower end. What is the shape of the path of this point? 4. A block X of mass 0.5 kg is held by a long massless string on a frictionless inclined plane of inclination 30° to the horizontal. The string is wound on a uniform solid cylindrical drum Y of mass 2 kg and of radius 0.2 m as shown in Figure. The drum is given an initial angular velocity such that the block X starts moving up the plane. (A) Find the tension in the string during the motion. (JEE, 1994) (B) At a certain instant of time the magnitude of the angular velocity of Y is 10 rad s�1 calculate the distance travelled by X from that instant of time until it comes to rest. 5. Two uniform thin rods A and B of length 0.6 m each and of masses 0.01 kg and 0.02 kg respectively are rigidly joined end to end. The combination is pivoted at the lighter end, P as shown in figure. Such that it can freely rotate PAB about point P in a vertical plane. A small object of mass 0.05 kg, moving horizontally, hits the lower end of the combination and sticks to it. (JEE, 1994) What should be the velocity of the object so that the system could just be raised to the horizontal position. 6. A wedge of mass m and triangular cross-section (AB = BC = CA = 2R) is moving with a constant velocity .vi� towards a sphere of radius R fixed on a smooth horizontal table as shown in Figu
re. The wedge makes an elastic collision with the fixed sphere and returns along the same path without any rotation. Neglect all friction and suppose that the wedge remains in contact with the sphere for a very short time. .t , during which the sphere exerts a constant force F on the wedge. (JEE, 1998)
ROTATIONALMECHANICS www.physicsashok.in 110 (A) Find the force F and also the normal force N exerted by the table on the wedge during the time .t . (B) Let h denote the perpendicular distance between the centre of mass of the wedge and the line of action of F. Find the magnitude of the torque due to the normal force N about the centre of the wedge, during the interval .t . 7. A uniform circular disc has radius R and mass m. A particle also of mass m, is fixed at a point A on the edge of the disc as shown in Figure. The disc can rotate freely about a fixed horizontal chord PQ that is at a distance R 4 from the centre C of the disc. The line AC is perpendicular to PQ. Initially, the disc is held vertical with the point A at its highest position. It is then allowed to fall so that it starts rotating about PQ. Find the linear speed of the particle as it reaches its lowest position. (JEE, 1998) 8. A man pushes a cylinder of mass m1 with the help of a plank of mass m2 as shown in Figure. There in no slipping at any contact. F M1 m2 The horizontal component of the force applied by the man is F. Find (A) the accelerations of the plank and the center of mass of the cylinder, and (B) the magnitudes and directions of frictional forces at contact points. (JEE, 1999) 9. A small ball of mass 2 x 10�3 kg having a charge of 1.C is suspended by a string of length 0.8 m. Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity which should be imparted to the lower ball so that it can make complete revolution. (JEE, 2001) 10. Three particles A, B and C, each of mass m, are connected to each other by three massless rigid rods to form a rigid, equilateral triangular body of side . . This body is placed on a horizontal frictionless table (x-y plane) and is hinged to it at the point A so that it can move without friction about the vertical axis through A (see figure). The body is set into rotational motion on the table about A with a constant angular velocity .. (JEE, 2002) (A) Find the magnitude of the horizontal force exerted by the hinge on the body. (B) At time T, when the side BC is parallel to the x-axis, a force F is applied on B along BC (as shown). Obtain the x-component and the y-component of the force exerted by the hinge on the body, immediately after time T. 11. A wooden log of mass M and length L is hinged by a frictionless nail at O. A bullet of mass m strikes with velocity v and sticks to it. Find angular velocity of the system immediately after the collision about O. (JEE, 2005) 12. A cylinder of massm and radius R rolls down an inclined plane of inclination . . Calculate the linear acceleration of the axis of cylinder. (JEE, 2005)
ROTATIONALMECHANICS www.physicsashok.in 111 13. Auniformcircular disc has radius Randmassm.Aparticle also ofmassmis fixed at a point A on thewedge of the disc as in fig. The disc can rotate freely about a fixed horizontal chord PQthat is at a distanceR/4 fromthe centreCof the disc.The lineAC is perpendicular to PQ. Initiallythe disc is held verticalwith the pointAat its highest R/4 CA position. It is then allowed to fall so that it starts rotating about PQ. Find the linear P Q speed of the particle at it reaches its lowest position. [JEE, 98] 14. Aman pushes a cylinder ofmassm1with the help of a plank ofmassm2 as shown. There is no slipping at anycontact.The horizontal component ofthe force applied by the man is F. Find [JEE, 99] F m2m1 (a) the accelerations of the plank and the centre ofmass of the cylinder, and (b) themagnitudes and directions of frictional forces at contact points. 15. ArodABofmassMand lengthLis lying on a horizontal frictionless surface.Aparticle ofmassmtravelling along the surface hits the end �A�of the rodwitha velocityv0 inthe direction perpendicular toAB.The collision is completely elastic.After the collision the particle comes to rest. (a) Find the ratio m/M. (b) Apoint P on the rod is at rest immediatelyafter the collision. Find the distanceAP. (c) Find the linear speed of the point P at a time .L/(3v0) after the collision. [JEE, 2000] 16. Two heavymetallic plates are joined together at 90º to each other.Alaminar sheet ofmass 30Kg is hinged at the lineAB joining the two heavymetallic plates. The hinges are frictionless. The moment of inertia of the laminar sheet about an axis parallel toAB and passing through its centre ofmas is 1.2Kg�m2. Two rubber obstacles P andQare fixed, one on eachmetallic plate at a distance 0.5mfromthe lineAB. This distance is chosen so that the reaction due to the hinges onthe laminar velocity1 rad/s and turns back. If the impulse onthe sheet due to each obstacle is 6 N�s. [JEE, 01] (a) Find the location of the centre ofmass of the laminar sheet fromAB. A Q P B (b) At what angular velocitydoes the liminar sheet come back after the first impact ? (c) After howmany impacts, does the laminar sheet come to rest ? 17. Two identicalladders, eachofmassMand lengthLare resting onthe rough horizontal surface as shown in the figure.Ablock ofmassmhangs fromP. If the systemis in equilibrium, find themagnitude and the directionof frictional force atAandB. [JEE, 2005] mP A B
v 18. There is a rectangular plate ofmassMkg of dimensions (a × b). The plate is held inhorizontalpositionbystriking n smallballs eachofmassmper unit area per unit time. These are striking in the shaded half region of the velocity v.What is v ? [JEE, 06] b a It is given n = 100, M= 3 kg, m= 0.01 kg; b = 2 m; a = 1m; g = 10 m/s2.
ROTATIONALMECHANICS www.physicsashok.in 112 ASSERTION-REASON Q. 1 2 3 4 5 6 Ans. A C A D B D MATCH THE COLUMN 13. A.Q, B.P, C.S, D.T 14. A.P, B.P, C.P, D.P 15. A.R, B.P, C.S, D.Q 16. A.S, B.R, C.Q, D.P 17. A.Q, B.P, C.S, D.R LEVEL � 1 Q. 1 2 3 4 5 6 7 8 9 10 Ans. C B A B B B C A C C Q. 11 12 13 14 15 16 17 18 19 20 Ans. A B A B B A A B C D Q. 21 22 23 24 25 26 27 28 29 30 Ans. B ABD BD BC BC AB AC D ACD BD Q. 31 32 33 34 35 36 37 38 39 40 Ans. A ABC A A B C C D B B Q. 41 42 43 44 45 46 47 48 49 50 Ans. C C B ABC ABD BD ABCD BD CD BC Q. 51 52 53 54 55 56 57 58 59 60 Ans. ABC AD BCD ABCD ABC D C A A A Q. 61 62 63 64 65 66 67 68 69 70 Ans. ABC C C C D A B B A CD Q. 71 72 Ans. AB D FILL IN THE BLANKS AND TRUE-FALSE 1. 23 mg 2. Angular momentum 3. 0 6 M M m . . 4. 5 2 3MRA. 5. �3 6. 6.94 x 1022 7. 6 mg , direction upward 8. , d x W xW d d . . . . . . . 9. 4.8 Ma2 10. F 11. F 12. F
ROTATIONALMECHANICS www.physicsashok.in 113 PASSAGE Q. 1 2 3 4 5 6 7 8 9 Ans. D D A AD B A AB ABCD C Q. 10 11 12 13 14 15 16 17 18 Ans. C A D D C B C A B Q. 19 20 21 Ans. C A B LEVEL � 2 1. (i) 3 kg (ii) 0.75 2. 2 ms.1 3. (i) 4 2kg ms.1 (ii) Zero (iii) 10 second (iv) 50 (Leftward) (v) 1250 radians (clockwise) (vi) 70 joule 4. 0.432, 2.2 m 5. (i) 3N (ii) 1.50 N (iii) 1.00 N (iv) 0.50 N 6. 0.2 7. dD . . 8. (a) 4.9 m/s2 (b) 6.53 m/s2 (c) 40.82 m 9. (a) 19 (b) 49 10. 3.45 N 11. 4 v 12. . . 1 2 M 2ag 2 1 m . . . . . 13. 0.398 14. 0 0 45 M m v r M g. . . . . . . . 15. 0 2 3 mv M . m 16. h = 42 m, J = 30 kg m/s. LEVEL � 3 1. . . . . . . 2 , m R r g R r m M m M m M . . . . 2. (a) 12 7VL (b) 3.5 ms�1 3. (i) Circular path of radius 2L (ii) 2 2 x y 1 L R r . . . . . . . . . . . . . . . 4. (i) 1.63 N, (ii) 1.22 m 5. 6.3 m/s 6. (a) . . 2 3� � , 2 � 3 3 mv i k mv mg k
t t . . . . . . . . . . , (b) 43mv h t . . 7. 5gR 8. (a) R. , 2R. (b) 1 1 1 2 1 2 3 , 3 8 3 8 FM FM M . m M . m 9. 5.86 m/s 10. (a) 3m..2 (b) . . net x 4F . F , . . 3 2 net y F . m.. 11. . . 3mv 3m M L . . . 12. axies a 2g sin 3 . . 13. v . 5gR 14. c . . 1 2 a 4F , 3m 8m . . p . . 1 2 a 8F , 3m 8m . . . . 1 1 1 2 f 3m F , 3m 8m . . . . 1 2 1 2 f m F 3m 8m . . 15. (a) m 1 M 4 . ; (b) x 2L 3 . ; (c) 0 v
2 2 16. (a) l = 0.1 m; (b) .´ = 1 rad/s ; (c) laminar sheet will never come to rest 17. f .M m. g cot 2 . . . 18. 10 m/s ��� X ��� X ��� X ���
S.H.M DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125
SIMPLE HARMONIC MOTION www.physicsashok.in 1 PERIODIC MOTION If a particle moves along X-axis, its position depends upon time t.We express this fact mathematically by writing x = f(t) or x(t) There are certainmotions that are repeated at equalinterval of time. Bythiswemean that particle is found at the same positionmoving in the same directionwith the same velocity and acceleration, after each period of time. Let T be the interval of time inwhichmotion is repeated. Then x(t) = x(t + T) The Tis theminimumchange it t. The function that repeats itself is known as a periodic function.During the period, its valuesmay remain finite. Such functions are bound functions. Periodicmotion of a particle is also bound functions. Periodicmotion ofa particle is also bound because itmust not go to infinityand retun back in one finite period. (That willmean infinite average speed). Periodicmotionsmay be oscillatory ornon oscillatory. Uniformcircularmotion, themotion of a planet around the sun, etc are periodic but not oscillatory.Also, an oscillatorymotionmaynot repeat its positionwith the old velocitydue to friction andwill be non-periodic. C1 : The position x of a particle is plottedwith time t as in the figure. Is themotion periodic ? O A B C t x Sol. Yes. The portionOAis repeated asAB, then as BC and so on. Hence themotion is periodic. C2 : Is the followingmotion periodic ? Is it oscillatory ? t x O A B Sol. Yes. The portionOAis repeated asAB, and so on.Themotion is oscillatoryalso because the velocitychanges its sign(direction) periodically. Period : Theminimumtime interval iswhich a particle repeats itsmotion is called the time period of themotion. The expression of �period� depends upon the type ofmotion. C3 : Aparticlemoves uniformly in a circle.Determine the time period of themotion. Sol. Let Vbe the speed and r be the radius of the circle. Then time taken to complete one revolution is the time period, T.Hence T 2 r V. . r V If . be the angular speed of the particle, its radial line turns by 2. radian in one time period. Hence T 2.
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SIMPLE HARMONIC MOTION www.physicsashok.in 2 C4 :Aballis dropped froma height h and falls under uniformacceleration g. It hits the ground elastically. Its velocity is reversed due to collision; the ball reaches the same starting position. Calculate the periodic time. Sol. Here time t of descent is h = 1/2gt2 or t = 2h g h This is also the time of ascend.Hence total time, which equals the periodic time T of thismotion, is 2t. Thus T 2t 2 2h g . . C5 : The position x of a particle depends upon as x = 4 (mm) sin4(s�1)t. Determine its time period. Sol. We knowthat sin . = sin (. + 2.) Hence sin 4t = sin (4t + 2.) We can expressRHS in the same formas LHS as sin 4t = sin4 t 24. .. . . . . . Comparing thiswithx(t) = x(t + T) We see that T 24. . Thus the period of the givenmotion is 24. s. Note : You may learn that period of sin.t or cos .t is 2. . . C6 : Determine the time period of x(t)motion. t (s) x(m) O 2 4 6 8 10 Sol. Here the repetition occurs at every 4s of interval. Hence T = 4 s. Frequency : Anyperiodicmotionthat repeats itselfat regular time intervaland continuesfor longtime is assigned a �frequency� of repetition ofmotion.We saythat one repetitionmeans occurrence ofmotion once.The number of times the motion is repeated per unit time is known as the frequencyof themotion. IfTbe the periodic time of amotion, the number of repetitions in unit time is 1/T. Hence frequency= 1/T. Frequency is denoted by n, f or v.Then v = 1/T The S.I. unit of frequency isHertz, denoted byHz. 1Hz = 1 cycle per second (1 cps)
SIMPLE HARMONIC MOTION www.physicsashok.in 3 Frequencyis a feature of anyperiodicmotion, not onlyofoscillations. Frequencyis also called �linear frequency�. Angular frequency is defined as 2. time the linear frequency. It is denoted by ..(unit : s�1) : . = 2 .v (Note that ..here is not angular velocity) C7 :Aball is dropped fromheight h on anelastic floor.Determine the frequencyof the periodicmotion executed by the ball. Sol. We have seen that periodic time of thismotion is T = 2 2h g Hence frequencyis given by v = 1T = 1 g 2 2h . Displacement as a function of time : Let the initial position of a particle be x0 and position at time t be x. Then its displacement relative to x0 is x � x0. This depends in somemanner on time t. we say that displacement varieswith time, or is a function of time t.We denote the function byf(t). Thus x � x0 = f(t) Here the expression for function depends upon type ofmotion. If themotion iswith uniformvelocity,we have x � x0 =Vt (linear function) Herewe saythat displacement x � x0 is proportional to t. Positionx = x0 +Vt is said to be a linear function of t, as power of t is 1. The graph of x(t) is a straight line. If themotion is along x-axiswithuniformacceleration the displacement is given by x � x0 = ut + 1/2at2 Herewe have quadratic function of t, as the highest power of t is 2. The graph of x(t) is a parabola. If acceleration is not uniform,we have infinite number ofways inwhich acceleration can change.Of these, one special case is veryimportant-arising froma periodic function of time. Periodic Function : Afunction that repeats itself at equal interval is called a periodic function. There are severalperiodic functions.All of themcan be symbolicallyexpressed as x(t) = x(t + T) where theminimumvalue ofTis known as t-period of the function x. C8 :What is the t-period of x =Asin (.t + .) Sol. It is knownfromtrigonometrythat sin . = sin(. + 2.)
SIMPLE HARMONIC MOTION www.physicsashok.in 4 Hence x =Asin(.t + ..+ 2.)Asin 2 t . . .. . ... . . . .. . . . . . x =Asin (.t´ + .), where t´ = t + 2. . . This shows that the function at t coincideswith the function at t´. The interval t´ � t is the period in t. This period is 2. . . C9 :What is the time period of x =Acos (.t´ + .) ? Sol. Fromtrigonometry, it is known that cos . = cos (. + 2.) Here x =Acos (.t + . + 2.) Acos 2 t . . .. . ... . . . .. . . . . . =Acos (.t´ + .), where t´ = t + 2./.. Thus, the function at t´ coincideswith that at t.The period is, thus, 2./.. C10 : Determine the period T of the periodic function x=Asin2.t. Sol. Using cos2..= 1 � 2 sin2 ., sin2. = 1 cos 2 2 . . The .�period of cos 2..is 2./2 i.e. .. Hence sin2.t has a similar treatment. . T = ./. NOTE : 1. Period is not changed by multiplying or dividing by (or by adding) a constant : If 0 x(t) x(t T) x(t) x m. . . has the same period T. 2. Period is reduced . times of t be multiplied by . : If x(t) = x(t + T) then x(.t) will have a period T. . 3. According to Fourier, the sum of periodic functions is also periodic. OSCILLATION If you displace a particle fromits equilibriumpositioninsuch awaythat its potential energyis increased, a force ariseswhich tries to bring the particle to the positionof lowest potential energy. If you release it, this forcewill bring it to the equilibriumposition. But itwillnot stop!Why? It has gained velocityand due to inertia ofmotion, it willgo on the other side. The restoring forcewill again be bringing to back to the equilibriumposition. Inertia ofmotionwillagainbe bringing it back to the equilibriumposition. Inertia ofmotionwillagain shift it to the other
side. This to and fromotionwillcontinue until the energyis lost in friction.
SIMPLE HARMONIC MOTION www.physicsashok.in 5 To and fromotion is called oscillation. It may be periodic or non-periodic. If an angle is used to describe oscillation, the oscillation is called angular. Iflinear quantities are used to describe oscillation, it is called linear oscillation. C11 :Ablock suspended froma spring performs oscillation. x C12 : Ablockmoving to and fro along smooth surface between two strings. x C13 : Afunction f(t) is said to be periodic of time period Tif f(t + T) = f(t) All sine or cosine functions of time are periodic. Thus, Y=Asin .t or Acos .t is periodic, of time period T = 2. . . C14. Find the period of the function, y = sin .t + sin 2.t + sin 3.t Sol. The given function can bewritten as, y = y1 + y2 + y3 Here y1 = sin .t, T1 = 2./. y2 = sin 2.t, T2 = 22. . . . . and y3 = sin 3.t, T3 = 2./3. . T1 = 2T2 and T1 = 3T3 So, the time period of the given function is T1 or 2./. Because in timeT= 2./., first function completes one oscillation, the second function two oscillations and the third, three. C15:What is the period ofmotion shown byposition-time graph ? 1 3 5 2 4 6 O t x Sol. Themotion is periodic, with period of 2s. C16: (a) Is themotion x = sin .t � cos .t periodic? simple harmonic ? (b) What type ofmotion is represented byx = sin3.t � periodic, non-periodic or simple harmonic ?
SIMPLE HARMONIC MOTION www.physicsashok.in 6 Sol. (a)We can always express x = a sin .t + b cos .t inthe formx = c sin (.t + .) where a = c cos ., b = c sin .. Hence themotion is simple harmonic and periodic. (b) sin 3.= sin(2.+ .) = sin2. cos. + cos 2. sin . = 2sin. cos2..+ (cos2. � sin2.)sin . = 2sin..(1 � sin2.) + (1 � sin2.) sin. = 3 sin. � 4 sin3. sin3 1 sin 3 3 sin 4 4 . . . . . . x 1 sin 3 t 3 sin t 4 4 . . . . This is a periodic function. v dx 1 cos3 t 3 cos t dt 4 4 . . . . . . 2 dv 9 3 2 a sin 3 t sin t dt 4 4 . . . . . . . . . �.2xThus, themotion is not SHM. SIMPLE HARMONIC MOTION : If the positionof a particle changes,we saythat it ismoving. then x= f(t). if the position changes periodically are call themotion �harmonic� or �periodic�. In studying vibration of harmoniumblades (musical instrument), sine and cosine functionswere used.We call sine and cosine functions a harmonic function.Themost simple function is linear, like �sin .�, �cos .�or asin. + bcos .. If the displacement of a particle froma fixed point be expressed using simple formof harmonic functions, we have x � x0 = a sin .t + b cos .t By using trigonometrywe canwrite this either as x � x0 =Asin(.t + .) or asAcos (.t + .). Hence themotion represented using simplest harmonic functionsmay be expressed as x � x0 =Asin(.t + .) Here x0 is a fixed co-ordinate andmaybe set equalto zero. Ifa particlemoves insuchawaythat its displacement relative to a fixed point is given by x =Asin(.t + .) themotion is called �simple harmonic� The velocityof the particle is given by v = dx dt = .A cos(.t + .)
SIMPLE HARMONIC MOTION www.physicsashok.in 7 The rate of change of velocityis acceleration �a�. Thus a = dV dt = �.2Asin (.t + .) = �.2x The negative sign shows that acceleration is directed opposite to displacement. Ifaparticlemoves insuchawaythat its accelerationis always directed towards a fixed point andis directlyproportional to themagnitude of displacement relative to the fixed point, the motion is called simple harmonic. a OP O P Fixed point Particle X In the figure a particle P is moving along X-axis. There is a fixed point O fromwhich the displacement is measured. OP .... is the displacement. If a. be the acceleration of the particle then, for SHM, (i) | a | . . OP . .... (ii) | a | .OP . .... a. = � (positive const.) OP .... Equation of SHM : Let x = 0 lies at �O� and �P� be at x. Then OP . (x . 0) �i . x �i .... x a . a �i . . 2 x a i� . .. x �i where, .2 = positive constant for real .. ax = �.2x (component form) If �O� lies at x0, thenax = �.2(x � x0) This is the kinematical formof equation ofSHM. The above formulation is valid in all frames of reference.However, its dynamical formusing force depends on reference frame. Dynamical Form :Multiplying ax = �.2x by the massmof the particle, max = �m.2x If the frame be inertial, Newton�s second lawofmotion is valid. Thenmax= Fx Then Fx = �m.2x = �kx where, k = m.2 Thus, SHMin an inertial frame is themotion inwhich the force acting onthe particle is directed opposite to the displacement froma fixed point and is proportional inmagnitude to the displacement fromthe fixed point. If the frame be non-inertial, then Fx will be the x-component of both real and pseudo forces.
SIMPLE HARMONIC MOTION www.physicsashok.in 8 Solving the differential equation of SHM : We know that ax = 2 2 d x dt , hence 2 2 d x dt = �.2x (Diff. eqn. of SHM) This is a differential equation in x. Let dx dt = Vx, Then 2 x x 2 x d x dV dx dV V dt dx dt dx . . (Chain rule) x 2 x V dV x dx . .. VxdVx = �.2x dx (variables separated) During oscillation, the particle returns at extreme points x = ±A(At x = ±A, the velocity becomes zero for a moment.) Then Vx x 2 x x 0 A V dV xdx . . . .. . 2 Vx 2 x x 2 0 A V x 2 2 . . . . . . . . .. . . . . . . Vx2 = �.2[x2 � A2] Vx2 = .2(A2 � x2) 2 2 x V . .. A . x ...(1) Equation (1) shows that � (i) Particle has opposite velocities for the same position. This is possible at different moments of time.
(ii) Particle has equalvelocities at two positions at equaldistance frommeanposition.Themaximumspeed occurs at x = 0, and is .A. (iii) Equation (1) can be rearranged inthe form 2 2 x 2 2 2 V x 1 A A . . . This shows that velocity-position graph is an ellipse. This is also true formomentum-position graph. The momentum-position graph is known as phase-space graph of the particle. Let 2 2 x V . . A . x at time t. Then dx A2 x2 dt . . . A � A �A A x Vx Separating the variables and integrating,we have
SIMPLE HARMONIC MOTION www.physicsashok.in 9 2 2 dx dt A x . . . . . sin 1 x t A . . . . . where, ..= constant of integration. xA = sin(.t + .) x =Asin(.t + .) ...(2) . is decided by the position x at a given time t. PHASE The general equation of SHMalongX-axismaybewritten as x =Asin(.t + .) Here (.t + .) contains all the informations ofdynamical state of oscillator. It indicateswhat portionofmotion is complete bythe time t and is called �phase� ofmotion. Characteristics ofSHMare the following : (i) Amplitude : themaximumdistance of the particle fromitsmean position is known as amplitude. Fromthe equation | x |max = A Thus, A= amplitude of SHM. (ii) Phase angle or phase : The portion ofmotion that is completed by a given time is represented by�phase� or �phase angle�. (.t + .) is �phase� at time t in the expression xA . . .. .. = sin(.t + .) (iii) Phase constant or epoch or initial phase : The phase at t = 0 is called �initial phase�, �phase constant�, or �epoch�. In the expression above, . = phase constant = epoch (iv) Angular frequency : 2. times frequency is called angular frequency. In the above expression, . is angular frequency:. = 2T. = 2.v S.I. Unit of ..is s�1. Special case (i) If the particle is at x = 0 at t = 0 Then 0 = . × 0 + . . = 0 . x =Asin .t (ii) If at t = 0, the particle is at rest (extreme), x = ±A. 0 = . × 0 + .
SIMPLE HARMONIC MOTION www.physicsashok.in 10 . = 0 . x =Asin .t For x = A For x = �A AA = sin(. × 0 + .) A A . = sin(. × 0 + .) 1 = sin . �1 = sin . . = sin�1 1 . = sin�1(�1) . = ./2 . = ./2 x =Acos .t x = �A cos .t Example 1: If x =A/2 at t = 0, find phase constant . in x =Asin(.t + .).At t = 0, a particle executing SHMis going along x-axis. Sol. Here A2 =Asin(. × 0 + .) A 2A = sin . or sin ..= 12 . = 30º or 5./6 Now we use the other condition : Vx < 0 at t = 0 We have Vx = .Acos(.t + .) Putting t = 0 and . = ./6, Vx = .A cos(..× 0 + ./6) x V 3 A 2 . . ,which is positive. Now . = 5./6 Vx = .A cos 56. = � 3 2 .A which is negative and satisfies the given condition. Hence the required .= 56. Example 2:Asmallmass executes linear SHMabout Owith amplitude a and period T. Its displacement fromO at time T/8 after passing throughOis : (A) a/8 (B) a / 2 2 (C) a/2 (D) a / 2 Sol: x =Asin .t x Asin 2 T T 8 . . . or x Asin A 4 2
.
. . Hence option (D) is correct.
SIMPLE HARMONIC MOTION www.physicsashok.in 11 Example 3: The figure shows the displacement-time graph of a particle executing SHM. If the time period ofoscillation is 2s, then the equation ofmotionis given by x = _______. t(s) 0 5 10 (mm) x Sol. In the case of SHM, x =Asin (.t + .) or 5 = 10 sin (. × 0 + .) (Here A= 10 mm= 0.01 m) . sin 12 . . 6. . . Here 2 2 T 2 . . . . . . . . x = 0.01 sin t 6. .. . . . . . . where xis inmetre. Example 4: Time period of a particle executing SHMis 8 sec.At t = 0 it is at themean position. The ratio of the distance covered by the particle in the 1st second to the 2nd second is : (A) 1 2 .1 (B) 2 (C) 12 (D) 2 .1 Sol: Here x =Asin .t x1 =Asin 28. × 1 x1 =Asin A 4 2 . . and 2 x Asin 2 2 A 8. . . . . The distance traveled in 2nd second is 2 2 1 x´ x x A A2 . . . . ( 2 1)A
2 . . . 12 Ratio x A/ 2 1 x´ ( 2 1)A ( 2 1) 2 . . . . . = 2 1 ( 2 1) ( 2 1)( 2 1) 2 1 . . . . . . = ( 2 .1) Hence Option (D) is correct.
SIMPLE HARMONIC MOTION www.physicsashok.in 12 C17: Aparticle executing SHMhas anamplitude of 1metre and time period 2 sec.What are the velocityand the accelerationwhen the displacement is 0.5metre ? Sol. 2 3.14 rad / sec. T. . . . Acceleration, f = �.2y = �(3.14)2 × 0.5 = �4.93 m/sec2. Velocity, V . . a2 . y2 . 3.14 (1)2 . (0.5)2 . 2.72 m/ sec. C18: In the above question, calculate themaximumvelocityandmaximumacceleration. Sol. Vmax = .a = 3.14 × 1 = 3.14 m/sec fmax = �.2a = �(3.14)2 × 1 = � 9.86 m/sec2. C19: Aparticle executing SHMof amplitude 10 cmhas speed 3.14 cm/sec. at the mean position; calculate its time period. Sol. At themean position, V a 2 a T. . . . . T 2 a 20 sec. V. . . C20: Aparticle is executing SHMoftime period 10 sec., and amplitude 9 cm.What would be the velocityof the particle 2.5 sec. after it passes through themean position ? Sol. V = a . cos .t., here t ismeasured fromthemean position. V a 2 cos 2 t T T . . . . . . . . . V 9 2 cos 2 2.5 10 10 . . . . . . . . . . . V 9 2 cos 0 10 2 . . . . . C21: Abody executing SHMhas its velocity10 cm/sec. and 7 cm/sec. when its displacement fromthe mean position are 3 cmand 4 cmrespectively. Calculate the length of the path. Sol. We have V . . a2 . y2 . 2 222 2 2 1 1 V a y V a y . . . or 22 49 a 16
100 a 9 . . . . a = 4.76 cm. Length of the path = 2a = 9.5 cm Example 5: Aparticle executes SHM on a straight line path. The amplitude of oscillation is 2 cm.When the displacement of the particle fromthemeanposition is 1 cm, thenumericalvalue ofmagnitude of acceleration is equal to the numericalvalue ofmagnitude of velocity. The frequencyof SHM(in second�1) is : (A) 2. 3 (B) 23. (C) 3 2. (D) 1 2. 3
SIMPLE HARMONIC MOTION www.physicsashok.in 13 Sol: . A2 . x2 . .2x .2 A2 . x2 . .4x2 or .2A2 ..2x2 . .4x2 or A2 � x2 = .2x2 or 22 � 12 = .2 × 12 or 3 = .2. . 3 or 2 3 T. . . T 23. . . f 1 3 T 2 . . . Hence option (C) is correct. Example 6:Astone is swinging in a horizontal circle 0.8min diameter at 30 rev/min.Adistance horizontal light beamcauses a shadow of the stone to be formed on a nearly verticalwall. The amplitude and period of the simple harmonicmotion for the shadowof the stone are (A) 0.4 m, 4 s (B) 0.2 m, 2 s (C) 0.4 m, 2 s (D) 0.8 m, 2 s Sol: A d 0.8 0.4m 2 2 . . . 30 2 60 . . . . . . or 2T. . . . T = 2 sec. Hence option (C) is correct. Example 7:Abodyundergoing SHMabout the origin has its equation is given byx =0.2 cos 5.t. Find its average speed fromt = 0 to t = 0.7 sec. Sol. v = Total distance Total time Here x = 0.2 cos .t Here . = 5. or 2 5 T. . . , . T 2 0.4 sec. 5 . .
Here 0 t nT t 4 . . t = n × 0.1 + t0 or 0.7 = n × 0.1 + t0 Fromthis, n = 7 and t0 = 0 But distance travelled inT/4 isA= 0.2 unit . Total distance = 0.2 × 7 = 1.4 unit. . v 1.4 2 unit 0.7 . .
SIMPLE HARMONIC MOTION www.physicsashok.in 14 Example 8:Aparticle perfroms SHMwith a period Tand amplitude a. Themean velocity of the particle over the time intervalduringwhich it travels a distance a/2 fromthe extreme position is (A) a/T (B) 2a/T (C) 3a/T (D) a/2T Sol: x = a cos .t a/2 = a cos .t0 or 0 cos cos t 3. . . or 0 t 3. . . . 0 t 3. . . v dx a sin t dt . . . . . . /3 0 /3 0 vdt v dt . . . . . .. But 2T. . . . v 3a T . Hence option (C) is correct. Example 9: The angular frequencyofmotionwhose equation is 2 2 d y 4 dt + 9y= 0 is (y= displacement and t = time) (A) 94 (B) 49 (C) 32 (D) 23 Sol: 2
2 d y 4 9y 0 dt . . or 22 d y 9 y dt 4 . . Comparingwith SHMequation 2 2 2 d y y dt . .. . 2 94 . . . 32 . . Hence option (C) is correct. Example 10: The acceleration-displacement (a � x) graph of a particle executing simple harmonicmotion is shown inthe figure. Find the frequencyof oscillation. a O x � � Sol. In SHM, . = �.2x . = �.2(�.) . . = .2. ...(i) (fromgraph) and . . . . or 2 f . . . . or f = 1 2 . . .
SIMPLE HARMONIC MOTION www.physicsashok.in 15 Example 11:Agraph of the square of the velocityagainst the square of the acceleration of a givensimple harmonic motionis : (A) v2 a2 (B) v2 a2 (C) v2 a2 (D) v2 a2 Sol: x =Asin .t v dx A cos t A2 x2 dt . . . . . . . a dv A 2 sin t dt . . . . . a dv 2x dt . . .. But 2 x . . a . . 2 2 4 v . . A . a. or 2 2 2 2 4 v A a . . . . . . . . . . Hence option (D) is correct. C22: Aparticle executing SHMhas its acceleration represented by the equation f = � 4.2y. Calculate its time period. Sol. f = � 4.2y = �.2y where, .2 = 4.2 . T 2 2 1 sec 2 . . . . . . . Example 12:Ablock is kept on a horizontal table. The table is undergoing simple harmonicmotion of frequency 3Hz ina horizontal plane. The coefficient of static frictionbetween block and the table surface is 0.72. Find the maximumamplitude ofthe table at which the block does not slip on the surface. Sol. amax = .2A
fmax = µSmg or mamax = µSmg or m.2A=µSmg . S2 2 µ g 0.72 10 A (2 f ) . . . . . 2 2 A 0.72 g 4 3 . . . . [. .2 = g] A = 0.02 m= 2 cm Example 13:Aplankwith a smallblock on top of its is under going verticalSHM. Its period is 2 sec.Theminimum amplitude at which the blockwill separate fromplank is : (A) 2 10 . (B) 2 10 . (C) 2 20 . (D) 10 .
SIMPLE HARMONIC MOTION www.physicsashok.in 16 Sol: Here g =A.2 . 2 2 2 2 A g g gT 2 4 T . . . . . .. . .. .. 2 2 A 10 4 10 m 4 . . . . . Hence (A) option is correct. Example 14: Aforce f = �10x + 2 acts on a particle of mass 0.1 kg, where k is in m and F in newton. If it is released fromrest at x = �2m, find : (a) amplitude; (b) time period ; (c) equationofmotion. Sol. f = �10x + 2 df 10 dx dt dt . . .f = �10 .x ma = � 10 .x . a 10 x 0.1 . . . or �.2.x = �100 .x or . = 10 rad/s (b) . T 2 2 sec. 10 5 . . . . . . . Let at x = x0,mean position is situated. . f = �10x + 2 0 = � 10x0 + 2 . 0 x 2 0.2 m 10 . . . f = �10(.x + 0.2) + 2 or f = �10.x � 2 + 2 ma = �10 .x or a 10 x 0.1 . . . or a = � 100 .x . .x = Asin (.t + .) or .x =Asin(10t + .) At t = 0, .x = � 2.2 metre . �2.2 =Asin .
Also, v . . A2 . .x2 or 0 . 10 A2 . (2.2)2 . A= 2.2 m . �2.2 = 2.2 sin .
SIMPLE HARMONIC MOTION www.physicsashok.in 17 . 2. . . . . x 2.2sin 10t 2 . .. . . . . . . . .x = �2.2 cos 10 t The displacement fromoriginal is x = � 2.2 cos 10 t + x0 x = 0.2 � 2.2 cos 10 t . x = 0.2 � 11 5 cos 10 t Example 15:Aparticle ofmass 4 kgmoves betweentwo pointsAand B on a smooth horizontal surface under the action of two forces such that when it is at a point P, the forces are 2PA N .. and 2PB N .. . If the particle is released fromrest at A, find the time it takes to travel a quarter of theway fromA to B. Sol. . F = 2(l � x) � 2x F = 2l � 2x � 2x = 2l � 4x or a 2 4x x 4 2 . . l . l . A B 2x P 2(l � x) l x or v v 0 0 vdv x dx 2 . . . . . . .. .. l or 2 2 x0 v x x 2 2 2 . . . . . . . . l or v2 x2 x 2 2 2 . . l or v2 = lx � x2 or v . lx . x2 or dx x x2 dt . l .
or t0 0 2 0 dx dt x x . . . . l l . 0 t second 3. . How to solve problems (Force method) : Step 1 : Write equilibriumequation. Step 2 : Displace the particle along x-axis fromequilibriumposition andwrite the forces in terms of x. Step 3 : Take x-component of all the forces and use step 1.Make approximation (like smallx, etc.)Compare with standard equation Fx = �m.2x to get ..
SIMPLE HARMONIC MOTION www.physicsashok.in 18 Oscillation of a spring : Ablock ofmassmis hanging verticallyfromthe lower end of an ideal springwhose upper end is fixed to the sealing. The stiffness of the spring is k. The block is now pulled downward and released.We can see the motion of the block is simple harmonic and find the angular frequencyof its oscillation. In equilibrium, the extension in the spring is x0. The spring force is kx0. This balancesmg. Hence mg = kx0 ...(1) Nowthe block is pulled by x. Then spring force is k(x0 + x) and ismore thanmg. taking x-component, Fx = mg � k(x + x0) Fx = mg � kx � kx0 (2) Using (1) Fx = �kx ...(3) This satisfies the equation of SHM. Hence themotion is SHM. The restoring force is kxwhere k is spring constant and x is extension or compression relative to equilibrium. Comparing Eq. (3) with standard equation of SHM, �m.2x = �kx; . = km This is the angular frequencyofthe oscillations. C23: A mass of 200 gm is connected to a light spring of force constant 5 N/m and is free to oscillate on a horizontal, frictionless surface. If themass is displaced 5 cmfromthe equilibriumposition and released from rest, thencalculate the period of itsmotion. Sol. We have . . K /m . 5/ 200 .10.3 . 5 rad / sec. . T = 2 ./. = 2./5 = 1.26 sec. C24:Aheavy brass sphere is hung froma light spring and is set in vertical small oscillationwith a period T.The sphere is nowimmersed in a non0-viscous liquidwith a density1/10th the densityof the sphere. If the system is nowset in vertical SHM, its periodwill be (A) (9/10)T (B) (9/10)2T (C) (10/9)T (D) T Sol: T 2 mk . . Hence option (D) is correct. Example 16: Two springs of spring constant K1 andK2 are suspended as shown inthe figure.AbodyofmassM is suspended as shown. (a) Calculate the effective spring constant of the system, and (b) The time period. Sol. (a) In this case same forcewill act on both the springs, but their displacement will not be the same. Let y1 and y2 be their displacement, and let F be the force, then y = y1 + y2 and F = �K1y1 . y1 = �F/K1 Also F = �K2y2 . y2 = �F/K2 IfKis the effective spring constant of the system, then
SIMPLE HARMONIC MOTION www.physicsashok.in 19 F = �Ky . y = �F/K K1 K2 M Since, y = y1 + y2 . 1 2 F F F K K K . . . . . 1 2 1 1 1 K K K . . or 1 2 1 2 K K K K K . . If K1 = K2 = K´, then K = K´ 2 (b) 1 2 1 2 2M M(K K ) T 2 2 K K K. . . . . If the two spring have same spring constant, i.e. ifK1 = K2 = K´, thenK= K´/2 T 2 2M K´ . . Example 17: Two spring of spring constant K1 andK2 are arranged as shown in the figure.Abody ofmassMis attached as shown. (a) Calculate the effective spring constant of the system, and (b) The time period. Sol. (a) In this case the displacement (y) in the two springswill be same, but different restoring forceswill act on them. Let F1 and F2 be the restoring forces acting on the two springs, then F = F1 + F2 Where F is the total force. Again, F1 = �K1y and, F2 = �K2y IfKis the effective force constant of the system, then M K1 K2 F = � Ky Since, F = F1 + F2 . �Ky = �K1y � K2y . 1 2 K . K . K If K1 = K2 = K´, then K . 2K´ (b) Time period, 1 2 T 2 M 2 M K K K . . . . .
If K1 = K2 = K´, then T 2 M 2K´ . . Example 18: Two spring of spring constant K1 andK2 are arranged as shown.AbodyofmassMis attached as showninthe figure. (a) Calculate the effective force constant of the system, and (b) The time period. Sol. (a) In this case same displacement �y�will take place in the two springs, but different forcewill act on them. If themassMis displaced in downward direction, then the other springwillbe compersed.
SIMPLE HARMONIC MOTION www.physicsashok.in 20 Since th forces are different, therefore, F = F1 + F2 and, Let ybe the common extension or compression, then F1 = �K1y, and F2 = �K2y IfKis the effective force constant of the system, then MK2 K1 F = �Ky Since, F = F1 + F2 . �Ky = �K1y � K2y . K . K1 . K2 If K1 = K2 = K´, then K . 2K´ (b) Time period T 2 MK . . or 1 2 M T 2 K K . . . If K1 = K2 = K´ then T 2 M 2K´ . . Example 19: Aspring of lengthL, force constant Khas time period T. it is cut intwo equal parts; calculate the time period of each part, if samemass is suspended fromeach part.Also calculate the spring constant of each part. Sol. Now, for one part 1 y 1 y 2 . and K1 = 2K Therefore, 1 1 T 2 M 2 M 1 T K 2K 2 . . . . . Time perid of each original Time period No. of parts . Example 20:Abodyis in SHMwith period Twhen oscillated froma freelysuspended spring. If this spring is cut in two parts of length ratio 1 : 3&again oscillated fromthe two parts separately, then the periods are T1&T2 then find T1/T2. Sol. Aswe know, spring constant is inverselypropertional to its natural length. . k . Cl 1 1 k . Cl
and 2 2 k . Cl . 1 2 2 1 k 3 k . . ll
SIMPLE HARMONIC MOTION www.physicsashok.in 21 . k1 = 3 k2 But 1 2 1 1 1 k k k . . or 2 2 2 1 1 1 1 3 k 3k k 3k. . . . . k 3k2 4 . . 2 k 4k 3 . . k1 = 3k2 = 4k But T 2 mk . . 1 1 T 2 mk . . and 2 2 T 2 mk . . . 1 2 2 1 T k 4k T k 3 4k . . . 12 T 1 1 : 3 T 3 . . C25: In the above question the two parts are connected inseries, if samemass is connected to one of them, then (a) calculate the spring constant ofthe system, and (b) the time period. Sol. (a) 1 2 1 2 K K K´ K K . . but K1 = K2 = 2K . K´ = K (b) T´ 2 M 2 M T
K´ K . . . . . C26: In the above questions. If the two parts are connected as showninthe figure, then(1) calculate the force constant of the system(2) time period of this system. M K1 T K1= 2K K1= 2K Sol. Clearly K´ = K M 1 + K2 = 2K1 = 4 K and time period. 1 T 2 M 2 M T K´ 4K 2 . . . . . Example 21: Find the ratio of time periods of two identical springs if theyare first joined in series&then in parallel &a massmis suspended fromthem: (A) 4 (B) 2 (C) 1 (D) 3
SIMPLE HARMONIC MOTION www.physicsashok.in 22 Sol: 1 eq T 2 m k . . (in series) eq 1 1 1 2 k k k k . . . . eq k k2 . . 1 T 2 2mk . . 2 T 2 mk´ . . (in parallel) But k´ = k + k = 2k 2 T 2 m2k . . . 12 2 2m T k 2 T m 2 2k . . . . Example 22: Two bodies P & Q of equal mass are suspended from two separate massless springs of force constants k1 &k2 respectively. If themaximumvelocity of themare equal during theirmotion, the ratio of amplitude ofPto Qis : (A) 12 kk (B) 21 kk (C) 21 kk (D) 12 kk Sol: 1 1 km
. . and 2 2 km . . . v1 max = v2 max or A1.1 = A2.2 or 1 2 2 2 1 1 A k A k . . . . Hence option (B) is correct. C27:Aspringmass systemis hanging fromthe ceiling of an elevator in equilibrium. The elevator suddenly starts accelerating upwardswith acceleration a, find k (a)the frequency; (b) the amplitude of the resulting SHM. m
SIMPLE HARMONIC MOTION www.physicsashok.in 23 Sol. Frequencywillremain unchanged onlymeanpositionwill change. f 1 k 2 m . . Amplitude : figure (a) : Free bodydiagramofblockw.r.t. ground is shown in figure. For equilibriumof block. kx =mg ...(i) kx mg (a) k(x + x ) mg + ma (b) 0 ma = pseudo force Refer figure (b) : Free body diagramof block w.r.t. elevator is shown in figur. For equilibriumof block, k(x + x0) = mg + ma ...(ii) Fromequations Eqs. (i) and (ii), x0 = amplitude = ma k Example 23. Apointmassmis suspended at the end of a masslesswire of length l and cross-sectionA. IfYis Young�smodulus ofelasticityforwire, obtain the frequencyofoscillation for simple harmonicmotion along the verticalline. Sol. The original lengthofwire = l The downward force acting on the block =mg IfTis the tension inthewire, then T =mg because the is in equilibrium According toHooke�s law y = Long stress T /A Long strain / . .l l where .l is the increase in length of thewire. . T =YA .l l Putting T = mg,mg =YA(.l / l) ...(1) Let the block be displaced downa little through a distance x during oscillators. The tension in thewire acting upwardswill be YA (. . x) . mg (. . x) . l l l l The downward force =mg . Resultant downward force acting on themasswill be F = � .mg (. . x) . mg. .. . .. l l F = �
. mg. x
.. .l .. or, F = � kx ...(3) Thus the net force is directlyproportionalto the displacement, but oppositelydirected.Hence themotion ofthe block is SHM. Comparingwith standard euation of SHMis F = �m.2x, ..(5) We get . . (k /m)
SIMPLE HARMONIC MOTION www.physicsashok.in 24 . T = 2./. = 2. m/ k = 2. ml / AY and frequencyof vibration = n= 1 1 AY T 2 m . . l ENERGY IN SHM Aparticle performing SHMhasmotion and hence, a kinetic energy the restoring force being a conservative force. It also has a potential energy. KINETIC ENERGY Letmbe themass of the simple harmonic oscillatormoving at a velocityVx. It has got a kinetic energydue to motion. It is given by 2xK 1 mv 2 . K 1 m 2 .A2 x2 . 2 . . . The plot ofK against x is a parabola. �A A x k(x) Ifwe have x =Asin(.t + .) We get V = .A cos(.t + .) This gives K = 12 m.2A2 cos2(.t + .) Its plot K(t) is given below: O t k(t) Note that its period its .. , the half of period of oscillation. C28 :Howmuchis the frequencyofvariation ofKE of a particlemoving simple harmonicallywith a time period of 2 s ? Sol. T = 1/2 × 2s = 1 s Potential energy of oscillator : If . F. dr . 0 . . . , thenF . is said to be conservative. The force acting on a simple harmonic oscillator is Fx = �m.2x. Here, fi x 2 x x . F . dr . . F dx . . .m. xdx . 0 . . . .
SIMPLE HARMONIC MOTION www.physicsashok.in 25 Thus, Fx(= �m.2x) is conservative. Since the force responsible for simple harmonic motion is conservative, a potential energy must exist corresponding to this force. IfUbe the potential energy thenthe decrement inUis equal to thework done by the conservative Fx. i.e., �dU = Fxdx = �m.2x dx dU = m.2x dx It can be integrated to get Uas a functionof x. On integrating,we get U = 12 m.2x2 + constant The value of constant depends uponour choice. By chosingU= 0 at x = 0, the equation becomes U = 12 m.2x2 �A A x Umax=1/2m A 2 2 The plot U(x) is a parabola as in the figure. The potential ismaximumat extremes. Mechanical energy of the particle in SHM : It is the sumof kinetic and potential energies.We have K = 12 m.2 (A2 � x2) or U = 12 m.2 x2 Hence ehmechanical energyE is given by . E = K + U E = 12 m.2A2 The mechanical energy of the oscillator remains constant. This is so because no non-conservative force is doingwork. The figure shows plot ofK(x),U(x) and E(x), whereU(0) = 0 is the reference value ofU. �A A x E 1/2m 2A2 E KU IfU(0) . 0, we shift U(x) and E(x) byU0. Then the graphs. E(x) and U(x) is shifted byU0 but Kremains unshifted. We note that energy taking part in oscillationis given by EOSC = 1/2m.2A2 UOSC = 1/2m.2x2 x = �A x=A 1/2m A 2 2 K(x) U(x) U0 E(x) Etot=Eosc=1/2m 2A2 KOSC = 1/2m.2 (A2 � x2).
SIMPLE HARMONIC MOTION www.physicsashok.in 26 C29: In SHM, F = �kx or a = �.2x, i.e., F�x graph or a�x graph is a straight line passing through originwith negative slope.The corresponding graphs are shown below. F F = � kxx (a) a F = � xx (b) 2 Slope = � k Slope = � 2 C30: Any function of t, sayy=y(t) oscillates simple harmonically if 2 2 d y y dt . . or we cansayif above condition is satisfied, ywilloscillate simple harmonically. C31: All sine and cosine functions of t are simple harmonic in nature. i.e., for the function y =Asin (.t ± .) or y =Acos (.t ± .) 2 2 d y dt is directlyproportional to �y.Hence, theyare simple harmonic in nature. C32: Howthe different physical quantities (e.g., displacement, velocity, acceleration, kinetic energy, etc.) vary with time or displacement are listed belowin tabular form. TABLE S.No. Name of the equation Expression of the equation Remarks 1. Displacement�time x =Acos(.t + .) x varies between+Aand �A 2. Velocity-time (v= dx/dt) v = �A.sin(.t + .) v varies between +A. and �A. 3. Acceleration-time (a = dv/dt) a = �A.2 cos(.t + .) a varies between +A.2 and �A.2 4. Kinetic energy-time (K= 1/2mv2) K = 1/2mA2.2 sin2(.t + .) Kvaries between 0 and 1/2mA2.2 5. Potential energy-time (U= 1/2m.2x2) K = 1/2m.2 A2 cos2(.t + .) U varies between 1/2mA2.2 and 0 6. Total energy-time (E =K+U) E = 1/2m.2A2 E is constant 7. Velocity�displacement v . . A2 . x2 v = 0 at x = ±A and at x = 0 v = ±A. 8. Acceleration�displacement a = �.2 x a = 0 at x = 0 a = ±..2A at x = . A 9. Kinetic energy�displacement K = 1/2m.2 (A2 � x2) K = 0 at x = ±A K = 1/2m.2A2 at x = 0 10. Potential energy-displacement U = 1/2m.2 x2 U = 0 at x = 0, U = 1/2m.2A2 at x = ±A 11. Total energy�displacement E = 1/2m.2A2 E is constant
SIMPLE HARMONIC MOTION www.physicsashok.in 27 C33: Fromthe above tablewe see that x, v and a are sine or cosine functions of time. So, theyalloscillate simple harmonicallywith same angular frequency .. Phase difference between x and a is ..and between any other two is ./2. C34: Kinetic energyversus time equation can also bewritten as K = 1/4mA2.2[1 � cos 2(.t + .)] This function is also periodic with angular frequency 2.. Thus, kinetic energy is SHMis also periodic with double the frequency then that of x, v and a. But these oscillations are not simple harmonic in nature, because 2 2 d K dt is not proportional to �K. But, K � 14 mA2.2 = � 14 mA2.2 cos 2(.t + .) = K0 (say) Here,K0 is simplya cosine function oftime. So,K0willoscillate simple harmonicallywithangular frequency2.. Same is the casewith potential energyfunction.Ualso oscillatewith angular frequency2.but the oscillations are not simple harmonic in nature. Totalenergydoes not oscillate. It is constant. Thus, x . oscillate simple harmonicallywith angular frequency. v . oscillate simple harmonicallywith angular frequency. a . oscillate simple harmonicallywith angular frequency. K . oscillatewithangular frequency2..but not simple harmoncally U . oscillatewithangular frequency2..but not simple harmoncally E . does not oscillate. C35: In the above discussion we have read that potential energy is zero at mean position and maximum at extreme positions and kinetic energy ismaximumat mean position and zero at extreme positions. But the correct statement is like this, At mean position .KismaximumandUisminimum(itmay be zero also, but it is not necessarily zero). at extreme positions .Kis zero andUismaximum. U(J) x(m) (a) U(J) x(m) (b) 5 2 U(J) x(m) (b) �2 Thus, infigure (a), oscillationswill take place about themean position x =0 andminimumpotentialenergy at mean positionis zero.
In figure (b)mean position is at x = 2mand theminimumpotentialenergy in this position is 5 J. In figure (c)mean position is at x = �2mand theminimumpotentialenergy in this position is again zero. THE CAUSES OF OSCILLATION Consider a particle free to move on x-axis, is being acted upon by a force given by, F = � kxn Here, k is a positive constant. Now, following cases are possible depending on the value of n.
SIMPLE HARMONIC MOTION www.physicsashok.in 28 (i) If n is aneven integer (0, 2, 4, ..... etc), force is always along negative x-axis,whether x is positive or negative. Hence, themotion of the particle is not oscillatory. If the particle is released fromany position on the x-axis (except at x= 0) a force innegative direction ofx-axis acts on it and itmoves rectilinearlyalong negative x-axis. (ii) If n is an odd integer (1, 3, 5, ..... etc), force is along negative x-axis for x > 0, along positive x-axis for x < 0 and zero for x= 0.Thus, the particlewilloscillate about stable equilibriumposition, x= 0. The force in this case is called the restoring force. Of these, if n = 1, i.e., F = � kx themotion is said to be SHM. Example 24. Describe themotion of a particle acted upon by a force (i) F = �2(x � 2)3 (ii) F = �2(x � 2)2 (iii) F = � 2(x � 2) Sol. (i) F = �2(x � 2)3 F = 0 at x = 2 Force is along negative x-direction for x > 2 and it is along positive x-direction for x < 2. Thus, the motion of the particle is oscillatory (but not simple harmonic) about x = 2. (ii) F = 0 for x = 2, but force is always along negative x-direction for any value of x except at x = 2. Thus, the motion ofthe particle is rectilinear along negative x-direction. (iii) Let, us take x � 2 = X, then the given force can bewritten as, F = � 2 X This is the equation of SHM.Hence, the particle oscillates simple harmonically about X= 0 or x= 2. Example 25: Abodyofmass 1kg is executing simple harmonicmotionwhich is given byy= 6.0 cos(100t + 4.) cm.What isthe (i) amplitude ofdisplacement, (ii)Angular frequency, (iii) initialphase, (iv)velocity, (v) acceleration, (vi)maximumkinetic energy? Sol. The given equationof SHMis y = 6.0 cos (100t + ./4) cm. Comparing it with the standard equation of SHM, y=Asin (.t + .),we have (i) AmplitudeA= 0.6 cm (ii) Angular frequency .= 100 sec�1. (iii) Initial phase . = ./4 (iv) Velocity v = . (A2 . y2 ) = 100 (36 . y2 ) cm/sec (v) Acceleration = �.2y = �(100)2y = � 104y (vi) Kinetic energy = 12 mv2 = 12 m.2(A2 � y2) The kinetic energy of a particle in SHMismaximum,when it passes itsmean position i.e. at y= 0 (K.E.)max = 2max 1 mv 2 = 12 mA2.2 = 12 × 1 × 104 × (0.06)2 = 18 joules
SIMPLE HARMONIC MOTION www.physicsashok.in 29 Example 26: Aparticle ofmass 0.8 kg is executing simple harmonicmotionwith an amplitude of 1.0 metre and periodic time 11/7 sec. Calculate the velocity and the kinetic energy of the particle at themoment when its displacement is 0.6metre. Sol. We knowthat, at a displacement y frommean position particle�s velocityis given as v = . (A2 . y2 ) or v 2 (A2 y2 ) T. . . v 2 3.14 [(1.0)2 (0.6)2 ] 3.2 (11/ 7) . . . . m/sec. Kinetic energyat this displacement is given by K = 12 mv2 K = 12 × 0.8 × (3.2)2 = 4.1 joule Example 27:Apersonnormallyweighing 60 kg stands on a platformwhich oscillates up and down harmonicallyat a frequency 2.0 sec�1 and an amplitude 5.0 cm. If amachine on the platformgives the person�sweight against time, deduce themaximumandminimumreading it will show, take g = 10m/sec2. Sol. As shown infigure, platformis executing SHMwith amplitude and angular frequencygiven us A= 5.0 cm and . = 2.n = 4. rad/sec [as n = 2 sec�1] +A �A weighing machine Hereweighingmachinewill showweightmore then that ofmanwhenit is belowits equailibriumpositionwhen the acceleration of platformis in upward direction. In this situation the free body diagramofman relative to platformis shown in figure. Herema is a pseudo force onman in downward direction relative to platform(orweighingmachine).Asweighingmachinewillread the normal reaction on it thus for equilibriumof anrelative to platform, mg + ma N we have N = mg + ma or N = mg + m(.2y) [as | a | = .2y] Where y is thedisplacement of platformfromitsmean position.Wewishto find themaximumweight shownby theweighingmachine,which is possiblewhenplatformis at its lowest extreme position as showninfigure, thus maximumreading ofweighingmachinewillbe N = mg + m.2A N = 60 × 10 + 60 × (4.)2 × 0.05
SIMPLE HARMONIC MOTION www.physicsashok.in 30 N = 600 + 480 N = 1080 Nt = 108 kg .t Similarlythemachinewillshowminimumreadingwhen it is at itsupper extreme positionwhenpseudo force on manwillbe in upward direction, thusminimumreading ofweighingmachinewillbe N = mg � m.2A N = 600 � 480 = 120 Nt = 12 kg .t Example 28: A20 g particle is oscillating simple harmonicallywith a period of 2 second and maximumkinetic energy2 J. The totalmechanicalenergy of the particle is zero.Write the following� (a) Amplitude ofoscillation (b) Potential energyas a function of displacement xrelative tomean position. Sol. (a) m = 20 g = 20 × 10�3 kg T = 2s Kmax = 2 J Using Kmax = 12 m.2 A2 we have 2 2 1 20 10 3 2 .A2 2 2 . . . . . .. ... . . 2 3 2 2 A 5 10. 4 . . . . A . 10 m . (b) Etot = EOSC + U0 0 = 2 + U0 U0 = �2 J . U(x) = 12 m.2 x2 � 2 2 U(x) 1 20 2 x2 2 2 1000 2. .. . . .. . . . . 2 u(x) x2 2 100 . . . (S.I. units) Example 29:Abody is executing SHMunder the action of forcewhose maximummagnitude is 50N. Find the magnitude of force acting on the particle at the timewhen its energyis half kinetic and half potential. Sol. Totalenergy E = 1/2 mA2.2 Up = Uk . 1 mA2 2 sin2 t 1 mA2 2 cos2 t 2 2 . . . . .
SIMPLE HARMONIC MOTION www.physicsashok.in 31 or tan2.t = 1 . t 4. . . . a = �.2Asin .t . F = ma = .2Asin 4. (in the sence ofmagnitude) . F 50 25 2 N 2 . . Example 30:Apoint particle ofmass 0.1 kg is executing SHMwith amplitude of0.1m.When the particle passes through themeanposition, itsK.E. is 8× 10�3 J.Obtainthe equation ofmotionof this particle ifthe initialphase of oscillation is 45º. Sol. 2 3 max 1 mv 8 10 2 . . . or 1 mA2 2 8 10 3 2 . . . . or 1 0.1 (0.1)2 2 8 10 3 2 . . . . . . or .2 = 16 . . = 4 rad/s . x =Asin t 4. .. .. . . . . . x = 0.1 sin 4t 4. .. . . . . . Example 31: In case of simple harmonicmotion (a)what fraction of total energy is kinetic andwhat fraction is potentialwhen displacement is one half of the amplitude, (b) at what displacement the kinetic and potential energies are equal? Sol. In case ofsimple harmonicmotion K 1 m 2 .A2 y2 . 2 . . . , U 1 m 2y2 2 . . and E 1 m 2A2 2 . . (a) So 2 K 2 f K 1 y 1 1 3 E A 4 4 . . . . . . . . . . . . . . . . . . as y A given
2 . . . .. .. 2 2 P 2 f U y (A / 2) 1 E A A 4 . . . . . . .. .. (b) According to given problem K = U, i.e., 12 m.2(A2 � y2) = 12 = m.2y2 i.e. 2y2 = A2 or y A 0.7 A 2 . .
SIMPLE HARMONIC MOTION www.physicsashok.in 32 Example 32:Aparticle starts oscillating simple harmonicallyfromits equilibriumposition then, the ratio of kinetic energy and potential energy of the particle at the time T/12 is : (T = time period) (A) 2 : 1 (B) 3 : 1 (C) 4 : 1 (D) 1 : 4 Sol: x =Asin .t v dx dt . =A. cos .t . 2 2 2 2 k U 1 mv 1 mA cos t 2 2 . . . . and 2 2 2 P U 1 mA sin t 2 . . . . 2 k 2 2 P U cos t cot t U sin t . . . . . k 2 P U 2 T cot U T 12 . . . k 2 2 P U cot cot 30º U 6. . . kP U 3:1 U . Hence option (B) is correct. Example 33:Abody performs simple harmonic oscillations along the straight lineABCDE withCas themidpoint ofAE. Its kinetic energies at B and D are each one fourth of itsmaximumvalue. IfAE = 2R, the distance between Band Dis A B C D E (A) 3 R
2 (B) R2 (C) 3 R (D) 2 R Sol: 2 2 2 k U 1 m (A x ) 2 . . . or 1 1 mA2 2 1 m 2 (A2 x2 ) 4 2 2 . . . . . or . . 2 A 2 2 A x 4 . . or 2 2 2 2 A 3A x A 4 4 . . . . x 3A 3 R 1 2 . . Hence option (A) is correct. Example 34: The potentialenergyof a simple harmonic oscillator ofmass 2 kg in itsmeanposition is 5 J. If its total energy is 9J and its amplitude is 0.01 m, its time periodwould be (A) ./10 sec (B) ./20 sec (C) ./50 sec (D) ./100 sec
SIMPLE HARMONIC MOTION www.physicsashok.in 33 Sol: 2max 1 mv 9 5 4 2 . . . or 2 mvmax . 8 or 2max 2 v . 8 or vmax = 2 m/s =A. . max v 2 200 A 0.01 . . . . or 2 200 T. . . T 2 200 100 . . . . Hence (D) option is correct. Example 35:Aspringmass systempreforms SHM. If themass is doubled keeping amplitude same, then the total energyofSHMwill become : (A) double (B) half (C) unchanged (D) 4 times Sol: E 1 mA2 2 2 . . E 1 mA2 k 2 m . . . .. .. E 1 A2k 2 . (Independent ofmass) Hence option (C) is correct. C36:Amass at the end of a spring executes harmonicmotionabout anequilibriumpositionwithan amplitudeA. Its speed as it passes through the equilibriumposition isV. If extended 2Aand released, the speed of the mass passing through the equilibriumpositionwillbe (A) 2V (B) 4V (C) V/2 (D) V/4 Sol: . v =A. . v´ = 2A. . v´ = 2v Hence option (A) is correct. C37: Alinear harmonic oscillator has a totalmechanical energyof200 J. Potential energyof it atmeanposition is 50 J. Find, (i) themaximumkinetic energy (ii) theminimumpotential energy (iii) the potentialenergyat extreme positions. Sol. Atmeanposition, potential energyisminimumand kinetic energyismaximum. Hence, Umin = 50 J (atmean position) and Kmax = E � Umin = 200 � 50 Kmax = 150 J (atmean position) At extreme positions, kinetic energyis zero and potential energyismaximum
. Umax = E Umax = 200 J (at extreme position)
SIMPLE HARMONIC MOTION www.physicsashok.in 34 C38: The potentialenergyof a particle oscillating on x-axis is given as U = 20 + (x � 2)2 HereU is in joules and x inmetres. Totalmechanical energyof the particle is 36 J. (a) Statewhether themotion of the particle is simple harmonic or not. (b) Find themean position. (c) Find themaximumkinetic energyof the particle. Sol. (a) F = � dU dx = � 2(x � 2) By assuming x � 2 =X, we have F = � 2X Since, F . � X Themotionof the particle is simple harmonic (b) The mean position of the particle is X= 0 or x � 2 = 0, which gives x = 2 m (c)Maximumkinetic energyof the particle is, Kmax = E � Umin = 36 � 20 = 16 J. Note : Umin is 20 J at mean position or at x = 2 m. Example 36. If a particlemoves in a potential energyfieldU=U0 � ax+ bx2, where a and b are positive constant, obtain an expression for the force acting on it as a function of position.Atwhat point does the force vanish? Is this a point of stable equilibrium? Calculate the force constant and frequencyof the particle. Sol. F dU a 2bx dx . . . . F = 0, at x = a/2b 2 2 d U 2b 0 dx . . i.e., x a 2b . is a point ofminimumpotential energy.Hence, the equilibriumis stable. k = 2b and f 1 k 1 2b 2 m 2 m . . . . Example 37:Aparticle ofmassmmoves in a one-dimensional potential energyU(x) = �ax2 + bx4, where a and b are positive constants. The angular frequencyof small oscillations about theminima of the potential energy is equal to (A) a 2b . (B) 2 am (C) 2a m (D)
a 2m Sol: U = �ax2 + bx4 . F dU 2ax 4bx3 dx . . . . .
SIMPLE HARMONIC MOTION www.physicsashok.in 35 Atmean position F = 0 . 2ax = 4bx3 or x2 2a a 4b 2b . . . 0 x a 2b . . . F = � 2a(x0 + x) + 4b(x0 + x)3 F = � 2ax0 � 2ax + 4bx03 3 0 1 x x . . . . . . . F = � 2ax0 � 2ax + 4bx03 0 1 3x x . . . . . . . F = �2ax0 � 2ax + 4bx03 + 30 0 12bx x x F = �2ax0 � 2ax + 4bx03 + 12bx02x F 2a a 2ax 4b a a 12b a x 2b 2b 2b 2b . . . . . F 2a a 2ax 2a a 6ax 2b 2b . . . . . ma´ = + 4ax . a´ 4a x m . . a´ . � x . a´ = .2x 4a 2 a m m . . . Example 38:Aparticle ofmassmmoves in the potential energy U shown above. The period of themotionwhen the particle has total energyEis U(x) U=mgx. x > 0 U= kx , x > 0 2 12 x (A) 2. m/ k . 4 2E /mg2 (B) 2. m/ k
(C) . m/ k . 2 2E /mg2 (D) 2 2E /mg2 Sol: For x < 0 F dU kx dx . . . . ma = � kx or a k x m . .
SIMPLE HARMONIC MOTION www.physicsashok.in 36 . 21x k x m .. . . . 1 km . . . 1 T 2 mk . . For x > 0 U=mgx . F dv mg dx . . . . But 20E 1 mv 2 . . 0 v 2E m . , it is speed at lowest point. . 0 2 2v 2 2E T g g m . . . 1 2 T m 2 2E T T 2 k g m . . . . . Hence option (C) is correct. PASSAGE (Q. 39 to 41) The graphs is figure showthat a quantityy varieswith displacement d ina systemundergoing simple harmonic motion. (I) y O d (II) y O d (III) y O d (IV) y O d Example 39: The total energy of the system
(A) I (B) II (C) III (D) IV Sol: In SHM, totalenergy remains constant. Example 40: The time (A) I (B) II (C) III (D) IV Sol: x =Asin .t Hence option (D) is correct. Example 41: The unbalanced force acting on the sytem (A) I (B) II (C) III (D) None Sol: a = �.2x . F = ma = � m.2x Hence option (D) is correct.
SIMPLE HARMONIC MOTION www.physicsashok.in 37 Energy method for SHMproblems : 1. Write the equilibriumequation. 2. Take displaced position andwrite the potential plus kinetic energies : E =K+U 3. Set dE dt = 0 and obtain dynamical equation. Example 42: The friction coefficient between the two blocks shown infigure is µ and the horizontalplane is smooth. If the systemis slightlydisplaced and released, (a) find the time period mM k (b) find themagnitude ofthe frictionalforce betweenthe blockswhenthe displacement fromthemean position is x, (c) find themaximumamplitude if the upper block does not slip relative to the lower block during oscillations. Sol. (a) For small amplitude, the two blocks oscillate together. The angular frequencyis of a single block spring system: . . [k /(M. m)] and so the time period is : T . 2. [(M.m) / k] (b) When the blocks are at distance x fromthemean position, their acceleration is : a = � .2x = � kx M. m The resultant force on the upper block is, therefore mf = � mkx M. m The force is prompted by the frictional of the lower block. Hence the magnitude of the frictional force is mk | x | M. m . (c) themaximumfrictionalforce required for simple harmonicmotion of the upper block ismKA/(M+m) at the extreme position.But themaximumfrictionalforce can onlybeµmg. Therefore mkA M. m =µmg or A= µ(M m)g k. Example 43:A2 kg mass is attached to a spring of force constant 600 N/mand rests on a smooth horizontal surface.Asecondmass of 1 kg slides along the surface toward the first at 6m/s. (a) Find the amplitude of oscillation if themassesmake a perfectlyinelastic collision and remain together on the spring.What is the period of oscillation? (b) Find the amplitude and period of oscillation if the collision is perfectly elastic. (c) For each case, write down the position x as a function of time t for themass attached to the spring, assuming that the collision occurs at time t = 0.
What is the impulse given to the 2 kgmass in each case ? Sol. (a) Fromconservationof linearmomentum, 1 × 6 = (1 + 2)v . v = 2 m/s
SIMPLE HARMONIC MOTION www.physicsashok.in 38 Nowfromconservationofmechanical energy, 1kg 6m/s 2kg 1 mv2 1 kA2 2 2 .A m v 3 2 k 600 . . . . . . . . . . . . . . . or A = 0.141 m= 14.1 cm T 2 m 2 3 0.44 s k 600 . . . . . (b) For perfectlyelastic collision, ´ 2 1 1 2 2 1 1 2 1 2 m m 2m v v v m m m m . . . . . . . . . . . . . . . . . or ´2 v 0 2 1 6 4 m/ s 1 2 . . . . . . . . . . . . 1/ 2 ´2 A m v 2 4 k 600 . . . . . . . . . . . . . . . or A = 0.23 m= 23 cm T 2 m 2 2 0.36 s k 600 . . . . . (c) In the first case, 2 2 14.28 rad / s T 0.44 . . . . . . and amplitude A= 14.1 cm . x =Asin .t = (14.1 cm) sin (14.28 t) In the second case, 2 2 17.45 rad / s T 0.36 . . . . . . and amplitude A= 23 cm . x = 23 sin (17.45 t) cm Impulse, J = .P = 2 × 2 = 4 N-s in the first case and 4 × 2 = 8 N-s in the second case. Example 44: Abody A of mass m1 = 1.00 kg and a body B of mass m2 = 4.10 kg are interconnected by a spring as shown in figure. The body A performs free vertical harmonic oscillationswithamplitude a=1.6 cmand frequency.=25 rad/s.Neglecting themass ofthe spring, find themaximumandminimumvalues of force that this system exerts on the bearing surface. A B
Sol. Nmax : k = .2m1 = (25)2 (1.0) = 6.25 N/m Acceleration ofcentre ofmass in extreme position, 2 1 1 CM 1 2 1 2 m a m A a (upwards) m m m m . . . . . Now 2 1 max 1 2 1 2 1 2 m A N (m m )g (m m ) (m m ) . . . . . . . m1 m2 a = A 2 . Nmax = (m1 + m2)g + m1.2A Nmax = (1.0 + 4.10)9.8 + (1.0)(25)2(1.6 × 10�2) = 59.98 newton
SIMPLE HARMONIC MOTION www.physicsashok.in 39 Example 45: In the figure shown, the blockAcollideswiththe blockBand after collision theystick together. Calculate the amplitude of resultant vibration. B A M u M Sol. Common velocity after collision = u2 Hence, K.E. = 2 1 (2M) u 1 Mu2 2 2 4 . . . .. .. It is also the totalenergyofvibrationbecause the stringis unstretched at thismoment, hence ifAis the amplitude, then 1 KA2 1Mu2 2 4 . or A M u 2K . . Example 46:Abody ofmassmfalls froma height h on to the pan of a spring balance. Themass of the pan isM and spring is massless. The force constant of the spring is k. The body gets struck to the pan and starts performing harmonic oscillations in the verticaldirection.Calculate the period, amplitude and energy of these oscillations. Sol. Let vbe the velocityimmediatelyafter impact. Then m 2gh = (M+ m)v or v m 2gh M m . . ...(1) The period of oscillation of a spring depends on themass attached and its force constant and it is given by T 2 mk . . . here T 2 M m k. . . or k M m . . . x´0´ x0 3 2 1 x´0
l0 l M m The bodyvibrates about the equilibriumposition. Let the spring be compressed by x´0 under theweight of the pan. ThenMg = kx´0. Let the spring be compressed by x´´0 fromthe previous position under theweight (M+m)g. Then (M+ m)g = k(x´´0 + x´0) This positive (position 2) is the equilibriumposition of the system (pan + body). Let x0 be the maximum compression of the spring relative to position1. This is the lower extreme position (position 3) of the system. Applying Conservation of energybetween positions 1 and 3 ´2 2 0 (M m)g 1 k x 1 (M m)v 2 2 . l . . . ´ 2 0 0 0 (M m)g( x ) 1 k(x x ) 0 2 . . l . . . .
SIMPLE HARMONIC MOTION www.physicsashok.in 40 2 ´ 2 0 0 0 0 m gh 1 (M m)g x k x x k x M m 2 . . . . . . 2 0 0 0 0 Mg x mg x k Mg x 1 k x k 2 . . . . . 2 200 2m gh kx 2mg x 0 M m . . . . 0 x mg mg 1 2kh k k M m . . . . The negative value is not acceptable . 0 x mg mg 1 2kh k k (M m)g . . . . Since the amplitude of SHMis the distance between the mean position and the extreme position, here a, amplitude ´´ 0 0 x x mg 1 2kh k (M m)g . . . . . and E 1 (M m)a2 2 2 . . .2 2 2 E 1 (M m) m g 1 2kh k 2 k (M m)g M m . . . . . . . . . . . 1 m2g2 2kh E 1 2 k (M m)g . . . . . . . . . Example 47: The spring shown in figure is kept in a stretched position x0. Assuming the horizontalsurface to be frictionless, calculate the frequency of oscillations after the systemis released. M m
k Sol. As there is no external force acting on the systemof two blocks plus the spring, the centre ofmass of the systemwill remain at rest. Themean positions of two simple harmonicmotion or two blocks occurwhen the spring becomes unstretched. If themassmmoves towards left through a distanceXarea themassMmoves towards left through a distance x before acquires natural length, x + X = x0 ...(1) where x andXwill represent the amplitudes of two blocksmandMrespectively. Because the centre ofmass of the systemmust not move during themotion, we canwtire mx =MX ...(2) From(1) and (2) x = 0 Mx M.m and X = 0 mx M.m ...(3) x andXalso give the amplitudes ofmandMduring oscillations.At the mean positions the two blocks have kinetic energies 12 m.2x2 and 12 M.2x2. Their summust be 12 kx02.
SIMPLE HARMONIC MOTION www.physicsashok.in 41 Now, from(3) X0 = M m X M. 12 kx02 = 12 .2(mx2 + Mx2) Using (3), angular frequency K(M m) Mm. . . and the frequency n K(M m) 2 Mm . . . . . Example 48: Two identical balls Aand B each ofmass 0.1 kg are attached to two identicalmassless springs.The spring-mass systemis constrained to move inside a rigid smooth pipe bent inthe formof a circle as shown infig. The pipe is fixed in a horizontalplane. The centres of the balls canmove in a circle of radius 0.06m. Each spring has a natural lengthof 0.06 .mand force constant 0.1N/m. Initially, both the balls are displaced by an angle . = ./6 radianwith respect to diameter PQ of the circle and released fromrest. /6 /6 A B P Q O y1 y2 (a) Calculate the frequency of oscillation ofthe ballB. (b) What is the total energy of the system? (c) Find the speed of the ballAwhenA and B are at the two ends of the diameter PQ. Sol. (a)As here two massesAand Bare connected by two springs, this problemis equivalent to the oscillation of a reduced massmby a spring of effective force constant keff given by 1 2 1 2 m m 0.1 0.1 m 0.05 kg m m 0.1 0.1 . . . . . . and keff = k1 + k2 = 0.1 + 0.1 = 0.2 N/m So f 1 keq 1 0.20 1 Hz 2 m 2 0.05 . . . . . . (b) As here one spring is compressed while the other is stretched by same amount (say y) and balls are at rest at AandB, so 2 2 2
1 2 E 1 k y 1 k y ky 2 2 . . . [as k1 = k2] But fromabove figure y = y1 + y2 = R.1 + R.2 = 2R. y = 2 × 0.06 × (./6) = 0.02 .m So E = (0.1)(0.02.)2 = 4.2 × 10�5 J (c) As at P andQsprings are unstretched so thewhole energy becomes kinetic of the ballsAand B, i.e. 2 2 2 5 1 1 2 2 1 m v 1 m v E 4 10 J 2 2 . . . . . . Here m1 = m2 = 0.1 kg and v1 = v2 = v So 0.1 v2 = 4.2 × 10�5, i.e., v = 2. × 10�2 m/s
SIMPLE HARMONIC MOTION www.physicsashok.in 42 Example 49: Figure shows a blockPofmassMrestingona horizontal smooth floor at a distance l froma rigid wall. Block is pushed toward right bya distance 3l/2 and released,when block passes fromitsmean position another block ofmassm1 is placed on it which sticks to it due to friction. Find the value ofm1 so that the combined block just collideswith the left wall. m k P l Sol. When block P is released fromrest froma distance 3l/2 toward right frommean position, this will be the amplitude ofoscillation, so velocityof blockwhen passing fromitsmean position is given as v A 3 k 2 m . . . l [as . = km ] Ifmassm1 is added to it and just after if velocityof combined block becomes v1, frommomentumconservation we have mv = (m+ m1)v1 or 1 . . 1 v m 3 k m m 2 m . . . . . . . . l If this is the velocity of combined block at mean position, it must be given as v1 = A.1 [now .1 = 1 k m. m ] WhereA1 and .1 are the newamplitude and angular frequencyofSHMof the block. It is given that combined block just reaches the left wall thus the newamplitude of oscillationmust be l sowe have 1 1 1 m . 3 k k (m m ) 2 m m m . . . l l or 1 3 m 1 2 m m . . or 9m= 4m+ 4m1 or 1 m 5 m 4 . Example 50: For the arrangement shown in figure, the spring is initially
compressed by3cm.When the spring is released the block collides with thewalland rebounds to compress the spring again. k = 104N/m m = 1 kg 4 cm (a) If the coefficient of restitution is 0.7, find themaximumcompression in the spring after collision. (b) If the time starts at the instant when spring is released, find theminimumtime after which the block becomes stationary. Sol. (a)Velocityof the block just before collision, 2 2 2 0 0 1 mv 1 kx 1 kx 2 2 2 . . or . 2 2 . 0 0 v k x x m . . here, x0 = 0.03 m, x = 0.01 m, k = 104 N/m, m= 1 kg . 0 v . 2 2 m/ s
SIMPLE HARMONIC MOTION www.physicsashok.in 43 After collision, v = ev0 = (0.7) 2 2 = 2 m/s Maximumcompression inthe spring is 2 2 2 m 1 kx 1 kx 1 mv 2 2 2 . . or 2 2 2 2 m 4 x x m v (0.01) 1(2) 2.23 cm k 10 . . . . . (b) In the cse of spring-mass system, since the time period is independent of the amplitude of oscillation, therefore0 1 T 1 T 2sin 2 3 . .. . . .. .. . . . . . . . . /2 v /2 O �1 �3 +3 x sin�1(1/3) sin�1(1/3) T m 2sin 1 1 k 3 . ... . . .. .. .. . . . . Example 51: The systemshown in the figure canmove on a smooth surface. the spring is initially compressed by 6 cmand then released. Find (a) time period k=800N/m 3kg 6 kg (b) amplitude of 3 kg block (c) maximummomentumof6 kg block. Sol. (a) T 2 µk . . Here 1 2 1 2 m m 3 6 µ 2 kg m m 9. . . . . . T 2 2 2 1 800 20 . . . . . . T second 10 .
. (b) Here x0 = 6 cm But x10 + x20 = x0 and m1x10 = m2x20 or 3x10 = 6x20 or x10 = 2x20 . x10 + x20 = x0 or 10 10 0 x x x 2 . . or 10 0 3 x x 2 . . 10 0 x 2 x 2 6 4 cm 3 3 . . . .
SIMPLE HARMONIC MOTION www.physicsashok.in 44 (c) Applying conservationprinciple ofmomentum 3 × v1 � 6 v2 = 0 . 3v1 = 6v2 . v1 = 2v2 Applyingmechanicalenergyconservation . 2 2 2 1 2 0 1 3 v 1 6 v 1 kx 2 2 2 . . . . . . or 2 2 2 1 2 3 v 3v 1 800 (0.06) 2 2 . . . . or 2 2 4 2 2 3 (2v ) 3v 400 36 10 2 . . . . . or 2 2 2 2 3 4v 3v 1.44 2 . . . or 9 v22 = 1.44 or 2 v 1.2 0.4 m/ s 3 . . . P2max = 6 × 0.4 = 2.4 kg m/s Example 52:A2Kg blockmovingwith10m/s strings a spring of constant .2 N/mattached to 2 Kg block at rest kept on a smooth floor. The time forwhich rearmoving block remainin contact withspringwillbe 2 kg 2 kg 10 m/s (A) 2 sec (B) 12 sec (C) 1 sec (D) 12 sec Sol: T 2 µk . . Here µ 2 2 2 2 . . . . reduced mass = 1 . 2 T . 2. 1 . 2 s. . Initially, systemis atmean position. Again blocks separatewhen spring comes in natural length (i.e.mean position). Hence, t = T/2 = 1 second.
Example 53:Two blocksA(2kg) andB(3kg) rest up on a smooth horizontalsurface are connected bya spring of stiffness 120N/m. Initiallythe spring is undeformed.Ais imparted a velocity of 2m/s along the line of the spring away fromB. Find the displacement of A t seconds later. 2m/s 2kg B A 3kg Sol. cm v 2 2 4 0.8 m/ s 2 3 5 . . . . . InC-frame, themotion is SHM.
SIMPLE HARMONIC MOTION www.physicsashok.in 45 But m1x1C = m2x2C But x0 = x0OC + x0OC . 2 0 1OC 1 2 x m x m m . . and 1 0 2OC 1 2 x m x m m . . Applying conservationprinciple ofmechanical energyinC-frame, 2 2 rel 0 1 µr 1 kx 2 2 . or 20 1 2 3 4 1 120 x 2 5 2 . . . . . or 206 4 x 5 120 . . . . 0 x 2 10 . . 2 0 1OC 1 2 3 2 m x 10 6 x m m 5 50 . . . . . 1OC x 12 0.12 m 100 . . Also, kµ . . here 1 2 1 2 m m 3 2 6 µ m m 5 5
.
. . .
.
. 120 10 rad / s 1.2 . . . . x1C = x1OC sin .t or x1 � xC = 0.12 sin 10t or x1 = xC + 0.12 sin 10t . x1 = 0.8t + 0.12 sin 10t Example 54. Two blocks ofmassesm1 andm2 are connected by a spring of stiffness k.Another block ofmassm1, sliding atV0 without friction,hits the set upelasticallyas showninthe figure. V0 m1 m1 k m2 Plot elastic energyof the spring as a functionoftime t after the collision. Sol. Due to one dimensional elastic collision of identicalbodies, velocity is transferred totallyto m1. m1 m2 V0 Let x1 and x2 be positions ofm1 andm2 at a time t. Then extension in the spring (with natural length = l) x = x2 � x1 � l ...(1)
SIMPLE HARMONIC MOTION www.physicsashok.in 46 The equation ofmotion ofm1 iswriten as 2 1 1 2 d x m kx dt . or 2 1 1 2 d x m kx dt . . Fromthese 2 2 2 2 1 2 2 1 2 2 d x d x d x (x x ) dt dt dt . . l . . 1 2 1 2 k x k x k k x m m m m . . . . . .. . . . .. . . . . . . 2 2 2 d x x dt . .. ...(2) where 2 1 k 1 1 m m . . . . . . . . . ...(3) Solving these, x =Asin (.t + .) At t = 0, x = 0, sin (.t + .) sin ..= 0 . = 0, . ...(1) At t = 0, 2 2 0 dx 0, dx V dt dt . . . 0 dx 0 V dt . . or .A cos . = � V0 ..= . ...(2) From(i) and (ii)
sin2. + cos2. = 0 + 2 0 2 2 V. A 0 A . V. . 0 0 x . V sin(.t . .) . . V sin.t . . Now we canwrite U(x) 1 kx2 2 . U(t) t kv0 2 22 Umax The plot of 2 0 2 2 kV U(t) sin t 2CO . . . . . . . . Circular Representation of SHM We have discussed that an oscillationcan be regarded as ShM if it satisfies the basic requirements to be SHM. Every SHM can be best represented as a projection of a particle in circular motion on its diameter. In fact themotion ofprojectionof a uniform circular motion on its diameter satisfies both the conditions required to beSHM. Let us analyze it from the figure shown. yy´ x´ x t = 0 yy´ x´CAx P 1 2´ 2 1´ P´ O t = 0 DB
SIMPLE HARMONIC MOTION www.physicsashok.in 47 AparticlePis executing a uniformcircularmotiononthe circlewitha uniformangular velocity.as shown.The particle is shown at its initialpositionAat t =0. It starts inanticlockwisedirectionwith uniformangular velocity .. Ifwe followthemotion of the projection of particle on its verticaldiameter in figure thenwe can see, along withP, its projectionP´ starts frommid point (centre of circle) inupward direction and the respective position ofP´ for position ofP at point 1 and 2 are 1´ and 2´ as shown in figure. When P reaches the topmost position, P´ will also be at this point and when P starts tracing the second quadrant of circle, P´ starts coming down towards pointO. Thuswe can see, as point Ptraces its circular path ABCD, its projectionondiameterYY´ follows oscillatorymotion alongOBODO.... and so on.This oscillatory motion ofP´ can be taken as SHMas it follows the conditions to be a SHM.We�ll prove it in next section. Heremotion ofparticle is in a straight linewith amplitude equal to the radius of circle.As shownin figure, the displacement ofP´ frompoint Oas a function of time can be given as y = R sin .t ...(1) As P´ is going up, its velocity can be given as v = dy dt =R..cos .t ...(2) x´ A x P´ y´ y t = 0 t R t = t y Its acceleration is a = dv dt = � R.2 sin .t Fromequation (1), we have a = � .2y ...(3) Here ..is a constant thus the acceleration of P´ is directly proportional to the displacement fromitsmean positionOandnegative sign in equation(3) shows that directionof acceleration is opposite to y that is towards mean positionO. Hence themotion of projection P´ can be regarded as SHM. Equation of SHM Equation ofan oscillation is themathematical expressiongiving the displacement of oscillating particle fromits mean position as a function of time. If a particle is executing SHMwith amplitude A, it can be regarded as the projection ofa circularmotionofrediusAas shown in figure, ifcircularmotion of point P is at a constant angular velocity ..then this is termed as angular frequencyof the point P´ whichis in SHM. C A
P P´ DB t = 0 P yP t R t = t Let us consider at t = 0, Pwas at point Aand starts in anticlockwise direction as shown in figure. Nowin time t, point P traversed by an angle .t (as shown) and P´ reaches to a displacement y as shown, can be given as y =Asin .t ...(4) This equation-(4) in this case is called as equation ofSHMof point P´ whichis in SHM.Here .t whichis the angular displacement of point P(in circularmotion) is called phase angle of point P´ in SHM. In previous articlewe�ve discussed the SHMof point P´ whichwas at itsmean position at t = 0. But it is not necessary that particle starts its SHMfromits mean position. It can start fromany point on its path, thus equation-(4) can not be accepted as a general equation of SHM, this being the equation of those all SHMs where particle starts (at t = 0) their SHMfromtheirmean position.
SIMPLE HARMONIC MOTION www.physicsashok.in 48 Nowconsider figurewhere a particle P´ starts its SHMwith a point having initial phase angle ..anticlockwise on the circle of point Pof which it is the projection. Let as t = 0, point P was at an angular displacement ..fromits reference point (pointA).Thus the point from whichP´willstart its SHMis shown in figure.At this position(project t = 0 P t = t B P Y D Y´ C X´ X A y P´ P´´ tion of P at t = 0 onYY´) ..is called as initial phase of point P´ in SHM, as ..is the initial angular displacement ofparticle Pwhich is in circularmotion. Nowafter time t, the angular displacement ofPis (. +..t) which is called the instantaneous phase of point P´ at time t = t and at this instant the displacement of point P´ frommean position is y =Asin (.t + .) ...(5) This equation-(5) is called as general equation of SHMofa particlewhich starts its SHMwith initialphase .. Here initialphase ofSHMimplies it is the initialangular displacement of the particlewhich is incircularmotion ofwhich the projection is executing SHM. As equation-(4) gives the equation ofSHMof those particleswhich starts their SHMfrommean position. Similarlywe can define an equation of SHMof those particles which start their SHMfromtheir extreme position bysubstituting .= ./2 inequation (5).As if a particle starts fromits extreme position,we can take its initial phase ./2 thus its equation ofSHMfromequation (5) can be given as y =Acos .t ...(6) In all type of problems inwhich a bodyor a particle execute SHM, we assume that this is the projection of an another perticlewho is in circularmotion and its projectionis executing SHM(that bodyor particlewhichis in SHM). Valocity and acceleration of a particle in SHM Equation (5) gives the general expressionfor displacement frommeanposition ofa particle executing SHMas a function of time. Thus velocityofthis particle as a function of time canbe given as v = dy dt =A. cos (.t + .) ...(7) To convert it in displacement function,we canwrite. v =A. 1. sin2 (.t . .) v =A.. 22 1 yA . [as y =Asin .t] v = .. A2 . y2 ...(8)
Equation (8) gives the velocityofa particle in SHMwith amplitudeA, and angular frequency..as a function of its displacement frommeanposition. Fromequation-(8)we canstate that inSHM, particle�svelocityismaximum when y= 0 i.e. at itsmean position and is given as at y = 0, vmax = A. At extreme positions of particlewhen y= ±A, its velocity is zerowhere it returns towards itsmean position. Fromequations-(5) and (8)we can plot the graphs ofdisplacement and velocityas a function of time as shown infigure.
SIMPLE HARMONIC MOTION www.physicsashok.in 49 O T 2T t y Asin T = 2 y = Asin ( t + ) (a) O T 2T t y T = 2 y = Asin t +A T/4 5T/4 T/2 3T/4 7T/4 �A (b) O T 2T t y y = A cos t T/4 3T/4 T/2 (c) Similarly acceleration ofparticle in SHMcanbe given as a = dv dt = �A.2 sin(.t + .) ...(9) or a = � .2y [as y =Asin (.t + .)] ...(10) Fromequation-(10)we cansee that atmeanposition(y=0)whenvelocityofparticle ismaximum, its acceleration is zero and at extremitieswhere y= ±A, accelration of particle ismaximumand itsmagnitude is given as amax = .2A (towardsmean position) It also shows that as particlemoves away frommean position, its acceleration continuously increases till it reaches its extreme position (at amplitude)when its velocity becomes zero and it returns. Equation-(10) canbe rewritten in differentialfromas 2 2 d y dt + .2y = 0 ...(11) This equation-(11) is called �BasicDifferentialEquation� ofmotion ofa particle in SHM.Everyexpression of displacement y as a function of timewhichsatisfies this equation can also be regarded as anequation of SHM. C39: Prove that yAei.t is an equation of SHM. Sol. According to given equation in problem, differentiatingwith respect to timewe get,
SIMPLE HARMONIC MOTION www.physicsashok.in 50 dy dt = iA.ei.t Differentiating againwith respect to time,we get 2 2 d y dt = � .2Aei.t = � .2y [as y =Aei.t] Thuswe have 2 2 d y dt + .2y = 0 This is the basic differential equation of SHMhence y=Aei.t is an equation of SHM. Example 55: The displacement of a body executing SHMis given by x =Asin (2.t + ./3). The first time from t =0when the velocityismaximumis (A) 1/12 sec. (B) 0.16 sec (C) 0.25 sec (D) 0.5 sec Sol: v dx 2 Acos 2 t dt 3. .. . . . . . . . . . or ± 2.A = 2.A cos (2.t + 3. ) or cos 2 t 1 3. .. . . . . . . . . Hence 2 3 . . . . . /3 t = 0 t = t0 3 2 Phasor diagram 6 . . . . . . . = .t0 or 0 2 t 6. . . . 0 t 1 sec. 12 12 . . . . Hence (A) option is correct Phase analysis of a particle in SHM We�ve alreadydiscussed about phase angle ofa particle in SHM. It is actuallythe angular displacement of that particlewho is in circularmotionandwhose projection is in SHM.At a general time t, the instantaneous phase of a particle in SHMcan bewritten as (.t + .) if ..is its initialphase (already discussed).
Phase difference in two SHM Case I : When two SHMs are of same agnular frequency Figure shows two particlesP´ andQ´ in SHMwithsame angular frequency.. PandQare the corresponding particles in circularmotion for SHMofP´ andQ´. Let P andQboth starts their circularmotion at the same time at t = 0 then at the same instant P´ andQ´ starts their SHMinupward direction as shown.As frequencyofboth are equal, bothwill reachtheir extreme position (topmost point) at the same time andwill again reach theirmean position simultaneouslyat time t = T/2. [T = Time period ofSHM= 2./.] andmove in downward direction together orwe can state that the oscillations of P´ andQ´ are exactlyparallel and at every instant the phase of bothP´ andQ´ are equal, thus phase difference in these two SHMs is zero. These SHMs are called same phase SHMs.
SIMPLE HARMONIC MOTION www.physicsashok.in 51 t P t = t P t = 0 ref. point P´ t Q t = t Q t = 0 ref. point Q´ A A Nowconsider figure,wherewe assume ifP´ starts its SHMat t = 0 butQ´will start at time t1. Inthis duriation fromt = 0 to t = t1, P´ willmove ahead in phase by .t1 radianswhileQ´ was at rest. NowQ´ starts at time t1 andmovewithsame angular frequency.. It cannever catchP´ asboth are oscillating at same angule frequency. Thus hereQ´ will always lag in phase by .t1 then P´ or we sayP´ is leading in phase by .t1 thenQ´ ans as . of both are constant their phase differencewill also remains constant so keep it inmind that in two SHMs of same angular frequency, if they have some phase difference, it always remains constant. t t = t Pt = 0 P´ Q´ Q A t = t1 1 1 Nowwe consider a spacial casewhen time lag between the starting of two SHMs is T/2 i.e. half of oscillation period. Consider figure, here ifwe assume, particle P´ starts at t = 0 andQ´ at t = T/2when P´ completes its half oscillation. Herewe can see that the phase difference in the two SHMis ..bywhichQ´ is lagging. Here whenQ´ starts its oscillationinupward directionP´moves indownward direction.As angular velocityofPand Q are same, both complete their quarter revolution in same time. Thuswhen P´ reaches its bottomextreme position,Q´will reach its upper extreme position and then afterQ´ startsmoving downward, P´ startsmoving upward and both of these will reach their mean position simultaneously but in opposite directions, P´ has completed its one oscillationwhere asQ´ is at half of its oscillation due to a phase lag of .. Pt = 0 Q´ Qt = T/2 t = T/2P´ P´ Thus ifwe observe both oscillations simultaneouslywe can see that oscillations of the two particls P´ andQ´
are exactlyantiparalleli.e.whenP´ goesup,Q´ comes downand at allinstants of time their displacements from mean position are equalbut in opposite direction if there amplitudes are equal. Such SHMs are called opposite phase SHMs. Case II : When two SHMs are of different angular frequency We�ve discussed that when two SHMs are of same angular frequency, their phase difference does not change with time.Consider two particles inSHMas shown infigure. Their corresponding particles for circularmotion areAandB respectivelyas shown. If bothA´ andB´ starts their SHMfrommeanpositionat t =0withangular frequencies .1 and .2, thenwe say at t = 0 their phase difference is zero but after time t, their respective phase
SIMPLE HARMONIC MOTION www.physicsashok.in 52 are .1t and .2t. Thus after time t, the phase difference in the two SHMs is . = (.1 � .2)t thus equation above shows that if angular frequencies of the two SHMs are different their phase difference continuously changeswith time. tt = tt = 0 A´ A´ 1 A1 t t = t t = 0 B´ B´ 2 B2 Same Phase SHMs As discussed above two particles execute SHMin such awaythat their oscillations are exactlyparallel to each other or their phase difference during oscillation is zero, theyare said to be in same phasewe�ve seen that this happenswhen both SHMs are started at same timewith same angular frequency. This canhappen also when the time lagbetween starting ofthe two SHMisTor anintegralmultiple oftime period ofSHM.Because ifone starts at t = 0 and other starts at t =T, in this duration first particlewillcomplete its first oscillation and is going to start its second oscillations and the second particlewill start insynchronizationwith the first.Hence the two oscillationwill stillbe parallel or in same phase. There phase diference in the two SHMswill be 2..Not only this even if the time lag in starting of the two SHMs is 2T, 3T..... nT or the phase difference in the two SHMs is 4., 6., 8. ...... 2n., then also these SHMs can be treated in same phase. Thus phase difference in two SHMs of same phase is . = 2., 4., 6. .......... 2n. Opposite Phase SHMs As discussed in previous article, two SHMs are said to be in opposite phase when their oscillations are antiparallel this happenswhen two SHMs of same angular frequency start with a time lag ofT/2 and phase difference among the two SHMs is .. By analyzing the situation it can also be stated that the same thing also happenswhen the time lag in starting of the two SHMs is 3T/2, 5T/2 ........ (2n + 1)T/2 or the phase difference between the two is 3., 5. ..........(2n + 1).. Thus phase difference in two SHMs of opposite phase is . = ., 3., 5. ........., (2n + 1). To understand the concept of phase and phase difference in SHM, we take fewexamples. Example 56: Two particles are in SHMin a straight line about same equilibriumposition.AmplitudeAand time period Tof both the particles are equal.At time t = 0, one particle is at displacement y1 = +Aand the other at y2 = �A/2, and they are approaching towards each other.Afterwhat time they cross each other ?
(A) T/3 (B) T/4 (C) 5T/6 (D) T/6 Sol: Fromphasor diagram, 0 0 t t 6 2 . . . . . . . or 0 2 t 3 4 2 6 2 6 6 3 . . . . . . . . . . . . . t0 t0 t0 6 t0 3 C2 A� 6 t = t0 t = t0 � . 0 t 2 2 T 6 6 2 / T 6 . . . . . . . . Hence option (D) is correct.
SIMPLE HARMONIC MOTION www.physicsashok.in 53 Example 57. Two particles execute SHMwith same amplitudeAand same angular frequency ..on same straight linewith samemean position.Given that during oscillation theycross each other in opposite directionwhen at a distanceA/2 frommean position. Find phase difference in the two SHMs. Sol. Figure shows that two respective particles P´ and Q´ in SHMalong with their corresponding particles in circularmotion.Let P´moves inupward directionwhen crossingQ´ atA/2 as shown in figure(a), at this instant phase of P´ is .1 = sin�1 12 . . .. .. = 6. ...(1) P´ P Q´ A 1 N2 N2 2 Q Similarlyas shown in figure(b) we can take particleQ´ ismoving in downward direction (oppositeP´) atA/2, this implies its circularmotion particle is in second quadrant thus its phase angle is .2 = . � sin�1 12 . . .. .. = . � 6. = 56. ...(2) As both are oscillating at same angular frequencytheir phase diff. remains constant which can be given from equation-(1) and (2), as .. = .2 � .1 = 56. � 6. = 23. Example 58:Aparticle executes SHMwith amplitudeAand angular frequency..At an instant when particle is at a distanceA/5 frommean position andmoving away fromit. Find the time afterwhich it will come back to this position agin and also find the time after which it will pass throughmean position. Sol. Figure shows the circularmotionrepresentationfor the particleP´giveninprolbem.
The initial situation of particle is shown in figure.AsPmoves, its projectionP´will go upand then come backto its initialpositionwhenP reaches to the corresponding position in second quadrant as shown. In this process P traversed an angular displacement .withangular veliocity., thus time taken in the process is P P A/5 BD C A 1 1 1 2cos (1/ 5) t [2cos (1/ 5)] . . . . . . . . . Nowwhen P reaches point C, P´will reach itsmean position. Thus time takenbyP fromits initialposition to point C is ( ) sin 1(1/ 5) t . .. . . . . . . . The same time P´will take fromA/5 position tomean position through its extreme position. Example 59: Two particles executing SHM with same angular frequency and amplitudes A and 2A on same straight line with same mean position cross each other in opposite direction at a distance A/3 frommean position. Find the phase difference in the two SHMs.
SIMPLE HARMONIC MOTION www.physicsashok.in 54 Sol. Figure shows the two corresponding particles of circularmotion for the twomentioned particles in SHM. Let particleP is going up and particleQ is going down. Fromthe figure shown, the respective phase differences of particles P´ and Q´ are Q´ N3 1 Q 2 P´ 2A .1 = sin�1 13 . . .. .. [Phase angle of P´] and .2 = . � sin�1 16 . . .. .. [Phase angle ofQ´] Thus phase difference in the two SHMs is .. = .2 � .1 = . � sin�1 16 . . .. .. � sin�1 13 . . .. .. C40:Aparticle stars its SHMfrommeanposition at t = 0. If its time period isT and amplitudeA. Find the distance travelled by the particle in the time fromt = 0 to t = 5T/4. Sol. We knowin one complete oscillation i.e. in period T, a particle covers a distance 4Aand in first one quarter of its period it goes fromitsmean positionto its extreme position as its starts frommean positionthus the distance travelled bythe particle in time 5T/4 is 5A. Example 60: Aspring block pendulumis shown in figure. The systemis hanging in equlibrium.Abullet ofmassm/2movingat a speed uhits theblock fromdownward direction and gets embedded in it. Find the amplitude of oscillation of the block now.Also find the time taken bythe block to reach its upper extreme position after hit bybullet. m h Nature length of spring Sol. If block is in equilibriumthen springmust be at some stretch if it is h, we have mg = kh If a bullet ofmassm/2 gets embedded in the block, due to this inelastic impact its newmass becomes 3m/2 and nowthe newmean position of the blockwill be say at a dept. h1 fromoldmean position then,wemust have 32 mg = k(h + h1) 3m2 g = k(h + h1)
m u v m/2 1 h mg 2k . (as mg = kh) Just after impact due to inelastic collision if the velocityof block becomes v,we have according tomomentumconservation m u 3m v 2 2 . or v u3 . Nowthe block executes SHMand at t = 0 block is at a distance 1 h mg 2k . above itsmean positionand having
SIMPLE HARMONIC MOTION www.physicsashok.in 55 a velocityu/3. If amplitude of oscillation isA,we have 2 u A2 mg 3 2k . . . . . .. .. or 2 2 u 2k A2 mg 9 3m 2k . . . . . . . . . . .. . . .. [As for this spring block system.= k (3m/ 2) ] mu2 mg 2 A 6k 2k . . . . .. .. Nowtime takenbyparticle to reachthe topmost point can beobtained bycircular motion representationas shown in figure.This figure shows the position of block P and its corresponding circularmotion particle P0 at the t =0. Block Pwill reach its upper extreme positionwhen particle P0will travers the angle . and reach the topmost point.As P0 moves at constant angular velocity ., it will take a time given as P P 1 A 1 h 1 1 1 t cos (h /A) 3m cos 3mg k 2k 2kA 3m/ 2 . . . . . . .. .. . . . Example 61: Figure(a) shows a spring block system, hanging in equilibrium. The block of systemis pulled downbya distance x and imparted a velocity v in downward direction as shown infigure(b).Find the time itwilltake to reachitsmeanposition. Also find themaximumdistance to which it willmove before returning back towardsmean position. equilibrium m position k m x (a) (b) Sol. As shown in figure (b),when the block is pulled down bya distance x and throwndownward, it will start executing SHM. It willgo further to a distanceA(amplitude) frommean position before returning back which can be found by using the velocityof block vat a displacement x fromitsmean position as v . . A2 . x2 or 2 . 2 2 . v k A x m . . [as for spring block system.= km ]
or 2 A mv x2 k . . Now time of motion of bob can be obtained by circular motion representation of the respective SHM. Corresponding circularmotion representation for this SHMis shown in figure at t =0.At t = 0, block P is at a distance x from its mean position in downward direction and it is moving downward so we consider its corresponding circularmotionparticle in III quadrant as shown in reference angular velocitywe consider Pwill reach itsmean positionwhen particle P0 reaches positionAby traversing an angle .. Shown in figure.Thus it will take a time given as
SIMPLE HARMONIC MOTION www.physicsashok.in 56 sin 1 .x /A. t km . . . . . . . A P0 P v x A B C D or 1 2 2 m x t sin k mv x k . . . . . . .. . . . . . . . . . Themaximumdistance towhich blockwillmove fromits initialpositionisA� xas it gas upto its lower extreme at a distance equal to its amplitudeAfromtomean position. Example 62: Figure shows a block ofmassmresting ona smooth horizontal ground attached to one end of a spring of force constant k in natural length. Ifanother block ofsamemass andmovingwitha velocityutoward right is placed on the blockwhich stick to it due to friction. Find the time it will take to reach its extreme position. Also find the amplitude of oscillations of the combinedmass 2m. m u m k smooth Sol. When secondmass sticks to the lowermass, due to such aninelastic collision, the velocityof combined block is reduced byhalf that is u/2 to conservemomentum.Nowatmean positionthe velocityof block can bewritten as u A 2 . . (ifAis the amplitude of oscillation) or u A k 2 2m . [As here for combined block newangular frequencyof SHMis . = k 2m ] or
A u 2m u m 2 k 2k . . As oscillation starts frommean position, in reaching its extreme position, particle has to cover a phase of ./2 radians, thus time taken by particle to reach its extreme position is t / 2 k 2 2m . . . . . Example 63: In previous example if block is pulled toward riht by a distance x0 and released, when the block passes through a point at a displacement x0/2 frommean position, another block ofsamemass is gentlyplaced on it which sticks to it due to friction. Find the newamplitude ofoscillation and find the time nowit willtake in reaching itsmean position and extreme positionon left side. Sol. When block was released at x0 frommean position, this will be the amplitude of oscillation and when it is passing through the positionof half amplitude x0/2, its velocitycan be given as 2 2 0 v A x2 . . . . . .. .. or 2 2 0 0 v k x x m 4 . . 0 v 3 x k 2 m .
SIMPLE HARMONIC MOTION www.physicsashok.in 57 When another block of samemass is added to it, due tomomentumconservation its velocitybecomes half and . of oscillation will, also change from km to k 2m . Again using the formula for velocity of SHMat a distance x0/2 frommeanposition,we get 2 v´ v A´2 x0 2 2 . . . .. . .. .. or 2 2 0 0 3 x k k A´ x 2 m 2m 2 . .. . .. .. [ifA´ is the newamplitude of oscillation] 2 2 2 0 2 0 0 3 x 5 A´ x x 8 4 8 . . . or 0 5 A´ x 2 2 . To find time ofmotion in SHMwe use circularmotion representationof the respectiveSHM.The figure shows the corresponding circularmotion. If at t = 0, the secondmass is added to the oscillating block, it was at a position x0/2 frommean position and moving towards it, and after adding the mass the new amplitude of oscillation changes toA´ and . changes from km to k 2m . Figure shows the corresponding position of particle in circularmotion at t =0 in II quadrant.Whenthis particle P0will reach the positionCafter traversing the angle ., particle P in SHMwill reach itsmean position and similarlywhen P0willreach positionD, P will reach the extreme position on other side. thus the time taken by P to reachmean position froma position of x0/2 frommean position is given as1 0 1 sin x t 2A´ ´ k 2m . . . . .. .. . .
. A t=0 A´ P0 A0 BPv´ D C x0 /2 or 1 1 1 sin 25 2m 2 t sin k k 5 2m . . . . .. .. . . . . . . . . Similarlytime taken by P to reach the left end extreme position is 1 2 t / 2 2m sin 2 ´ k 2 5 . . . . .. . . . . . . . . . . . .. . . .. Example 64: Figure shows a block P of mass m resting on a smooth horizontal surface, attached to a spring of force constant kwhich is rigidlyfixed on thewallon left side, shown in figure.At a distance l to the right of block there is a rigidwall. Ifblock is pushed toward lift so that spring is compressed by a distance 5l/3 and released, it will start m k P l its oscillations. If collision of blockwith thewall is considered to be perfectly elastic. Find the time period of oscillations of the block. Sol. As shown infigure, as the block is released fromrest at a distance 5l/3 fromitsmean position, thiswill be the amplitude of oscillation. But on other side ofmean position block canmove onlyupto a distance l frommean
SIMPLE HARMONIC MOTION www.physicsashok.in 58 position and then it will returnfromthis point withequal velocitydue to elastic collision. Consider figure. If no right wall is present during oscillation block Pwill executing complte SHMon right side ofmean position also up to its amplitude 5l/3. Thus we can observe the block P at a point X at a distance l frommean position (where in our casewall is present), if block passes this position at speed v(which is the speedwithwhich block hits thewallinour cse), after reaching its extreme positionY, it will returnand during return pathitwill cross the positionXwith the same speed v (as displacement frommean position is same). mP 5l/3 l 5l/3 Yv v X (a) Thuswe canstate that inour case of givenproblem, block P is executing SHMbut it skips a part XYXon right half of itsmotion due to electric collisionof the blockwith thewall. IfT0 is the time period of this oscillationwe look at figure (b),which shows the corresponding circularmotion representation. Here during oscillation of point P, particle P0 covers its circularmotion alongECDAF, and fromF it instantlyjumps to E (due to elastic collision of P withwall at X) and again carry out ECDAF and so on, thus we can find the time of this total motionas 2 2sin 1(3/ 5) t km. . . . . . . . . t m 2sin 1 3 k 5 . ... . . .. .. .. . . . . ...(1) l P XY B FA 5l/3 D CEP0 5l/3 (b) We can also find the time period of thismotion bysubtracting the time of FYE fromthe total time period as
2 2cos 1 3t 5 . . . .. .. . . . . or t m 2 2cos 1 3 k 5 . .. . . . .. .. .. . . . . ...(2) these equation-(1) and (2)will result same numericalvalue. Example 65: Figure shows a spring block systemhanging in equilibrium. Ifa velocityv0 is imparted to the block in downward direction. find the amplitude ofSHMof the block and the time afterwhich it will reach a point at half of the amplitude of block. m k h Natural length P Mean position Sol. Initially inequilibriumif block is at a depth hbelowthe natural length of spring then we have mg = kh If at mean position block is imparted a velocity v0, thiswould be themaximumvelocity of block during its oscillation. Ifits amplitude of oscillation isA, then it is given as v0 =A. [where .= km ] or 0 0 A v v mk . . . Now to find the time taken by block to reach its half of amplitude point we
SIMPLE HARMONIC MOTION www.physicsashok.in 59 consider the corresponding circularmotion ofthe SHMas shownin figure. E A/2 v0 A AP P0 C DB Here block Pwill reachto half of its amplitudewhen particle P0willreach point E shown in figure at an angular displacement . frommean position relative to point A, thus time taken by it is t sin 1 .1 2. m k m 6 k . . . . . . . Angular Harmonic Motion : If the angular acceleration of a systembe directlyproportional inmagnitude to the angular displacement and be directed opposite to angular displacement relative to fixed position, itsmotion is said to be angular simple harmonic. Z Z In the figure rotationof the body about z-axis is such that angular acceleration .z is given by .z = � .2. where .2 is a positive constant. Themotion is then angular SHM.As 2 z 2 ddt. . . , the differential equationmaybewritten as 2 2 2 ddt. . .. . Multiplying both the sides by themoment of inertia of the body about the z-axis (I), we get .z = � I.2. where .z is the z-component of net torque . . acting on the body. Table : SHM Features Quantity Linear SHM Angular SHM Equation m 2 2 d x dt + kx = 0 I 22 d
dt. + C. = 0 Angular frequency . . k /m . . C/ I Displacement x =Asin(.t + .) . = .0sin(.t + .) Phase .t + . .t + . Phase constant . . Amplitude A .0 Totalenergyof oscillation 1/2m.A2 = 1/2kA2 1/2I.2.02 = 1/2C.02 Potentialenergy 1/2m.2x2 1/2I.2.2 KineticEnergy 1/2I.2(A2 � x2) 1/2I.2(.02 � .2) Simple Pendulum (mathematical pendiulum) Aheavy pointmass suspended froma rigid support using inextensible, elastic andmassless thread and free to oscillatewithout friction is called a simple pendulum. All these conditions are ideal and cannot be realised completely in particle. Hence such a pendulumis also called mathematical pendulum. In laboratory a thread, a bob, a split cork and stand are arranged to get an approximate simple pendulum.
SIMPLE HARMONIC MOTION www.physicsashok.in 60 Splitcork lhr L= l + h + r friction Zero rigidmassless, inextensible, elastic cord heavy point mass Motion of simple pendulum Let the thread be deflected byan angle . about z-axis as in the figure. The torque abotu z-axis is given by = 0 mgZ-axis l sin l .z = �mgl sin . For small angle, sin . . .. (see table) Table : Some values of sin . ..(degree) ..(radian) sin . 1 0.01745 0.01745 2 0.03490 0.03490 3 0.05235 0.05234 4 0.06981 0.06976 5 0.08727 0.08716 6 0.10472 0.10453 7 0.12217 0.12187 8 0.13963 0.13917 9 0.15708 0.15643 10 0.17453 0.17365 11 0.19199 0.19081 12 0.20944 0.20791 13 0.22689 0.22495 14 0.24435 0.24192 15 0.26180 0.25882 (Values reounded at 5th place) Then, .z = �mgl. (angular SHM) Now, .z = 22 d I dt. = ml2 22 ddt. . = ml2 22 d
dt. = � mgl. 2 2 d g dt. . . . .. . . . l . Comparing thiswith standard equation,we get .2 . gl
SIMPLE HARMONIC MOTION www.physicsashok.in 61 Thus, the angular frequency .of the simple pendulumis given by . . gl The time period of the pendulumis given by T = 2. . . T = 2. gl This expression contains four informations known as laws of simple pendulum. Laws of simple pendulum : (i) T . l (lawof length) (ii) T 1g . (lawof acceleration due to gravity) (iii) T doesn�t depend uponmass (lawofmass) (iv) T doesn�t depend upon amplitude (lawof amplitude) Second�s Pendulum: The simple pendulumhaving time period of 2s is called second�s pendulum.Apendulum clock generallyuses this pendulum. Simple pendulum in accelerated frame : Case I : Let the simple pendulumbe located in a lift. The accelerationDthe lift is a0. a0 . upward : In this casema0 andmg are downward giving effective-g geff = (g + a0) ma0 mg a0 m(g+a0) . eff 0 T 2 2 g g a . . . . . l l Time period has decreased.Apendulumclock in such a lift willbe running fast. Case II : a. directed downward : In this case pseudoforce ma0 acts upward. The net downward force ismg � ma0. This gives effective-g equal to �g � a0�. Then 0 T 2 g a . . .l mg ma0 a0 m(g+a0) < g
SIMPLE HARMONIC MOTION www.physicsashok.in 62 Case III : a0 . is horizontal : In this case ma0 . and mg . are at right angles. They give as effective - g 2 2 eff 0 g . g .a Hence the period is ma0 a0 mg mg eff= m g2+ a 2 0 2 20 T 2 g a . . . l Simple pendulum in a fluid (non-visicous) : In a non-visicous fluid, the vertical buoyant force (m/.).g acts opposite to mg . . The net downward force is mg � (m/.).g. The effective-g is 1 g . .. . . . . .. . mg F = m b g m = Volume displaced m = mass of bob = density of bob = fluid density < eff mg � mg = mg 1 � g g 1 . .. . . . . . .. . T 2 g 1 . . . .. . . . . .. l C41: The length of a simple pendulumis increased by 4%; calculate the percentage increase inits time period. Sol. Here, 1 1 2 1 L L 4L 104L 100 100 . . . Therefore, 2 2 1 1 1 1 T L 104L 104 4 1 T L 100L 100 100 . . . . . . . .. .. 1/2
21 T 1 4 1 2 102 1.02 T 100 100 100 . . . . . . . . . . .. .. .. .. 2 1 T . 1.02 T Increase [or change] in time period = T2 � T1 = T1(1.02 � 1) = 0.02 T1 %Increase (or change) in time period = 1 1 0.02 T 100% 2% T . . C42: Calculate the length of a second pendulumon the surface of themoon (given gm = 1/6ge). Sol. T 2 L 8m . . or 2 2 L 8m . . . m2 2 L 8g 9.8 16.5 cm 6 (3.14) . . . . .
SIMPLE HARMONIC MOTION www.physicsashok.in 63 C43: The length of a second pendulumis increased by 4%, thenwhat will be its newtime period ? Sol. Here 1/2 2 2 1 1 T L 4 1 1.02 T L 100 . . . . . . .. .. . T2 = 1.02 × T1 = 2.04 sec. C44: Calculate howmuch time a second pendulumwill lose in a day, if its length is increased by4%? Sol. If the length is increased 4%then the time period of the seconds pendulumwill become 2.04 sec; hence it will lose 0.04 sec in every 2.04 sec. Therefore it will lose. 0.04 86, 400 2.04 . sec in a day = 1694 sec in a day (nearly). C45: Calculate the frequencyofa secondspenduluminanelevator,whichis accelerated upwardwithanacceleration of 49.0m/sec2. Sol. 2 2 2 1 1 n g 49 9.8 58.8 6 n g 9.8 9.8 . . . . . . . . . . . . 2 1 n n 6 1 6 1.224 per sec. 2 . . . . C46:Asimple pendulum50 cmlong is suspended fromthe roof of a cart accelerating in the horizontal directionwith a = 7m/s2 (figure). Find the period ofsmall oscillations of the pendulumabout its equilibrium angle. 50 cm m a = 7 m/s2 Sol. 2 2 2 2 2 eff g . a . g . (7) . (9.8) . 12.0 m/ s eff T 2 2 0.5 1.28 s g 12 . . l . . . Example 66. The lengthof a simple pendulumis 1m. The bob of the pendulum(mass = 0.10 kg) is releasedwhen the string is horizontal. Calculate in the lowest point ofits path : (i) its kinetic energy, (ii) the tensionin the string. Sol. The amplitude .0 = ./2 is large.We cannot use SHMequations (i) When the bob is released fromhorizontal, the loss in its potential energy at the lowest point =mgl, where l is the lengthofthe pendulum. Let the velocityof bob at the lowest point is v.Then gain inK.E. = 1/2mv2 Fromconservationofmechanical energy mgl = 1/2 mv2
Hence, K.E. = 1/2 mv2 =mgl = 1/100 × 9.8 × 1 = 0.098 J (ii) Tension in the string T �mg =mv2/l Using (i) v . 2gl . 2 . 9.8.1 . 19.6 = 4.42 m/s . T = 1 9.8 1 19.6 100 100 1 . . . = 0.294 N.
SIMPLE HARMONIC MOTION www.physicsashok.in 64 Example 67. The figure shows a simple pendulumwhose bob is an elastic ball. It collides elasticallywith the slant wall. Calculate the period of the pendulumoscillations for small . and . as shown in the figure. >l Sol. Let the stringmake and angle .withthe downward vertical. Let t = 0 at . = .. Then . = . cos .t where . . gl = = � At thewall . = �. �. = . cos .t t 1 cos.1 . ... . . . . . . . The period is double of this value. Hence T 2 cos 1 g . . .. . . . . . . . l Example 68: Figure shows two identical simple pendulums of length l. One is tilled at an angle ..and imparted an initial velocity v1 toward mean position and at the same time other one is projected awayfrom mean positionat a velocityv2 at aninitial angular displacementB. Find the phase difference in oscillations ofthese two pendulums. A B v1 v2 l l Sol. It is given that first pendulumbob is given a velocity v1 at a displacement l. frommean position, using the formula for velocitywe can find the amplitude of its oscillations as 2 2 1 1 V . . A . (l.) [IfA1 is the amplitude of SHMof this bob] As for simple pendulum..= gl we have 2 2 2 1 1 V . g [(A ) . (l.) ] l 2 2 2 1 1 v A g . . . l l ...(1) Similarly ifA2 is the amplitude of SHMof second pendulumbobwe have 2 2 2 2 V . . A . (l.) or 2 2 2 2 2
v A g . . . l l ...(2) Nowwe represent the two SHMs bycircularmotion representation as shown in figure.
SIMPLE HARMONIC MOTION www.physicsashok.in 65 B A A 2 t=0 1 t=0 0 l 2 B I SHM l A1 A0 (a) II SHM (b) In first pendulumat t = 0 the bob is thrown froma displacement l. frommean positionwith a velocity v1 toward mean position.As it ismoving toward mean position, in figure(a), we consider the corresponding circular motion particleA0 of the bobAis second quadrant, as the reference direction of ., we consider anticlockwise.As shown in figure initialphase of bobAis given as 1 1 1 sin A . . . . . . . . . . . . l ...(3) Similarly for second pendulumbobB, its corresponding circular motion particle B0 at t = 0 is considered as shown in figure (b) its initialbase is given as 1 2 2 sin A . . . . . . . . . . l ...(4) As both the pendulums are identical, their angular frequencyfor SHMmust be same, so their phase difference will not changewith time, hence their phase difference can be given as 1 1 2 1 2 1 sin sin A A . . . . . . .. .. . . . . . . . . . . . . . . . . l l Example 69: In previous question if second pendulumbob is thrown at velocityv2 at an angle . frommeanposition but on other side ofmean position. Find the phase difference inthe two SHMs nowas shown in
figure. v A B 1 v2 l l Sol. In this case still amplitudes of the two SHMswill remain same and are given byequation (1) and (2) but when we represent the two SHMs on their corresponding circularmotions, the position of the particle in circular motion in second pendulumis nowdifferent as shown in figure. As shown in figure (a) and (b) the initial phases of the two pendulumbobs are .1 = . � sin�1 1 A . .. .. .. l A1 t=0 A1 l A0 II SHM (a) 2 A2 l B0 B II SHM (b) and .2 = . � sin�1 2 A . . . .. .. l As . for both SHMare same, their phase difference remains constant so it is given as .. = .2 � .1 = sin�1 2 A . . . .. .. l + sin�1 1 A . .. .. .. l
SIMPLE HARMONIC MOTION www.physicsashok.in 66 Example 70: Two pendulums have time periods T and 5T/4. They starts SHMat the same time fromthemean position.After howmany oscillations of the smaller pendulumtheywillbe again in the same phase : (A) 5 (B) 4 (C) 11 (D) 9 Sol: n1T1 = n2T2 or 1 2 n T n 5T 4 . or 12 n 5 n 4 . . n1 = 5 Hence, option (A) is correct. C47: Asimple pendulumhaving a bobofmassmswingswith an angular amplitude of 40º. Showthat the tension in the string is greater thanmg cos 20ºwhen its angular displacement is 20º. Sol. The tension is the string at an angle 20º is T = mg cos 20º + mV2 L or T � mg cos 20º = mV2 0 L . . T >mg cos 20º. COMPOUND PENDULUM This ismadewhenever a rigid bodyis hanging freelyfroma horizontally pivoted axis, as showninfigure.Abodyofmassmispivotedat pointOthrougha horizontal axisAA´.The bodyis hanging freelyunder gravity and inequilibriumposition its centre ofmassCis vertically belowthe suspension point O, at a distance l from O. O A´ A l l c c mg If the body is slightly tilted fromits equilibriumposition by an angle .,mg will exert a restoring torque on it in opposite direction to restore the equilirbium position. Thus restoring torque on bodyin dotted position after tilting is .R = �mg. l sin . [�ve signfor restoring nature] .R = �mgl. [for small ., sin . . .] If its angular acceleration is .,we have I. = �mgl. [Here I is themomentumofinertia ofbodyabout axisAA´] or . = � mgI l . ...(1) Comparing equation(1)with standard differential equation of angular SHMwe get
mg I . . l Thus its time period of oscillation is T 2 2 I mg . . . . . l
SIMPLE HARMONIC MOTION www.physicsashok.in 67 EQUIVALENT LENGTH OF A SIMPLE PENDULUM We knowthat the time period of oscillations of a simple pendulumis given by 2 g . l ...(3) Similarlywe�ve also discussed that the time period of a compound pendulumis given as c 2 I Mg . l ...(4) Where I is themoment of inertia of the rigid body about the axis of rotation and lc is the distance of centre of gravity of body fromthe suspension point (Axis of rotation). ifwe consider a simple pendulumof length leq which has a bob of samemassMas that of rigid bodyof compound pendulumand has the time period same as that of the compound pendulumthen this length of simple pendulumleq is called �Equivalent length of simple pendulum for the given compound pendulum�. For the time periods of the two pendulums to be equal, we have c 2 2 I g mg . l . . l ...(5) Ifk be the radius of gyration ofthe rigid bodyabout an axis passing throughits centre ofmass then themoment of inertia of the rigid bodyabout point of suspension is given as I = Mk2 + Mlc2 ...(6) Nowfromequation-(5), we have2 2 eq c c Mk M g mg. . l l l or 2 eq c c k l . l . l ...(7) Equation-(7) gives the equivalent length ofsimplependulumfor the givencompoundpendulum.One important point to be noted here is if in equation-(7)we replace lc by k2/lc,we get 2 2 eq 2 c c k k k
l . . l / l 2 eq c c k l . . l l ...(8) Which is same as that of equation-(7).Thuswe can saythat if the same rigid body,which is suspended froma point, situated at a distance lc fromcentre of gravityof body,we suspend if froma point at a distance k2/lc from centre ofgravityof the bodyand oscillate like a compound pendulum, its equivalent lengthof simple pendulum remains same or the time period ofoscillation of bodyremains same.Consider figure, a rigid bodyis suspended at a pointOthrougha horizontal axisAA´. HereCis the centre of gravityofthe body. If it oscillates then the time period of smalloscillations can be given as
SIMPLE HARMONIC MOTION www.physicsashok.in 68 2 c c k T 2 g . . . l l ...(9) As discussed inlast sectionwe cansaythat if the same bodyis suspended either fromanypoint on circular arcs PQor RS [as shown in figure(b)] of radius lcwith centre at C(circleM1) or any point on the circle of radius k2/lcwith centre at C(circleM2), the time period of smalloscillations of the bodywill remain same. 1 2 34 R S B´ P Q B M1 M2 lcc k2 /lc (a) (b) O A A´ lc C Here we can develop one property of the bodywhen it is used as a compound pendulum. This is, when a straight line is drawn passing through the centre of gravity of bodyas in figure (b) a line BB´ is drawn. There axis four points on this line (as here points 1, 2, 3 and 4) about which if bodyis suspended, the time period of small oscillation of body remains same. This we can also prove graphically as ifwe plot the time period of oscillation, the curve looks like as shown in figure. As shown in firure, if the body is suspended fromC, from equation-(9), we can see that if lc = 0, time period becomes ., and at lc = k i.e. if the body is suspended froma point at a distance equal to radius of gyration of bodyfromC, the time period of oscillation isminimumand at all other suspension points the time period is higher and ifwe drawa horizontal line in graph at time period T1which ismore thanminimum period, it cuts the graph at four points as shown,whichverifies the statementwe´ve discussed earlier. 1 2 3 4 k k TminT1
T x lc = k lc = 0 lc = k C Fromequation-(9)we can also find the value of lc atwhich this time period has aminimumvalue by equating c dT dl = 0, as 2 2 2 c c c c dT 2 1 k . 1 0 d g k 2 . . . . . . . . . . . l l l l or lc = ± k Which also verifies the experimental result obtained bygraph shownin figure.
SIMPLE HARMONIC MOTION www.physicsashok.in 69 Example 71:Aring of diameter 2moscillates as a compound pendulumabout a horizontal axis passing through a point at its rim. It oscillates suchthat its centremove in a planewhich is perpendicular to the plane of the ring. The equivalent length of the simple pendulumis (A) 2m (B) 4m (C) 1.5 m (D) 3m Sol: I 3mr2 T 2 mg 2mgr . . . l or T 2 3r 2m . . or 2 2 3r g 2g . l . . . 3 r 3 1 1.5 m 2 2 l . . . . Hence option (C) is correct. C48: Calculate the time period of ametre scale if it is free to rotate about a horizontal axis passingthrough its one end. Sol. Total length of themetre scale = 1metre Here, K= (radius of gyration) = 112 l = distance of the C.G. fromone end = 0.5 m T = 2. K2 2 g. l l = 1.6 sec (nearly). C49: Calculate the time period of smalloscillation ofa uniformmetre stick, ifit is suspended through 10 cmmark. Sol. For a metre stick, K2 = L 1 12 12 . . L = 1 metre Hence, K2 1 0.208 12 0.4 . . l . K2 / 0.208 0.4 T 2 2 g 9.8 . . . . l l . . . 1.56 sec (nearly) C50: Auniformcircular disc ofmassmand radius R is suspended froma small hole in the disc. Calculate the minimumpossible time period fromsmalloscillations. Sol.
k2 / T 2 g . . . l l ; T will beminimumwhen (k2/l + l) isminimum,when k / l . l . 0 ;or l . k . R 2 C51: Calculate the lengthofa simplependulumwhose time period is equalto that ofaparticular physicalpendulum. Sol. We have T 2 L 2 I g mg . . . . l or L I m . l
SIMPLE HARMONIC MOTION www.physicsashok.in 70 Example 72: Consider a pulleyofmoment of inertia I supporting a block ofmassm, tied to an idealspring of stiffness k. The block is slightly pulled downward and released. Determine its period of oscillation. R Im k Sol. Inequilibrium mg = T . T = kx0 Also, . mg = kx0 ...(i) Let the block be displaced downward by x andmovingwith velocitydx/dt. Then 0 U 1 k(x x ) mgx 2 . . . 2 K 1 m dx 1 I 2 2 dx 2 . . . . . .. .. or 2 2 2 K 1 m dx 1 I dx 2 dt 2 R dt . . . . . . .. .. .. .. . 2 2 0 2 E 1 k(x x ) mgx 1 dx m I 2 2 dt R . . . . . . . . . .. .. . . . . . Differentiating it and setting to zero, dE 0 dt . k(x + x0) 2 2 2 dx mg dx dx m I d x dt dt dt R dt . . . . . . . .. .. . . . . = 0 k(x + x0) 2 2 2 mg d x m I 0 dt R . . . . . . .. .. ...(ii) Using (i) and (ii),we get 2 2 2 d x k x dt m I /R . . . Comparing it with standard equation ofSHM,
2 2 2 d x x dt . .. , 2 2 k m I /R . . . . m I /R2 T 2 K . . . Example 73:An L-shaped bar ofmassMis pivoted at one of its end so that it can freely rotate in a vertical plane, as shown inthe figure. (a) Find the value of .0 at equilibrium. 0 L B L C A (b) If it is slightlydisplaced fromits equilibriumposition, find the frequency of oscillation. Sol. (a) TakingB as the origin, the co-ordinates of its c.m. are C M L 2 2 L x M M 4 2 2 . . . and C M L y 2 2 L M 4 . . xC M/2 L M/2 c.m. yC B LA C
SIMPLE HARMONIC MOTION www.physicsashok.in 71 0 tan L/ 4 1 3L/ 4 3 . . . 1 0 tan 13 . . . .. .. . . (b) The frequencyof oscillation for a compound pendulumis f 1 mgd 2 1 . . where d = distance of theCOMfromthe point of suspension. I =moment of inertia about the point of suspension. 2 2 d 3L L L 10 4 4 4 . . . . . . . .. .. .. .. 0 L/4 B C A 3L/4 L/4 2 2 2 2 I M L M L M L2 L ML 2 3 2 12 2 2 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . .. . 2 Mg L 10 1 4 f 2 ML 3 . . or f 1 (2.37) g 2 L . . Example 74:Asystemof two identical rods (L�shaped) ofmassmand length l are resting on a peg Pas shown in the figure. If the systemis displaced in its plane by a small angle ., find the period of oscillations : l P l (A) 2 2 3g . l (B) 2 2 2 3g . l (C) 2 23g . l
(D) 3 3g . l Sol: T 2 I mgL . . Here m 2 m 2 2mg 2 I 3 3 3 . . . l l l l/2 l/2 L 45º 45º G Fromfigure : sin 45º L/ 2 . l . L 2 2 . l . 2m 2 2 2 T 2 2 3 mg 3g 2 2 . . . . . l l l Hence option (C) is correct.
SIMPLE HARMONIC MOTION www.physicsashok.in 72 Example 75: Asolid cylinder is attached to ahorizontalmassless spring so that it can rollwithout slipping along a horizontal surface. The spring constant k is 3 N/m. If the systemis released fromrest at a point inwhich the spring is stretched by 0.25m, find (a) the translationalkinetic energyand (b) rotational kinetic energyof the cylinder as it passes the equilibriumposition. (c) Showthat under these conditions the centre ofmass ofthe cylinder executes SHMwith time period T . 2. (3M/ 2k) Sol. If at any position the stretch in the spring is x and the velocityof centre ofmass of the cylinder isV, then as U 1 kx2 2 . , 2 T K 1 MV 2 . and 2 2 R K 1 I 1 MV 2 4 . . . [as L = 12 Mr2 and . = V/r] So totalmechanicalenergy of the system E = KT + KR + U i.e. E = 12 MV2 + 14 MV2 + 12 kx2 i.e. E = 34 MV2 + 12 kx2 ...(1) According to given problemV = 0 if x = 0.25 so 2 E 1 (3) 1 3 J 2 4 32 . . . . . . .. .. Now at equilibriumpositionU = 0 [as x = 0]. so 3 MV2 3 J 4 32 . , i.e. MV2 1 J 8 . So at equilibriumposition : (a) translationKE = (1/2)MV2 = (1/16) J (b) RotationalKE = (1/4)MV2 = (1/32)J (c) As in SHMenergy is conserved, i.e., dE/dt = 0. Fromeqn. (1), we have 3 M. 2V dV 1 k . 2x dx 0
4 dt 2 dt . . or 2 2 3 d x 1 M kx 4 dt 2 . . 2 2 as dx V and dV d x dt dt dt . . . . . . . . or 2 2 2 d x x dt . .. with 2 2k 3M . . This is the standard equation ofSHMwith time period T = (2./.). So here T 2 3M 2k . . Example 76: Acylinder of radius r andmassmrests on a curved path of radiusRas shown in Fig. Showthat the cylinder can oscillate about the bottompositionwhen displaced and left to itself. Find the period of oscillation. Assume that the cylinder rollswithout slipping.
SIMPLE HARMONIC MOTION www.physicsashok.in 73 Sol. Restoring torque acting oncylinder after a smalldisplacement ., about anaxis through point ofconstant Owith the curved path, . = � mg sin . r . mgr . (as for small ., sin . = .) Angular accelerationof cylinder, 2 2 ddt. . . r. = (R � r). (R r) r. . . . R A Ar m Consequently 2 2 2 2 d (R r) d dt r dt . . . . . . Moment of inertia about contact point, 3mr2 I 2 . 2 2 2 3 (R r) d �mgr mr 2 r dt . . . . or 2 2 d g 3 0 dt (R r) 2 . . . . . Hence g 3 (R r) 2 . . . or 3 (R r) 6(R r) T 2 2 g g
. .
. . . . Example 77. Two identical cylinders C1 and C2 are placed with their axes horizontal and in the same horizontal plane. Theyare rotated uniformly about their once axes inclockwise andanticlockwise directionrespectively. AuniformplankAB ofmassMis kept restingof these rotating cylinders. C1 C2 A B 2L mg If the plank is displaced slightlyfromits equilibriumpositionthe demostrate that it performs simple harmonic oscillations. Calculate the time period of these oscillations. The axes of the cylinders are separated bya distance 2Land the co-efficient of friction betweenthe plank and thewheels is µ. Sol. Let the plank be displaced to the right (x-axis) through a small distance x and released. The normal forces executed bythe two cylinders not equal. C1 C2 A B (L+x) mgx X(L�x) G G´ Let thembeN1 andN2.The frictionalforces are µN1 and µN2 to the right and left respectively. Considering the vertical forces, N1 + N2 = mg ...(1) Also, considering the rotational equilibriumof the plank about C. N1(L + x) = N2 (L � x)
SIMPLE HARMONIC MOTION www.physicsashok.in 74 (N1 � N2) L = � (N1 + N2)x ...(2) Hence resultant force acting on the plank along x-axis : Fx = µN1 � µN2 = µ(N1 � N2) Fx = �µMgx/L [using (1) and (2)] This is the restoring force, acting towards the centre. Comparingwith standard equation of SHM Fx = M.2x Here time-period of SHMis T 2 2 L mg . . . . . . TORISIONAL PENDULUM In aTorisionalPendulumanobject is suspended fromawirewith rigiditycoefficient C. Ifsuch awire is twisted by an angle ., due to its elasticity it exerts a restoring torque . = C. on the twisted object attached to it. Ageneral torisional pendulumis shown in figure. Here a discDof radiusRandmassMis attached to a stiff wirewhose other end is suspended fromceiling as shown. Fromthe equilibriumposition of this disc if it is twisted byan angle . as shown, thewire applies a restoring torque on it,which is given as .R = �C. [�ve sign, for restoring nature] R M D If during restoring motion the angular acceleration of disc is ., we can write . = I.where I is themoment of inertia ofdisc about its central axis, thuswe have I. = �C. or . = � CI . ...(1) Equation-(1) resembleswith the basic differential equation of SHMin angular formthus we can state the angular frequencyofthis SHMis . . C/ I ...(2) Thus the period of SHMis T 2 2 I C . . . . . In the above cases of some pendulumswe�ve discussed, howto find the angular frequency and time period of a bodyin SHM. Nowwe take some similar examples of physical situations inwhich an object is in SHM. C52: Acircular disc ofmass 1.5 kg and radius 15 cmis suspendedwiththe help of awire,whose one end is fixed to the centre of the disc.When a torque of 15 × 10�2 Nm/rad is applied, then the disc begins to oscillate. Calculate the time period for small displacement. Sol. M.O.I. of the dis, I 1 Mr2 2 .
and 2 22 T 2 I 2 Mr 2 (1.5)(0.15) 2.107 sec. C 2C 2 15 10. . . . . . . . . .
SIMPLE HARMONIC MOTION www.physicsashok.in 75 Example 78: Figure shows a torisional pendulumconsists of a uniformdiscDofmassM and radiusRattached to a this rod oftorisional constant C. Find the amplitude and the energyofsmalltorisionaloscillations ofthedisc, ifinitiallythe discwas imparted angular speed .0. D R Sol. We knowfor a torisionalpendulumangular frequencyof smalloscillations is given as . . C/ I Where I = 1/2MR2, thus 2 2C MR . . At it is given that frommean position the disc is imparted as\an angular speed .0, if the angular amplitude of oscillations of disc is .,we have .0 = .. or 2 2C MR . . Thus angular amplitude is given as 2 0 MR 2C . . . For angular SHMthe totaloscillation energyis given as 2 2 T E 1 I 2 . . . [If Umin = 0] Thus here oscillation energyof the disc is given as 2 2 2 T 2 0 E 1 1MR 2C MR 2 2 MR 2C . . . . . . . . . . . . . . . . . . . . or 2 2 T 0 E 1 MR 4 . . SUPERPOSITION OF PERPENDICULAR SHM�s : Let us consider two SHM�s acting on the same particle along perpendicular directions.The resultingmotion of the particlewill be obtained by eliminating t fromthe expressions for xand y. Let x =A1 sin .t ...(1) y =A2 sin(.t + .) ...(2) From(2) y x P(x, y)
y/A2= sin .t cos . + cos .t sin . 2 y A = 1 x A cos . + 221 1 xA . sin . [using (1)] 2 2 2 2 2 1 1 y x cos 1 x sin A A A . . . . . . .. . . . . . . . . .
SIMPLE HARMONIC MOTION www.physicsashok.in 76 2 2 2 2 2 2 2 2 2 2 1 1 2 1 y x 2xy x cos cos sin sin A A A A A . . . . . . . . 2 2 2 2 2 2 1 1 2 y x 2xy cos sin A A A A . . . . . This is anequation to an ellipse. Special cases : Case (i) . = 0. In this case we have 2 2 2 2 2 1 1 2 y x 2xy 0 A A A A . . . 2 2 1 y x 0 A A . . . . . . . . . 21 A y x A . The pathis a straight line segment (see figure) A1 A2 �A2 �A1 21 A tan A . . Thismotion isSHMwith amplitude 2 2 1 2 A . A Case (ii) . = .. In this case, we have 2 2 2 2 2 1 1 2 y x 2xy 0 A A A A
. . . A1 A2 �A2 �A1 Y X tan = � A2A1 21 A y x A . . The path is a straight line segment inclined at and .where the amplitude of oscillationis 2 2 1 2 A . A . Case (iii) . = ./2. In this case, we have 2 2 2 2 2 1 x x 1 A A . . �A1 �A2 X Y A2 (x, y)A1 This is an ellipse. Case (iv) . = ./2, A1 =A2. In this case, we have x2 + y2 = A2 This is a circle. A y x
SIMPLE HARMONIC MOTION www.physicsashok.in 77 Table : Superposition of SHMs Superposition Superposition features Same line x = x1 + x2 =Asin (.t + .); A = 2 2 A1 . A2 . 2A1A2 cos. tan . = 2 1 2 A sin A A cos . . . Perpendicular lines 2 2 2 2 2 2 1 1 2 y x 2xy cos sin A A A A . . . . . Lissajous figure : This figure is the locus of particle inmotion under action ofmutually perpendicular SHM�s having different frequencies. It is a closed path known asLissajous figure. Example 79: Find the amplitude and initial phase of a particle inSHM, whosemotion equation is given as y =Asin .t + B cos .t Sol. Here inthe given equationwe canwrite A = R cos . ...(12) and B = R sin . ...(13) Thus the givenequation transforms to y = R sin(.t + .) ...(14) Equation-(14) is a general equation of SHMand here R is the amplitude of given SHMand ..is the initial phase of the oscillating particle at t = 0. HereRis given by squaring and additing equation-(12)&(13) R = A2 . B2 Initial phse ..can be given by dividing - (13) &(12) tan ..=B/A . = tan�1 (B/A) Note : Equations y =Aei(.t + .) and y =Asin .t + B cos .t, we can also represent the general equation of SHM. Example 80. Two simple harmonicmotions are represented bythe following equations y1 = 10 sin(./4) (12t + 1) y2 = 5 (sin 3.t + 3 cos 3.t) Here t is in seconds. Find out the ratio of their amplituds.What are the time periods of the two motions ? Sol. Given equations are y1 = 10 sin(./4) (12t + 1) ...(1) y2 = 5(sin 3.t + 3 cos 3.t) ...(2) We recast these in the formof standard equation ofSHMwhich is y =Asin (.t + .) ...(3) Equation (1)maybewritten as y1 = 10sin[(12.t/4) + (./4)]
SIMPLE HARMONIC MOTION www.physicsashok.in 78 y1 = 10sin[(3.t + ./4)] ...(4) Comparing eqn. (4) with eqn. (3)we have Amplitude of first SHM=A1 = 10 cms�1 and .1 = 3. . Time period of firstmotion T1 = 2./.1 = 2./3. = (2/3) s Eq. (2)mayalso bewritten as y2 = 5sin 3.t + 5 3 cos 3.t Let us put 5 = A2 cos . ...(5) and 5 3 =A2 sin . ...(6) Then y2 =A2cos . sin 3.t +A2sin . cos3.t y2 =A2 sin(3.t + .) ...(7) Squaring (5) and (6) and adding,we have 2 2 A2 . [(5) . (5 3) ] .10 cm i.e. amplitude of second SHM= 10 cm and time period of second SHM= T2 2./.2 = 2./3. = 2/3 s Thus, the ratio of amplitudes A1 :A2 = 1 : 1 and periodic times are T1 = T2 = (2/3) s Example 81:The resulting amplitudeA´ and the phase of the vibrations . S =Acos(.t) + A2 cos t 2. .. .. . . . . + A4 cos(.t + .) + A8 cos t 32. .. .. . . . . =A´ cos (.t + .) are _______ and ________ respectively. Sol. . x a A A 3A 4 4 . . . and y a A A 4A A 3A 2 8 8 8 . . . . . . 2 2 2 2 x y A a a 9A 9A 16 64 . . . . x ya2 = A2 a a1= A 3 = A4 a4 = A82 2
2 2 x y A a a 36A 9A 64. . . . A= 45A 3 5 A 8 8 . and yx a 1 tan a 2 . . . tan 1 12 . . . .. .. . .
SIMPLE HARMONIC MOTION www.physicsashok.in 79 Example 82:Aparticle is subjected to twomutuallyperpendicular simple harmonicmotions such that its xand y coordinates are given by x = 2 sin .t ; y = 2 sin t 4 . .. .. . . . . The parthof the particlewillbe : (A) an ellipse (B) a straight line (C) a parabola (D) a circle Sol: Hence sin t x2 . . . 4 x2 cos t 2. . . . y 2sin t cos 2 cos t sin 4 4 . . . . . . or y 2 sin t 2 cos t 2 2 . . . .x 4 x2 y 2 2 2 2. . . . or x 4 x2 y 2 2 . . . or 2 y . x . 4 . x2 or 2y2 + x2 � 2 2 xy = 4 � x2 or 2y2 + 2x2 � 2 2 xy = 4 or x2 + y2 � 2 xy = 2 This is an equation of ellipse. Hence option (A) is correct. Example 83: The amplitude of the vibrating particle due to superposition of two SHMs, y1 = sin t 3. .. .. . . . . and y2 = sin .t is : (A) 1 (B) 2 (C) 3 (D) 2 Sol: x a 1cos 1 3 3 2 . . . . y 1 a a sin 1 3 3 3 2 2 . . . . . y x
a1 = 1 a2 = 1 /3 . 2 2 2 2 x A a ay 3 3 2 2 . . . . . . . . . . . . . . . . 9 3 3 4 4 . . . Hence option (C) is correct.
SIMPLE HARMONIC MOTION www.physicsashok.in 80 Example 84: Two simple harmonic motions y1 =Asin .t and y2 =Acos .t are supre imposed on a particle of massm. The totalmechanical energy of the particle is : (A) 12 m.2A2 (B) m.2A2 (C) 14 m.2A2 (D) zero Sol: y1 =Asin .t y2 =Asin (.t + ./2) ax = A ay = A . a = Resultant amplitude = 2 2 ax . ay . A2 . A2 . 2 A 90º y x a2= A a1= A . E = 1/2ma2.2 E = 1/2 m2a2.2 or E . mA2.2 Example 85:Verticaldisplacement of a plankwith a body ofmassmon it is varying according to lawy= sin .t + 3 cos .t. Theminimumvalue of .forwhich themass just breaks off the plank and themoment it occurs first after t = 0 are given by : (y is positive vertically upwards) (A) g , 2 2 6 g . (B) g , 22 3 g. (C) g , 2 2 3 g . (D) 2g , 23g. Sol: amax = .2A or g = .2A or g = .2 2 2 1 2 A . A or g = .2 12 . 3 or g = 2.2 . g2 . . and if occurs at extreme at upper extreme position. . A= sin .t + 3 cos .t or 2 = sin .t0 + 3 cos .t0 or 2 = 2sin(.t0 + .) But tan 3 3 1
. . . . 60º 3. . . . or 0 2 2sin t 3. .. . .. . . . . or 0 t 3 2 . . . . . or 0 t 2 3 6 . . . . . . . or 0 t 2 6 6 g . . . . . Hence option (A) is correct.
SIMPLE HARMONIC MOTION www.physicsashok.in 81 THINKING PROBLEMS 1. Which ofthe following functions are (a) aperiodic (b) periodic but not simple harmonic (c) simple harmonic : (i) sin 2.t (ii) 1 + cos 2.t (iii) a sin .t + b cos .t (iv) sin .t + sin 2.t + cos 2.t (v) sin3 .t (v) log (1 + .t) (vii) exp (�.t) ? 2. Can everyoscillatorymotion be treated as simple harmonicmotionin the limit ofsmall amplitude ? 3. What would happen to themotionof an oscillating systemif the sign of the force termin equation F = � kx is changed ? 4. (a) Can amotion be periodic but not oscillatory ? (b) Can amotion be oscillatory but not simple harmonic ? If your answer is yes, give an example and if not, explainwhy ? 5. Can a bodyhave accelerationwithout having velocity ? 6. Determine whether or not the following quantitie scan be in the same directionfor a simple harmonicmotion : (a) displacement and velocity, (b) velocityand acceleration, (c) displacement and acceleration. 7. (a) Canwe ever construct a simple pendulumstrictlyaccording to its definition ? (b) Is themotion of a simple pendulumlinear SHMor angular SHM? 8. Agirl is swinging in a sitting position.Howwill the period of swing be affected if : (a)The girl stands upwhile swinging ? (b)Another girl of samemass comes and sits next to her ? 9. Ahollowmetal sphere is filledwithwater and a small hole ismade at its bottom. It is hanging bya long thread and ismade to oscillate.Howwill the period ofoscillation change ifwater is allowed to flowthrough the hole till the sphere is empty ? 10. The resultant of two simple harmonicmotions at right angles and of the same frequency is always a circular motion.True or false?Explain. SOLUTION OF THINKING PROBLEMS 1. (a) Functions (vi) log (1 + .t) and (vii) exp (�.t) increase (or decrease) continuouslywith time and can never repeat themselfso are aperiodic. (b) Function (ii) (1+ cos 2.t), (iv) sin .t + sin 2.t +cos 2.t and (v) sin3 .t are periodic [i.e., f(t + T) = f(t)] with periodicity (./.), (2./.) and (2./.) respectively but not simple harmonic as for these functions (d2y/dt2) is not . � y. (c) Functions (i) sin 2.t and (iii) a sin .t + b cos .t, i.e., (a2 + b2)1/2 sin [.t + tan�1 (b/a)] are simple harmonic [with time period (./.) and (2./.) respectively] as for these (d2y/dt2) ..� y. 2. No, not at all. In the limit of small amplitudes onlythose oscillatorymotions can be treated as simple harmonic for which the restoring force (or torque) becomes linear. For example, the oscillatorymotion of a simple or spring pendulumormotion of atoms inamolecule becomes simple harmonic in the limit ofsmall amplitude as restoring force (or torque) becomes linear while in case of oscillatorymotion of a ballbetween two inclined planes or two perfectly elasticwalls, themotion does not become simple harmonic evenfor vanishingysmall amplitude.
3. If the sign is changed in the force equation, accelerationwill not be opposite to displacement and hence the particlewillnot oscillate, butwillaccelerate inthedirectionofdisplaement. So themotionwillbecome accelerated translatory. However, equations ofmotion cannot be applied to analyse themotion as acceleration (= .2y) is not constant. Mathematical analysis shows that in this situation both velocity and displacement will increase exponentially withtime.
SIMPLE HARMONIC MOTION www.physicsashok.in 82 4. Yes;Uniformcircularmotion, (b)Yes;When a ball is thrown froma height on a perfectlyelastic plane surface themotion is oscillatory but not simple harmonic as the restoring force F =mg = constt. and not F . � x] 5. Yes; In SHMat extreme position velocityis zero but acceleration ismaximum= .2A. 6. (a)Yes;whenthe particle ismoving fromequilibriumposition to extreme position, (b)Yes;whenthe particle is moving fromextreme to equilibriumposition, (c)No; as inSHMdisplacement is alwaysopposite to acceleration.] 7. (a) No, (b)Angular SHM. 8. (a) decreases, (b) unchanged 9. Time periodwill first increase, reaches amaximumand thenwill decrease. 10. False. The generalmotion is elliptical.Acircle is a particular case of ellipsewhen itsmajor axis is equal to its minor axis. In general, themotion is described by 2 2 2 2 2 d y 2xy cos sin a b ab . . . . . When a = b and . = ./2, x2 + y2 = a2 ASSERTION-REASON TYPE Astatement of Statement-1 is given and a Corresponding statement of Statement-2 is given just belowit of the statements,mark the correct answer as � (A) If both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation ofStatement-1. (B) If both Statement-1 and Statement-2 are true and Statement-2 isNOT correct explanation ofStatement-1. (C) If Statement-1 is true but Statement-2 is false. (D) If both Statement-1 and Statement-2 are false. (E) IfStatement-1 is false but Statement-2 is true. 1. Statement-1 : The force acting on a particle moving along x�axis is F = �kx + v0t, where k is a constant. Statement-2 :To an observermoving along x-axiswith constant velocity v0, it represents SHM. 2. Statement-1 : Aparticle executing simple harmonicmotioncomes to rest at the extreme positions. Statement-2 :The resultant force on the particle is zero at these positions. 3. Statement-1 : Soldiers are asked to break stepswhile crossing the bridges. Statement-2 :The frequencyofmarchingmaybe equalto the naturalfrequencyof bridge andlead to resonance which can break the bridge. 4. Statement-1 : The functionY= cos2 .t + sin .t does not represent a simple harmonicmotion. Statement-2 : Sumof two harmonic functionmaynot be harmonicmotion. 5. Statement-1 : Ablock ofmassmis attached to a spring inside a trolley of mass 2mas shown in figure, all surfaces are smooth. Spring is stretched and released. Both trolleyand block oscillate simple harmonicallywith same time period and same amplitude. m 2m Statement-2 : Inabsence ofexternalforce,centreofmass ofsystemofparticles
does not accelerate.
SIMPLE HARMONIC MOTION www.physicsashok.in 83 MATCH THE COLUMN 1. Asimple harmonic oscillator consists ofa block attached to a spring with k= 200N/m.Theblock slides ona frictionless horizontal surface, with equilibriumpoint x = 0.Agraph of the block�s velocity v as a function of time t is shown. Correctlymatch the required information in the column Iwith the values given in the column II. (use .2= 10) 2 �20 0.10 0.20 t(s) V(m/s) Column-I Column-II (A) The block�smass in kg (P) �0.20 (B) The block�s displacement at t = 0 inmetres (Q) �200 (C) The block�s acceleration at t = 0.10 s inm/s2 (R) 0.20 (D) The block�smaximumkinetic energyin Joule (S) 4.0 2. Matchthe following Column-I Column-II (A) Linear combination of two SHM�s (P) T = 2. Rg (R is radius of the earth) (B) y =Asin .1t +Asin(.2t + .) (Q) SHMfor equal frequencies and amplitude (C) Time periodof a pendulumofinfinite length. (R) Superpositionmay not be a SHMalways. (D) Maximumvalue of time period ofan oscillating (S) amplitude will be 2 A for .1 = .2 and pendulum. phase difference of ./2. 3. Matchthe following Column-I Column-II (A) Aconstant force acting along the line of (P) The time period SHMaffects (B) Aconstant torque acting along the arc of (Q) The frequency angular SHMaffects. (C) Aparticle falling on the block executing SHM (R) themeanposition when the later crosses themean position affects (D) Aparticle executing SHMwhen kept on a (S) The amplitude uniformly accelerated car affects.
SIMPLE HARMONIC MOTION www.physicsashok.in 84 LEVEL � 1 1. Two particlesP andQdescribe SHMof same amplitude a, same frequencyf along the same straight line.The maximumdistance between the two particles is a 2 . The phase difference between the particle is : (A) zero (B) ./2 (C) ./6 (D) ./3 2. Arod of length l is inmotion such that its endsA and B aremoving along x-axis and y-axis respectively. it is given that d 2 dt. . rad/s always. P is a fixed point on the rod. LetMbe the projection ofPon x-axis. For the time interval inwhich . changes from 0 to ./2, choose the correct statement, y x P M l A B (A) The acceleration ofMis always directed towards right (B)Mexecutes SHM (C)Mmoveswith constant speed (D)Mmoveswith constant acceleration 3. The coefficient of frictionbetween block ofmassmand 2mism= 2 tan .. There is no friction between blockofmass 2mand inclined plane. Themaximumamplitude of two block systemfor which there is no relativemotion between both the blocks. k m2m (A) g sin . km (B) mg sin k . (C) 3mg sin k . (D) None of these 4. Graph shows the x(t) curves for three experiments involving a particular spring-block systemoscillating in SHM. The kinetic energy of the systemismaximumat t = 4 sec. for the situation : 1 2 4sec. t(in sec) 3 0 x (A) 1 (B) 2 (C) 3 (D) Same in all 5. Aparticle ofmassm= 2 kg executes SHMin xy-plane between pointsAand B underaction of force x y
F . F �i . F �j . .Minimumtime taken by particle to move fromAto B is 1 sec.At t = 0 the particle is at x =2 and y=2.Then Fx as function of time t is A(2, 2) y x B(�2, �2) (A) �4.2 sin .t (B) �4.2 cos .t (C) 4.2 cos .t (D) None of these 6. The speed v of a particlemoving along a straight line,when it is at a distance (x) froma fixed point of the line is given by v2 = 108 � 9x2 (allquantities are in cgs units) : (A) themotionis uniformlyaccelerated along the straight line (B) themagnitude of the acceleration at a distance 3cmfromthe point is 27 cm/sec2 (C) themotionis simple harmonic about the given fixed point. (D) themaximumdisplacement fromthe fixed point is 4 cm.
SIMPLE HARMONIC MOTION www.physicsashok.in 85 7. The time period of an ideal simple pendulumis given by : T = 2. l / g The time period of actual simple pendulum(T´) is slightlydifferent due to smalldamping (friction). Then (A) T´ > T (B) T´ < T (C) T = T´ (D) None 8. Asmallbob of a simple pendulumcontainswater. Its periodic time isT.During oscillation, the temperature of surrounding is lowered so that water gets freezed. The newtime period is T´. Then : (A) T > T´ (B) T < T´ (C) T = T´ (D) None 9. Two simple pendulaAand B are shown in the figure. The phase difference between A and B is : (A) . (B) ./2 l l (C) 0 (D) ./4 A B 10. Abodyexecutes simple harmonicmotionunder the action of a force F1witha time period (4/5) second. If the force is changed to F2 it executes SHM with a time period (3/5) second. If both the forces F1 and F2 act simultaneously in the same direction on the body, its time period in second is : (A) 12/25 (B) 24/25 (C) 35/24 (D) 25/12 11. What should be the displacement of a simple pendulumwhose amplitude is A, at which potential energy is 1/4th of the total energy ? (A) A/ 2 (B)A/2 (C)A/4 (D) A/ 2 2 12. The potential energyUof a particle is given byU= {20 + (x � 4)2} J.Totalmechanical energy of the particle is 36 J. Select the correct alternative(s) (A) The particle oscillates about point x= 4m (B) The amplitude of the particle is 4m (C) The kinetic energy of the particle at x = 2mis 12 J (D) Themotion of the particle is periodic but not simple harmonic 13. Arod ofmassMand lengthL is hinged at its centre ofmass so that it can rotate in a vertical plane. Two springs each of stiffness k are connected at its ends, as shown in the figure. The time period of SHMis L, M k Hinge k (A) 2 M6k . (B) 2 M3k . (C) 2 ML k . (D) M6k . 14. Aparticlemoves along the x-axis according to the equation x =Asin2 .t (A) The particle oscillates about the origin (B) The particle oscillates about t
he point x =A (C) The particle oscillateswith a period T = ./. (D) The particle oscillateswith amplitudeA/2 15. In the arrangement shown in figure the pulleys are smooth and massless. The spring k1 and k2 aremassless. The time period of oscillation of themassmis m k k1 (A) 2 1 2 2 m k k . . (B) 1 2 2 2m k k . . (C) 1 2 1 2 m(k k ) 2 2k k. . (D) 1 2 1 2 m(k k ) 2 k k. .
SIMPLE HARMONIC MOTION www.physicsashok.in 86 16. The equationofSHMofa particle oscillating alongthe x-axis is givenbyx=3 sin t 6 . .. . . . . . . cm.The acceleration of the particle at t = 1 s is (A) �1.5 .2 cms�2 (B) 2.6 .2 cms�2 (C) �2.6 .2 cms�2 (D) 1.5 .2 cms�2 17. For the systemshown in the figure, initiallythe spring is compressed bya distance a fromits natural lengthandwhenreleased, itmoves to a distance b fromits equilibriumposition. The dicrease in amplitude for one half - cycle (�a to +b) is k µ m a b (A) µmg k (B) 2µmg k (C) µmg 2k (D) none of these 18. Themotion of a particle is given x =Asin .t +B cos .t. Themotion of the particle is (A) not simple harmonic (B) simple harmonicwith amplitudeA+B (C) simple harmonicwith amplitude (A+B)/2 (D) simple harmonicwith amplitude A2 . B2 19. Two massesm1 and m2 are suspended together by amassless spring of force constant K.When the masses are in equilibriumis removedwithout disturbing the system. The amplitude of oscillation is (A)m2g/K (B)m1g/K (C) (m1+m2)g/K (D) (m1�m2)g/K 20. Two bodiesMandNof equalmasses are suspended fromtwo seperatemassless springs of spring constants k1 and k2 respectively. If the two bodies oscillate vertically such that theirmaximumvelocities are equal, the ratio of the amplitude of vibration ofMto that onNis : (A) k1/k2 (B) 1 2 k / k (C) k2/k1 (D) 2 1 k / k 21. Aparticle executing SHMwhilemoving fromone extremityis found at distance x1, x2, x3 fromthe centre at the ends of three successive seconds, the period is (A) . (B) ./2. (C) 2./. (D) 1/. where ..= cos�1(x1 + x3)/2x2 22. One end of a spring of force constant k is fixed to a verticalwall and the other to a body ofmassmresting on a smooth horizontal surface. There is anotherwall at a distance x0 fromthe body. The spring is then compressed by 2x0 and released. The time taken m x0 to strike thewall is (A) 1 k 6 m . (B) mk (C) 2 m 3 k . (D)
k 3 m . 23. In the above problemthe velocitywithwhich the body strikes the other wall is (A) 0 x mk (B) 0 3k x m (C) 0 k x m (D) 0 2x mk 24. Aparticlemoves intheX�Yplane according to the equation: r . .2�i . 4�j. . sin .t. Themotion ofthe particle is: (A) parabolic (B) circular (C) straight line (D) None of these
SIMPLE HARMONIC MOTION www.physicsashok.in 87 25. The speed v of a particlemoving along a straight line, when it is at a distance x froma fixed point on the line is given byV= (108 � 9x2)1/2.All quantities are in SI units. (A) Themotionis uniformlyaccelerated along the straight line. (B) Themagnitude of acceleration at a distance 3mfromthe fixed point is 27m/s2. (C) Themagnitude of acceleration at a distance 3mfromthe fixed point is 9m/s2. (D) Themaximumdisplacement of the particle fromthe fixed point is 4m. 26. The potential energyUof a particle is given byU= (2.5X2 + 100) Joule. If themass of the particle is 0.2 kg, then: (A) themotion of the particle is SHM. (B) themean position isX= 0. (C) angular frequency of the oscillation is 5 rad/s (D) the time period of oscillation is 1.26 sec. 27. In SHM: (A) displacement and velocitymaybe in the same direction. (B) displacement and velocity can never be in the same direction. (C) velocity and accelerationmay be in the same direction. (D) displacement and acceleration can never be in the same direction. 28. Aparticlemoves in theX�Yplane according to the equation r . .3�i . �j.cos5 t . . Themotionof theparticle is: (A)Along a straight line (B) along an ellipse (C) periodic (D) along a parabola 29. The potential energyof a particle ofmass 2 kg,moving alongthe x�axis is given byU(x) = 16(x2 � 2x) J,where x is inmetres. Its speed at x = 1 mis 2 ms�1 : (A) Themotion of the particle is uniformlyaccelerated (B) Themotion of the particle is oscillatory fromx = 0.5mto x= 1.5m. (C) Themotion of the particle is simple harmonic (D) The period of oscillation of the particle is ./2 s. 30. If a SHMis given by y= (sin .t + cos .t)m,which of the following statements is/are true ? (A) The amplitude is 1m (B) The amplitude is 2 m. (C) Particle starts itsmotion fromy= 1m. (D) Particle starts itsmotion fromy= 0m. 31. Three simple harmonic motions in the same direction having the same amplitude a and same period are superposed. If each differs in phase fromthe next by45º, then : (A) the resultant amplitude is (1 + 2 ) a. (B) the phase of the resultantmotion relative to the first is 90º. (C) the energy associatedwith the resultingmotion is (3 + 2 2 ) times the energy associatedwith any single motion. (D) the resultingmotion is not simple harmonic. 32. Alinear harmonic oscillator offorce constant 2 × 103N/mand amplitude 0.01mhas a totalmechanical energy of 160 J. Its : (A)maximumpotentialenergyis 100 J. (B)maximumkinetic energy is 100 J. (C)maximumpotential energy is 160 J. (D)minimumpotentialenergyis zero.
SIMPLE HARMONIC MOTION www.physicsashok.in 88 33. Auniformcylinder of lengthLandmassMhaving cross-sectional areaAis suspended,with its lengthvertical, froma fixed point by amassless spring, such that it is half-submerged in a liquid of density . at equilibrium position.When the cylinder is given a small downward push and released it starts oscillating verticallywith small amplitude. If the force constant of the spring is k, the frequencyof oscillation of the cylinder is : (A) 1/ 2 1 k A g 2 M . . . . . .. .. (B) 1/ 2 1 k A g 2 M . . . . . .. .. (C) 1 k gL2 1/ 2 2 M . . . . . .. .. (D) 1/ 2 1 k A g 2 A g . . . . . .. . .. 34. A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is V(x) = k|x|3where k is a positive constant. If the amplitude of oscillations is a, then its time period Tis (A) proportional to 1/ a (B) independent of a [JEE, 98] (C) proportional to a (D) proportional to a3/2 35. Aparticle free to move along the x-axis has potential energy given by U(x) = k[1 � exp(�x2)] for �. < x < +., where k is a positive constant of appropriate dimensions. The (A) at point awayfromthe origin, the particle is in unstable equilibrium. (B) for any finite nonzero value of x, there is a force directed awayfromthe origin. (C) ifits totalmechanical energyis k/2, it has itsminimumkinetic energyat the origin. (D) for smalldisplacements fromx= 0, themotion is simple harmonic. [JEE, 99] 36. Three simple harmonic motions in the same direction having the same amplitude a and same period are superposed. If each differs in phase fromthe next by45º, then [JEE, 99] (A) the resultant amplitude is (1. 2) a (B) the phase of the resultantmotion relative to the first is 90º (C) theenergyassociatedwiththeresultingmotionis (3. 2 2) timesthe energyassociatedwithanysinglemotion. (D) the resultingmotion is not simple harmonic. 37. The period of oscillation of simple pendulumof lengthL suspended fromthe roof of a vehiclewhichmoves without friction down an inclined plane of inclination . is given by [JEE, 2000] (A) 2 L gcos . . (B)
2 L gsin . . (C) 2 Lg . (D) 2 L g tan . . 38. Aparticle executes simple harmonicmotionbetween x= �Aand x= +A.The time taken for it to go from0 to A/2 is T1 and to go fromA/2 toAisT2. Then [JEE(Scr), 01] (A) T1 < T2 (B) T1 > T2 (C) T1 = T2 (D) T1 = 2T2 39. Aparticle is executing SHMaccording to y = a cos .t. Thenwhich of the graphs represents variations of potential energy: [JEE(Scr), 03] I II i P.E. III IV P.E. x (A) (I)&(III) (B) (II)&(IV) (C) (I)&(IV) (D) (II)&(III)
SIMPLE HARMONIC MOTION www.physicsashok.in 89 40. Ablock Pofmassmis placed on a frictionless horizontal surface.Another blockQ of same mass is kept on P and connected to the wallwith the help of a spring of spring constant k as shownin the figure. µS is the coefficient of friction QP k µS between P andQ. The blocksmove together performing SHMof amplitudeA. Themaximumvalue of the friction force between P andQis [JEE, 04] (A) kA (B) kA/2 (C) zero (D)µSmg 41. Asimple pendulumhas time period T1.When the period of suspensionmoves vertically up according to the equation y= kt2where k = 1m/s2 and t is time then the time period of the pendulumis T2 then (T1/T2)2 is [JEE(Scr), 05] (A) 5/6 (B) 11/10 (C) 6/5 (D) 5/4 42. Function x =Asin2.t + B cos2.t + C sin .t cos .t represents SHM [JEE, 06] (A) for any value ofA, B and C (expect C = 0) (B) ifA= �B; C = 2B, amplitude = | B 2 | (C) ifA= B; C = 0 (D) ifA= B; C = 2B, amplitude = | B | 43. Astudent performs an experiment for determination of g = (4.2l/T2)l = 1mand he commits anerror of .l.For the takes the time on n oscillationswiththe stopwatch of least count .T and he commits a humanerror of 0.1 sec. For which of the following data, themeasurement of gwillbemost accurate ? [JEE, 06] . l .T n Amplitude of oscillation (A) 5mm 0.2 sec 10 5mm (B) 5mm 0.2 sec 20 5mm (C) 5mm 0.1 sec 20 1mm (D) 1mm 0.1 sec 50 1mm 44. Thex�tgraphofaparticleundergoing simpleharmonicmotionis shown below. the acceleration of the particle at t = 4/3 s is 1 �1 0 4 12 t(s) x(cm) 8 (A) 3 / 32 .2cm/s2 (B) �.2/32 cm/s2 (C) .2/32 cm/s2 (D) � 3 / 32 .2cm/s2 [JEE, 09] 45. Auniformrod of lengthLandmassMis pivoted at the centre. Its two ends are attached to two springs of equal spring constants k. The springs are fixed to rigid supports as shown in the figure, and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle . in one direction and released. the frequencyof oscillation is [JEE, 09] (A) 1 2k 2. M (B) 1 k 2. M (C) 1 6k 2. M (D) 1 24k 2. M
46. ThemassMshownin the figure oscillates in simple harmonicmotionwith amplitudeA.The amplitude of the point Pis [JEE, 09] P M (A) k1 k2 12 k A k (B) 21 k A k (C) 1 1 2 k A k . k (D) 2 1 2 k A k . k
SIMPLE HARMONIC MOTION www.physicsashok.in 90 PASSAGE Passage � 1 Ablock ofmassmis suspended fromone end of light spring as shown. The originO is considered at distance equal to natural length of spring fromceiling and vertical downward direction as +veYaxis.When the sytemis in equilibriuma bullet ofmass m/3moving in verticalupward directionwith velocityv0 strikes the block and embeds into it.As a result theblock (withbullet embeddedinto it)moves upand starts oscillating. v0 m m/3 YO k l Based onabove information, answer the following equations : 1. Mark the correct statement(s). (A) The block-bullet systemperforms SHMabout y=mg/k. (B) The block-bullet systemperforms oscillatorymotionbut not SHMabout y=mg/k. (C) The block-bullet systemperforms SHMabout y = 4mg/3k. (D) The block-bullet systemperformoscillatorymotion but not SHMabout y= 4mg/3k. 2. The amplitude ofoscillationwould be (A) 2 204mg mv 3k 12k . . . .. .. (B) 2 20mv mg 12k 3k . . . .. .. (C) 2 20mv mg 6k k . . . .. .. (D) 2 2 mv0 4mg 6k 3k . . . .. .. 3. The time taken by block-bullet systemto move fromy=mg/k (initial equilibriumposition) to y= 0 (natural length of spring) is [Arepresents the amplitude ofmotion] (A) 4m cos 1 mg cos 1 4mg 3k 3kA 3kA .. . .. .. . . .. .. .. . . . . . . (B) 3k cos 1 mg cos 1 4mg 4m 3kA 3kA .. . .. .. . . .. .. .. . . . . . . (C) 4m sin 1 4mg sin 1 mg 6k 3kA 3kA .. . .. .. . . .. .. .. . . . . . . (D) None of the above Passage � 2 A platform is executing SHM in a vertical direction, with an amplitude of 5 cm and a frequency of 10/. vibrations/sec.Ablock is placed on the platformat the lowest point of its path.
[Take g = 10m/s2] Answer the following questions based on above information : 4. At what point will the block leave the platform? (A) 2.5 cmfrommean positionwhen acceleration is acting down and velocity is in upward direction. (B) 2.5 cmfrommean positionwhen platformismoving up. (C) 2.5 cmabovemean positionwhen platformismoving down. (D) 2.5 cmbelowthemean position. 5. Mark the correct statement(s). (A) Normal contact force between the platformand block is constant. (B)As platformapproachesmean position frombottom, the normal contact force between the block and platformincreases. (C)As platformmoves up awayfrommean position, the normalcontact force between the block and platform decreases. (D) Both (B) and (C) are correct. 6. At what point, the block returns to the platform? (A) 1.3 cmabove equilibriumposition (B) 1.3 cmbelowequilibriumposition (C) 4.3 cmabove equilibriumposition (D) 4.3 cmbelowequilibriumposition
SIMPLE HARMONIC MOTION www.physicsashok.in 91 LEVEL � 2 1. Aringofmassmcanfreelyslide ona smoothverticalrod.Thering is symmetrically attached with two springs, as shown in the fig., each of stiffness k. Each spring makes an angle . with the horizontal. If the ring is slightly displaced vertically, determine its time period. k k mrod 2. Ablock B ofmassm= 0.5 kg is attachedwith upper end of a vertical spring of force constant K= 1000Nm�1 as shown in fig.Another identical block A falls froma height h = 49.5 cmon the block B and gets stuckwith it. The combined body starts to perform verticaloscillations. h AB Calculate amplitude of these oscillations. (g= 10ms�2) 3. One end of an ideal spring is fixed to awall at originOand axis of spring is parallel to x-axis.Ablock ofmass m= 1 kg is attached to free end of the spring and it is performing SHM. Equation of position of the block in co-ordinate systemshown in Figure is x = 10 + 3 sin(10t), where t is in second and x in cm. Another block ofmassM=3 kg,moving towards the originwith velocity 30 cm/sec collides with the block performing SHM at t = 0 and gets stuck to it. Calculate 1 kg 3 kg x (i) newamplitude of oscillations, O (ii) newequation for position of the combined bodyand (iii) loss of energyduring collision.Neglect friction. 4. Two identical blocks Aand B ofmassm= 3 kg are attached with ends of an ideal spring of force constant K = 2000 Nm�1 and rest over a smooth horizontal floor.Another identical block C moving with velocity v0 = 0.6ms�1 as shown in fig. 100 strikes the blockA and gets stuck to it. Calculate for subsequent motion m V K 0 (i) velocityofcentre ofmass of the system, m m (ii) frequencyof oscillations of the system, (iii) oscillation energyof the system, and (iv)maximumcompressionof the spring. 5. In the arrangement shown in fig, pulleys are small and light and springs are ideal. K1, K2, K3 and K4 are force constants of the springs. Calculate period ofsmall vertical oscillations of block ofmassm. K2 K4 K1 K3m 6. Fig., shows a particle ofmassm= 100 gm, attachedwith four identical springs, eachoflengthl =10 cm. InitialtensionineachspringisF0 =25 newton.Neglecting gravity, calculateperiodofsmalloscillationsoftheparticle alonga line perpendicular to the plane of the figure. p
A m B C D 7. In the arrangement shown in Fig., bodyB is a solid cylinder of radius R = 10 cm withmassM= 4 kg. It can rotatewithout friction about a fixed horizontal axisO Ablock A ofmass m = 2 kg suspended by an inextensible thread is wrapped around the cylinder.Ahorizontal light spring of force constant K= 100 Nm�1 fixed at one end keeps the systemin static equilibrium. Calculate A R O(i) initial elongationinthe spring and B (ii) period of small vertical oscillations of the block. (g = 10ms�2)
SIMPLE HARMONIC MOTION www.physicsashok.in 92 8. Arigid rod ofmassmwith a ballofmassMattached to the free end is restrained to oscillate in a verticalplane as shown in the figure. Find the naturalfrequencyof oscillation. M 3/4L L/4 k m 9. Find the period of small oscillations in a vertical plane performed by a ball ofmassmfixed at themidle of a horizontally stretched string of length l. The tension of the string is assumed to be constant and equal to F >> mg. 10. Aparticlemoves along the x-axis according to the lawx = a cos .t. Find the distance that the particle covers during the time interval t =0 to t. 11. Find the amplitude, frequencyand epoch ofthe simple harmonicmotionrepresented byx=3 sin .t +4 cos .t. 12. Ahelical spring elongates 10 cmwhen subjected to a tension of 5N. Find themasswhich should be attached to the bottomof the spring so that when pulled down and released themasswillvibrate twice per second. Find also itsmaximumvelocitywhen the amplitude of vibration is 1 cm. 13. Find the frequency of small oscillations of the arrangement illustrated in figure. The radius of the pulley isR, itsmoment of inertia relative to the axis of rotation is I, the mass of the body ism, and the spring stiffness is k. The mass of the thread and the spring are negligible, the thread does not slip over the pulleyand there is no friction in the axis of the pulley. mR 14. Aplankwith a body ofmassmon it executes simple harmonicmotionof cyclic frequency .= 11 rad s�1 between the levels 1 and 2 separated by a distance 2a as shown infig. Find : (a) the force that the body exerts on the plank as it moves fromlevel 1 to 2 when a = 4 cm; a m a O2(b) theminimumvalue of awhenthe bodystarts falling behind the plank; 1 (c) the amplitude of oscillation at whichthe body jumps up to a height h= 50 cmrelative to level 1. 15. (a)Ablock ofmassmis tied to one end of a stringwhich passes over a smoothfixed pulleyAand under a light smoothmovable pulleyB. Theother end of the string is attached to the lower end ofa spring ofspring constant k2. Find the period of smalloscillations ofmassmabout its equilibriumposition. mA B k2
k1 Figure (a) m A B k2 k1 Figure (b) (b)Ablock ofmassmis attached to one end ofa light inextensible string passing over a smooth light pulleyB and under another smooth light pulleyAas shownin the figure.The other end of a string is fixed to a ceiling.A and B are held by springs of spring constants k1 and k2. Find angular frequency of small oscillation of the system.
SIMPLE HARMONIC MOTION www.physicsashok.in 93 16. Find the angular frequencyofmotion of blockmfor smallmotion of rodBDwhenwe neglect the inertial effects of rod BD. Spring constant are k1 and k2.Neglect friction forces. BD C k b 1 k2 m 17. Find the angular frequency for small oscillation of block ofmassmin the arrangement shown in the figure. Neglect themass of the rod. l/2 l/2 k2 k1 m k3 18. ArodAB ofmassMis attached as shown belowto a spring of constant k.A small block ofmassmis placed on the rod at its free endA. (i) if endAismoved down througha smalldistance d and released, determine the period of vibration. b m k B A l (ii) determine the largest allowable value of d if the blockmis to remain at all times in contactwith the rod. 19. A rod ofmass mand length l is hinged at its upper end and carries a block of mass M at its lower end. Two springs having the spring constants k1 and k2 respectively are attached to the rod at the distances b and c fromthe hinge as shown in the figure. Find the angular frequencyofvibrationfor smalloscillationof the system. M l b m k1 k2 C 20. Auniformbroad of lengthL andweightWis balanced on a fixed semi-circular cylinder of radiusR as shown in the figure. If the plank is tilted slightlyfromits equilibriumposition, determine the period ofits oscillations. l r 21. Aplankwith a body ofmassmplaced on it start moving straight up according to the lawy = a(1 � cos .t), where y is the displacement fromthe initial position, .= 11 rad/s. Find : (a) The time dependence of the force that the body exerts on the plank. (b) Theminimumamplitude of oscillation ofthe plank at which the bodystarts falling behind the plank. 22. Aparticle ofmassmfree to move in the x�y plane is subjected to a forcewhose components are Fx =�kx and
Fy = �ky, where k is a constant. The particle is released when t = 0 at the point (2, 3). Prove that the subsequentmotion is simple harmonic along the straight line 2y � 3x = 0. 23. In the shown arrangement, both the springs are in their natural lengths. The coefficient of friction betweenm2 andm1 isµ. There is no friction betweenm1 and the surface. If the blocks are displaced slightly, they together perform simple harmonicmotion.Obtaion. mm12 k1 k2 (a) Frequencyof such oscillations. (b) The conditionif the frictional force on blockm2 is to act in the direction of its displacement frommean position. (c) If the condition obtained in (b) ismet, what canbemaximumamplitude of their oscillations ?
SIMPLE HARMONIC MOTION www.physicsashok.in 94 LEVEL � 3 1. Statewhether true or false �Two simple harmonicmotions are represented by the equations x1 = 5sin [2.t + ./4] and x2 = 5 2 (sin 2.t + cos2.t) their amplitudes are in the ratio 1 : 2� [REE,96] 2. Ablock is kept on a horizontal table. the table is undergoing simple harmonicmotion of frequency 3Hz in a horizontalplane.The coefficient of static frictionbetweenblock and the table surface is 0.72. Find themaximum amplitude of the table at which the block does not slip on the surface. [REE,96] 3. Abob ofmassMis attached to the lower end of a vertical string of lengthL and cross sectional areaA.The Young�smodulus of thematerial of the string isY. If the bob executes SHMin the verticaldirection, find the frequencyofthese oscillations. [REE, 2000] 4. Adiatomicmolecule has atoms ofmassesm1 andm2. The potentialenergyof themolecule for the interatomic separation r is given byV(r) = �A+B(r � r0)2,where r0 is the equilibriumseparation, andAandBare positive constants. The atoms are compressed towards eachother fromtheir equilibriumpositions and released.What is the vibrational frequencyof themolecule ? [REE, 01] 5. Two massesm1 andm2 connected bya light spring ofnatural length l0 is compressed completelyand tied bya string. This systemwhilemoving with a velocity v0 along +ve x�axis pass through the origin at t = 0.At this position the string snaps. Position ofmassm1 at time t is given bythe equation. x1(t) = v0t �A(1 � cos .t) Calculate : [JEE, 03] (a) Position of the particlem2 as a function of time. (b) l0 in terms ofA. 6. Asmallbodyattached to one end of a verticallyhanging spring is performing SHMabout it�s mean position with angular frequency . and amplitude a. If at a height y* fromthe mean position the bodygets detached fromthe spring, calculate the value of y* so that the height m y0 H attained by the mass ismaximum. The body does not interact with the spring during it�s subsequent motion after detachment. (a.2 > g). [JEE, 05]
SIMPLE HARMONIC MOTION www.physicsashok.in 95 Answer Key ASSERTION-REASON TYPE Q. 1 2 3 4 5 Ans. A C A B D MATCH THE COLUMN 1. [(A�R), (B�P), (C�Q), (D�S)] 2. [(A�QR), (B�QRS), (C�P), (D�P)] 3. [(A�RS), (B�RS), (C�PQS), (D�RS)] LEVEL � 1 Q. 1 2 3 4 5 6 7 8 9 10 Ans. B B C A B BC A B A A Q. 11 12 13 14 15 16 17 18 19 20 Ans. C ABC A CD D D B D B D Q. 21 22 23 24 25 26 27 28 29 30 Ans. C C B C B ABC ACD AC BCD B Q. 31 32 33 34 35 36 37 38 39 40 Ans. AC BC B A D AC A A A B Q. 41 42 43 44 45 46 Ans. C ABD D D C D PASSAGE Q. 1 2 3 4 5 6 Ans. C B A A C D LEVEL � 2 1. T 2 m2k . . 2. 5 cm 3. (i) 3 cm; (ii) x = 10 + 3sin(5t + .)cmor x = 10 � 3sin(5t) cm; (iii) 0.135 Joule 4. (i) 0.2m/sec.; (ii) 5 10 Hz . ; (iii) 0.09 Joule; (iv) 3 10 mm. 5. 1 2 3 4 T 4 m 1 1 1 1 k k k k . . . . . . . . . . . 6. 0.02 . sec. 7. (i) 20 cm; (ii) 0.4 . sec. 8. f 1 3k 2 27M 7m . . . 9. T mF . . l
SIMPLE HARMONIC MOTION www.physicsashok.in 96 10. S = a[n + 1 � cos (.t � n2. )] when n is even; S = [n + cos (.t � n2. )] when n is odd. 11. A= 5, f = ./2. and epoch = 53º 12. m= 310 g, vmax = 12.5 cm/sec. 13. 2 f 1 k 2 m 1/R . . . 14. (a) a 2 N mg 1 cos t g . . . . . . . . . . ; (b) min 2 a . g . ; (c) a = 0.2 m. 15. (a) 1 2 1 2 m(k k ) T 2 k k. . . ; (b) 1 2 1 2 1 k k 2 m(k k ) . . . 16. . . 2 1 2 2 2 1 2 k k C m k C k (a b) . . . . 17. 2 3 1 2 3 1 4k k k m k 4k . . . . . . . . . . 18. (i) 2 2 T 2 (m M/ 3) kb . . . l ; (ii) 2
max 2 g (m M/ 3) d kb. . l 19. 2 2 1 2 2 (k b k c ) (M/m/ 2)g (M m/ 3) . . . . . l l 20. T 2 L 3 gr 4 . . 21. (a) N = m(gta.2 cos.t); (b) 8 km. 23. (a) 1 2 1 2 1 k k 2 m m . . . ; (b) 1 1 2 2 k m k m . ; (c) 1 2 2 1 2 2 1 µ(m m )m g m k m k . . LEVEL � 3 1. True 2. 2cm 3. 1 YA 2. ML 4. 1 2 1 2 1 m m 2 f 2B(m m ) . . . 5. (a) 1 2 0 2 m x v t A(1 cos t) m . . . . ; (b) 1 0 2 m 1 A m . . . . . . . . l 6. 2 y mg g a
k . . . .
OPTICS
DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125
OPTICS www.physicsashok.in 1 Review of Concepts (a) Due to reflection, none of frequency, wavelength and speed of light change. (B) Law of reflection : (i) Incident ray, reflected ray and normal on inident point are coplanar. (ii) The angle of incidence is equal to angle of reflection. Some important points : In case of plane mirror : (i) For real object, image is virtual. (ii) For virtual object, image is real. (iii) Image size = Object size. (iv) The converging point of incident beam behaves as object. (v) If incident beam on optical instrument (mirror, lens etc) is converging in nature, object is virtual.(vi) If incident beam on the optical instrument is diverging in nature, the object is real. (vii) The converging point of reflected or refracted beam from an optical instrument behaves a image.(viii) If reflected beam or refracted beam from an optical instrument is converging in nature, image is real. Real Image P Virtual Object P nn (ix) If reflected beam or refracted beam from an optical instrument is diverging in nature, P P� Real Object Virtual Object nn . ... image is virtual. (x) For solving the problem, the reference frame is chosen in which optical instrument (mirror, lens, etc.) is in rest. (xi) The formation of image and size of image is independent of size of mirror. (xii) Visual region and intensity of image depend on size of mirror.
OPTICS www.physicsashok.in 2 (xiii) If the plane mirror is rotated through an angle . , the reflected ray and image is rotated through an angle 2 . in the same sense. (xiv) If mirror is cut into a number of pieces, then the focal length does not change. (xv) The minimum height of mirror required to see the full image of a man of height h is h/2. Image Rest Object (xvi) v Image Rest Object (xvii) v sin. v sin. v cos. v cos. . Image Rest Object (xviii) v vm 2 vm - v Object Image In rest (xix) vm 2 vm Image Object (xx) v vm 2 vm + v (C) Number of images formed by combination of two plane mirrors : The images formed by combination of two plane mirrors are lying on a circle whose centre is at the meeting points of mirrors. Also, object is lying on that circle. Here, . . n . 360 where . = angle between mirrors. (i) If . 360. is even number, the number of image is n�1. (ii) If . 360. is odd number and object is placed on bisector of angle between mirrors, then number of images is n�1. (iii) If . 360. is odd and object is not situated on bisector of angle between mirrors, then the number of images is euqal to n. (D) Law of reflecteion in vector form : Let e�1 = unit vecotr along incident ray.
Let e� 2 = unit vector along reflected ray n n� 1 �.2 �. n� = unit vector along normal on point of incidence Then, e� 2 . e�1 . 2 .e�1.n� .n� (e) Spherical mirrors : (i) It is easy to solve the problems in geometrical optics by the help of co-ordinate sign convention.
OPTICS www.physicsashok.in 3 x� x y� y x� x y� y x� x y� y x� x y� y x� x y� y (ii) The mirror formula is . . . . 1 u1 1 Also, R = 2 . These formulae are only aplicable for paraxial rays. (iii) All distances are measured from optical centre. It means optical centre is taken as origin. (iv) The sign convention are only applicable in given values. (v) The transverse magnification is object size u image size .. . . . 1. If object and image both are real, . is negative. 2. If object and image both are virtual, . is negative 3. If object is real but image is virtual; . is positive. 4. If object is virtual but image is real, . is positive. D F d Sun . . 5. Image of star; moon or distant object is formed at focus of mirror. If y = the ddistance of sun or moon from earth. D = diameter of moon or sun�s disc. . = focal length of the mirror d = diameter of the image . = the angle subtended by sun or moon�s disc Then tan . = . = . . d yD Here, . is in radian. Laws of Refraction 1. (a)The incident ray, the refracted ray and normal on incidence point are coplanar. (B) .1 sin.1 . .2 sin.2 ...cons tant . .1
.2
.2
.1 (C) Snell�s law in vector form: .1 .2 2 �. 1 �. n� Let, e�1 = unit vector along incident ray e� 2 = unit vecotr along refracted.
OPTICS www.physicsashok.in 4 n� = unit vector along normal on incidence point. Then .1 .e�1 . n� . . .2 .e� 2 . n� .. Some important points : (i) The value of absolute retractive index . is always greater or equal to one. (ii) The value of refractive index depends uponmaterial of medium, colour of light and temperature of medium. (iii) When temperature increases, refractive index decreases. (iv) Optical path is defined as product of geometrical path and refractive index. i.e., optical path = .x (v) For a given time, optical path remains constant. i.e., .1 x1 . .2 x2 . . cons tant . dt dx dt dx 2 2 1 .1 . . . .1 c1 . .2 c2 . 21 12 c. c .. i.e., c. . 1 (vi) The frequency of light does not depend upon medium. . c1 . ..1 , c2 . . .2 . 12 12 21 cc .. . . .. . . . . 1 2. (a) When observer is in rarer medium and object is in denser medium: Then apparent depth . . real depth (B) When object is in air and observer is in Air Observer Denser medium
(.) P� Object Apparent depth P Real depth denser medium: real position . . apparent position (C) The shift of object due to slab is . .. . . .. . . x . t 1. 1 (i) This formula is ony applicable when observer is in rarer medium. (ii) The object shiftness does not depend upon the P P� Q Object shiftness = x t position of object. (iii) Object shiftness takes place in the direction of incidence ray. (D) The equivalent rerfractive index of a combination of a number of slabs for normal incidence is i ii tt . .. . . Here, .ti = t1 + t2 + ....... t1 .1 .2 t2 .. . . . . . . 2 2 1 1 i ti t t
OPTICS www.physicsashok.in 5 (e) The apparent depth due to a number of media is i ti . . . i t r d . (f) The lateral shifting due to a slab is d = t sec r sin (i � r). 3. (a) Cricital angle : When a ray passes from denser medium ( .2 ) to rarer medium ( .1), then for 90° angle of refraction, the c Rarer .1 Denser .2 90° corresponding angle of incidence is critical angle. Mathematically, 2sinC 1 .. . (B) (i) When angle of incidence is lesser than critical angle, refraction takes place. the corresponding deviation is sin sini i 11 2 . . .. . . .. . .. . . . for i = C Rarer medium (. ) 1 i r c i i i < C i = C Denser medium (. ) 2 (ii) When angle of incidence is greater than critical angle, total internal reflection takes place. the corresponding deviation is . . . . 2i when i > C 4. The . . i graph is (i) Critical angle depends upon colour of light, material of medium, c i . ./2 and temperature of medium.
(ii) Critical angle does not depend upon angle of incidenct. PRISM (a) Deviation produced by prism is ..i.i..A. (B) r + r� = A (C) For grazing incidence, i = 90° (D) For grazing emergence, i� = 90° (e) For not transmitting the ray from prism, . > cosec 2A A B C n n� r r� i i� (f) For limiting angle of prism, i = i� = 90°, the limiting angle of prism = 2C where C is critical angle. If angle of prism exceeds the limiting values, then the rays are totally reflected. (g) . . i graph for prism: (h) For minimum deviation, (i) i = i� and r = r� (ii) 2sin A2 A sin m ... ... . . . . i . .m
OPTICS www.physicsashok.in 6 In the case of minimum deviation, ray is passing through prism symmetrically. (i) For maximum deviation ..max .. i = 90° or i� = 90° (j) For thin prism, . . .. .1. A (k) Angular dispersion, D . ... . .r . A (l) Angular deviation, .y . ..y .1.A (m) dispersive power = . . .. . . .. . .. . . . . . y 1r (n) .. . .. . . . . . . . 2 r y (o) For dispersion without deviation, ..y . 0 (p) For deviation without dispersion, .D . 0 x� A x B O C .1 .2 Refractive surface formula, u r 2 1 .2 . .1 . . . . . Here, . = image distance, u = object distance, r = radius of curvature of spherical surface (a) For plane surface, r = . (B) Transverse magnification, object size u m Image size 21 . . . . . (C) Refractive surface formula is only applicable for paraxial ray. LENS 1. Lens formula : . . . . 1 u1 1
(a) Lens formula is only applicable for thin lens. (B) r = 2 . formula is not applicable for lens. (C) object size u m image size . . . (D) Magnification formula is only applicable when object is perpendicualr to optical axis. (e) Lens formula and the magnification formula is only applicable when medium on both sides of lenses are same. (f) f(+ve) (i) f(-ve) (ii) f(-ve) (iii) f(+ve) (iv) f(-ve) (v) f(+ve) (vi) (g) This lens formula is applicable for converging as well diverging lens. .1 .1 .Thin lens maker�s formula : 2 . .. . . .. . . . .. . . .. . . . . . . . 1 1 2 2 1 r1 r1 1
OPTICS www.physicsashok.in 7 2. (a) Thin lens formula is only applicable for paraxial ray. (B) This formula is only applicable when medium on both sides of lens are same. (C) Intensity is proportional to square of aperture. (D) When lens is placed in a medium whose refractive index is greater than that of lens. i.e., .1 . .2 . Then converging lens behaves as diverging lens and vice versa. (e) When medium on both sides of lens are not same. Then both focal lengths are not same to each other. (f) If a lens is cut along the diameter, focal length does not change. (g) If lens is cut by a vertical, it converts into two lenses of different + f f1 f2 focal lengths. i.e., 1 2 1 1 1 . . . . . (h) If a lens is made of a number of layers of different refractive index (shown in figure). Then number of images of an object formed by the lens is + + + + + +.1.2 .3 .4 .5 .6 equal to number of different media. (i) The minimum distance between real object and image in is 4 . . (j) The equivalent focal elngth of co-axial combination of two d<f1 d<f2 o1 o2 f1 f2 d lenses is given by 1 2 1 2 1 1 d F1 . . . . . . . (k) If a number of lenses are in contact, then .. . . . .
1 2 1 1 F1 (l) (i) Power of thin lens, FP . 1 (ii) Power of mirror is FP . . 1 (m) If a lens is silvered at one surface, then the system behaves as an equivalent mirror, whose power P = 2PL + Pm Here, PL = Power of lens = . .. . . .. . . . .. . . .. . . . . . 1 1 2 2 1 r1 r1 Pm = Power of silvered surface Fm. . 1 Here, 2 F r2 m . , where r2 = radius of silvered surface. FP . . 1 Here, F = focal length of equivalent mirror.
OPTICS www.physicsashok.in 8 ASSERTION & REASION THE NEXT QUESTIONS REFER TO THE FOLLOWING INSTRUCTIONS A statement of assertion (A) is given and a Corresponding statement of reason (R) is given just below it of the statements, mark the correct answer as � (A) If both A and R are true and R is the correct explanation of A. (B) If both A and R are true but R is not the correct explanation of A. (C) If A is true but R is false. (D) If both A and R are false. (E) If A is false but R is true. 1. Assertion (A) : A single ray can�t be isolated from a source however small it may be. Reason (R) : The concept of single ray is hypothetical. (A) (B) (C) (D) (E) 2. Assertion (A) : Virtual images can be photographed. Reason (R) : Rays from virtual images are diverging. (A) (B) (C) (D) (E) 3. Assertion (A) : Virtual object can�t be seen by human eye. Reason (R) : Virtual object is formed by converging rays. (A) (B) (C) (D) (E) 4. Assertion (A) : A Convex mirror is used as rear view mirror. Reason (R) : The Convex mirror always forms virtual, erect and diminished image. (A) (B) (C) (D) (E) 5. Assertion (A) : The behavior of any lens depends on surrounding medium. Reason (R) : A lens can be looked upon as a collection of small prism with varying prism angle. (A) (B) (C) (D) (E) 6. Assertion (A) : Human eye can see virtual object. Reason (R) : Virtual object is formed by apparent intersection of incident rays. (A) (B) (C) (D) (E) 7. Assertion (A) : Real image is formed by real intersection of reflected or refracted rays. Reason (R) : Real image can�t be obtained on screen. (A) (B) (C) (D) (E) 8. Assertion (A) : If a portion of lens or mirror is blocked or removed, then intensity of image reduces. Reason (R) : As every portion of lens or mirror forms image, hence blocking or removing a portion will result in intensity reduction. (A) (B) (C) (D) (E) 9. Assertion (A) : A rectangular glass slab produces no deviation and no dispersion. Reason (R) : Dispersive power of glass is zero. (A) (B) (C) (D) (E) 10. Assertion (A) : A double convex lens .. .1.5. has focal length 10 cm. When immersed in water 43 ... . .. . . , its focal length becomes 40 cm. Reason (R) : 1 2 1 1 1 l m m f R R . .
.
. . .
. . . .
. . (A) (B) (C) (D) (E)
OPTICS www.physicsashok.in 9 11. Assertion (A) : A convex lens of glass .. .1.5. behave as diverging lens when immersed in carbon disulphide of higher refractive index .. .1.65.. Reason (R) : A diverging lens is thinner in the middle and thicker at the edges. (A) (B) (C) (D) (E) 12. Assertion (A) : A biconvex lens of focal length 10 cm is split into two equal parts by a plane parallel to its principal axis. The focal length of each part will be 20 cm. Reason (R) : The focal length depends on how many parts the convex lens has been split. (A) (B) (C) (D) (E) 13. Assertion (A) : Radius of curvature of a convex mirror is 20 cm. If a real object is placed at 10 cm from pole of the mirror, image is formed at infinity. Reason (R) : When object is placed at focus, its image is formed at infinity. (A) (B) (C) (D) (E) 14. Assertion (A) : For a prism of refracting angle 60° and refractive index 2 , minimum deviation is 30°. Reason (R) : At minimum deviation, 1 2 30 2r . r . A . . (A) (B) (C) (D) (E) 15. Assertion (A) : Image formed by concave lens is not always virtual. Reason (R) : Image formed by a lens if the image is formed in the direction of ray of light with (A) (B) (C) (D) (E) 16. Assertion (A) : Minimum deviation for a given prism does not depend on the refractive index . of the prism. Reason (R) : Deviation by a prism is given by . . 1 2 . . i . i . A and does not have the term . . (A) (B) (C) (D) (E) Level # 1. Objective Type Question Multiple Choice Question with ONE correct answer : 1. Two plane mirrors M1 and M2 are inclined to each other at 70°. A ray incident on the mirror M1 at an angle . falls on M2 and is then reflected parallel to M1 for (A) . = 45° (B) . = 50° (C) . = 55° (D) . = 60° 2. A light ray is incident on a horizontal plane mirror at an angle of 45°. At what angle should a second plane mirror be placed in order that the reflected ray finally be reflected horizontally from the second mirror, as shown in figure. (A) . = 30° (B) . = 24° (C) . = 22.5° (D) . = 67.5° 3. A plane mirror is placed in y-z plane facing towards negative x-axis. The mirror is moving parallel to yaxis with a speed of 5 cm/s. A point object P is moving infront of the mirror with a velocity (3 cm/s)i�+ (4 cm/s) j � + (5 cm/s)k�. Find the velocity of image with respect to mirror (A) (�3 cm/s)i�+ (4 cm/s) j � + (5 cm/s)k�(B) (3 cm/s)i�
+ (4 cm/s) j � + (5 cm/s)k� (C) �(3 cm/s)i�� (4 cm/s) j � � (5 cm/s)k�(D) none of the above.
OPTICS www.physicsashok.in 10 4. The size of the face of a dancer is 24 cm x 16 cm. Find the minimum size of a plane mirror required to see the face of dancer completely by (i) one eyed dancer. (ii) two eyed dancer. (Distance between the eyes is 4 cm.) (A) (i) 12 x 8 cm2 (ii) 12 x 6 cm2 (B) (i) 8 x 10 cm2 (ii) 12 x 2 cm2 (C) (i) 10 x 12 cm2 (ii) 9 x 8 cm2 (D) (i) 12 x 2 cm2 (ii) 6 x 13 cm2 5. A bullet of mass m2 is fired from a gun of mass m1 with horizontal velocity v. A plane mirror is fixed at gun facing towards bullet. The velocity of the image of bullet formed by the plane mirror with respect to bullet is (A) . .. . . .. . . 12 m 1 m (B) . . .. . . .. . .1 1 2 mm m (C) . . . .1 2 1m2 m m (D) none of these 6. In the given figure, the angle of reflection is (A) 30° (B) 60° (C) 45° (D) none of these. 7. Two plane mirrors A and B are aligned parallel to each other, as shown in figure. A light ray is incident at an angle of 30° at a point just inside one end of A. The plane of incidence coincides with the plane of figure. The 0.2 m AB 30° 2 3 m maximum number of times the ray undergoes reflections (excluding the first one) before it emerges out is (A) 28 (B) 30 (C) 32 (D) 34 8. A point source of light B is placed at a distance L in front of the centre of a mirror of width d hung vertically on a wall as shown. A man walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown. d B
L 2 L The greatest distance over which he can see the image of the light source in the mirror is (A) ½ d (B) d (C) 2d (D) 3d 9. A plane mirror having a mass m is tied to the free end of a massless spring of spring constant k. The other end of the spring is attached to a wall. The spring with the mirror held vertically to the floor can slide along it smoothly. When the spring is at its natural length, the mirror Wall is found to be moving at a speed of v cm/s. The separation between k the images of a man standing before the mirror, when the mirror is in its extreme positions (A) kv m (B) km 2v (C) k2v m (D) k4v m 10. Two spherical mirrors M1 and M2, one convex and other concave having same radius of curvature R are arranged coaxially at a distance 2R (consider their pole separation to be 2R). A bead of radius a is placed at the pole of the convex mirror a shown. The M1 M2 ratio of the size of the first three images of the bead is (A) 1 : 2 : 3 (B) 1 : 3: 1 21 (C) 41 : 1 11 : 1 31 (D) 3 : 11 : 41 11. An object is placed in front of a convex mirror at a distance of 50 cm. A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and the plane mirror is 30 cm, there is no parallax between the images formed by the two mirrors. The radius of curvature of the convex mirror (in cm) is (A) 60 (B) 50 (C) 30 (D) 25
OPTICS www.physicsashok.in 11 12. A rectangular glass slab ABCD of refractive index n1, is immersed in water of refractive index n2 (n1 > n2). A ray of light is incident at the surface AB of the slab as shown. The maximum value of the angle of incidence .max , such that the ray comes out only from the other surface CD is given by (A) .. .. . .. .. . . . .. . . .. . .. . . .. . . . 11 2 21 1 ncos sin n nsin n (B) .. .. . .. .. . . . .. . . .. . .. . . .. . . . 2 1 1 1 nsin n cos sin 1 (C) . .. . . .. . . 21 1 nsin n (D) . .. . . ..
. . 11 2 nsin n 13. Two thin slabs of refractive indices .1 and .2 are placed parallel to each other in the x-z plane. If the direction of propagation of a ray in the two media are along the vectors j�b i�a r 1 . . and j�d i�r 2 . c . x y .2 .1 then we have (A) .1a . .2b (B) 2 2 2 2 2 1 c d c a b a . . . . . (C) .1 (a2 + b2) = .2 (c2 + d2) (D) none of these 14. A man stands on a glass slab of height . and inside an elevator accelerated upwards with �a�. The bottom of the slab appears to have shifted with respect to the man by a distance (if the R. I. of the glass is .g) (A) less then . .g (B) greater than . .g (C) equal to . .g (D) can�t be said. 15. A ray of light travels from a medium of refractive index . into air. If the angle of incidence at the plane surface of separation is . and the corresponding angle of deviation is D, the variation D with . is shown correctly by the figure. C C C D D D D D1 D1 D1 D2 D2 D2 (0, 0) . ./2 (0, 0) . (0, 0) . ./2 (0, 0) . ./2 (A) (B) (C) (D) 16. An observer can see through a pin hole at the top end of a thin rod of height h placed as shown in the figure. Beaker height is 3h and its radius is h. When the beaker is filled with a liquid up to a height 2h, he can see the h 2 h
3 h Eye lower end of the rod. Then refractive index of the liquid is (A) 23 (B) 23 (C) 25 (D) 25 . (Assume that the distance between rod and the wall is negligible). 17. A glass sphere of radius 5 x 10�2 m has a small bubble 2 x 10�2 m from its centre. Bubble is viewed along the diameter of the sphere, from the side on which it lies. If refractive index of glass is 1.5 then how far from the surface will the bubble appear? (A) 2.1 cm (B) 2.5 cm (C) 1.5 cm (D) 2.0 cm
OPTICS www.physicsashok.in 12 18. A ray of light travelling in a transparent medium falls on a surface separating the medium from air at an angle of incident 45°. The ray undergoes total internal reflection. If . is the refractive index of the medium with respect to air, select the possible value(s) of . from the following : (A) 1.3 (B) 1.4 (C) 1.5 (D) 1.7 19. A tank contains a transparent liquid of refractive index n the bottom of which is made of a mirror as shown. An object O lies at a height d above the mirror. A person P vertically above the object sees O and its image in the mirror and d OP finds the apparent separation to be (A) 2nd (B) n 1 2d . (C) n2d (D) .1 n. nd . 20. A fish looks up at the surface of a perfectly smooth lake. The surface appears dark except a circular area directly above it. The plane angle . that this illuminated region subtends is (A) 48.6° (B) 24.3° (C) 97.2° (D) 12.15° 21. A ray of light enters an anisotropic medium from vacuum at grazing incidence. If . is the angle made by the reflected ray inside the medium with the interface and n( . ) is the refractive index of the medium then, (A) n( . ) sin . = 1 (B) n( . ) cos . = 1 (C) 1 sin n( ) . . . (D) 1 cosn( ) . . . 22. The slab of a material of refractive index 2 shown in figure has a curved surface. APB of radius of curvature 10 cm and a plane surface CD. On the left of APB is air and on the right of CD is water with refractive indices as given in figure. An object O is placed at a distance of 15 cm from pole P as shown. The distance of the final image of O from P, as viewed from the left is (A) 20 cm (B) 30 cm (C) 40 cm (D) 50 cm 23. An object is placed at a distance of 12 cm from a convex lens on its principal axis and a virtual image of certain size is formed. On moving the object 8 cm away from the lens, a real image of the same size as that of virtual image is formed. The focal length of the lens in cm is (A) 15 (B) 16 (C) 17 (D) 18 24. A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive
index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at the point O and PO = OQ. The distance PO is equal to (A) 5 R (B) 3 R (C) 2 R (D) 1.5 R 25. A lens of focal length . is placed in between an object and screen fixed at a distance D. The lens forms two real images of object on the screen for two of its different positions, a distance x apart. The two real images have magnifications m1 and m2 respectively (m1 > m2). (A) m1 m2 x. . . (B) m1 m2 = 1 (C) 4DD2 . x2 . . (D) all the above
OPTICS www.physicsashok.in 13 26. A liquid of refractive index 1.33 is placed between two identical plano-convex lenses, with refractive index 1.50. Two possible P Q arrangement P and Q are shown. The system is (A) divergent in P, convergent in Q. (B) convergent in P, divergent in Q. (C) convergent in both (D) divergent in both. 27. A lens of refractive index . is put in a liquid of refractive index .. . If the focal length of the lens in air is . , its focal length in liquid will be (A) . . .. . .. ... . .1 (B) . . .. .. .1. . .. . . (C) . . . ... . .. .. . .1 (D) .. . ... ... . 28. A convergent lens is placed inside a cell filled with a liquid. The lens has a focal length +20 cm when in air and its material has a refractive index 1.50. If the liquid has a refractive index 1.60, the focal length of the system (A) �160 cm (B) � 24 cm (D) �80 cm (D) + 80 cm 29. A double convex lens, made of glass of refractive index 1.5, has focal length 6 cm. The radius of curvature of one surface is double than that of other surface. The small radius of curvature has value (A) 4.5 cm (B) 6 cm (C) 4 cm (D) 9 cm 30. If the distance between a projector and screen is increased by 1%, then illumination on the screen decreases by (A) 1 % (B) 2 % (C) 3 % (D) 4 % 31. A lens forms a sharp image of a real object on a screen. On inserting a parallel slide between the lens and the screen with its thickness along the principal axis of the lens it is found necessary to shift the screen parallel to itself �d� away from the lens for getting image sharply focused on it. If the refractive index of the glass relative to air is ., the thickness of slab is (A) .d (B) .d (C) 1 d . . . (D) . . . . .1 d 32. A thin convex lens in used to form a real image of a bright point object. Th
e apeture of the lens is small. A graph, shown is obtained by plotting a suitable O X Y-1 . parameter Y against another suitable parameter x. If . = the focal length of the lens u = object distance v = image distance and Real Positive Convention is used then (A) (uV) . x; (u + V) . y (B) (u + V) . x; (uV) . y (C) u . x; vu . y (D) u1 . x; v1 . y
OPTICS www.physicsashok.in 14 33. Which of the following best represents object distance u vs image distance v graph for a convex lens. (A) y . (B) y . (C) y . (D) y . 34. Three thin prisms are combined as shown in figure. The refractive indices of the crown glass for red, yellow and violet rays are .r, .y and .v respectively and those for the flint glass are .�r, .�y and .�v respectively. The ratio A�/A for which there is no net angular dispersion. (A) . . 1 2 1 yy. .. . (B) y yy y 2 .. . .. . . (C) . . . . y y y y 1 1 . . . . . . . . (D) y y2 y . y.. . . .. . 35. A point object is placed at distance of 0.3 m from a convex lens of focal length 0.2 m cut into two equal halves, each of which is displaced by 0.0005 m, as shown
in figure. If C1 and C2 be their optical centres then, (A) an image is formed at a distance of 0.6 m from C1 or C2 along principal axis. (B) two images are formed, one at a distance of 0.6 m and other at a distance O C1 C2 of 1.2 m from C1 or C2 along principal axis. (C) an image is formed at a distance of 0.12 m from C1 or C2 along principal axis. (D) two images are formed at a distance of 0.6 m from C1 or C2 along principal axis at a separation of 0.003 m. 36. A glass prism of refractive index 1.5 is immersed in water (refractive index 4/3). A light beam incident normally on the face AB is totally reflected to reach on the face BC if (1983) A B . (A) sin 89 . . (B) 2 sin 8 3 9 . . (C) sin 23 . . 37. A ray of light from a denser medium strike a rarer medium at an angle of incidence i (see Figure). The reflected and refracted rays make an angle of 90° with each other. The angles of reflection and refraction are r and r� The critical angle is i r r' (A) sin.1 .tan r . (B) sin.1 .tani. (C) sin.1 .tan r.. (D) tan.1 .sin i.
OPTICS www.physicsashok.in 15 38. Two coherent monochromatic light beams of intensities . and 4 . are superposed. The maximum and minimum possible intensities in the resulting beam are (A) 5 . and . (B) 5 . and 3 . (C) 9 . and . (D) 9 . and 3 . 39. An isosceles prism of angle 120° has a refractive index 1.44. The parallel monochromatic rays enter the prism parallel to each other in air as shown. The rays emerge from the opposite faces 120° (A) are parallel to each other (B) are diverging (C) make an angle 2 [sin�1 (0.72) � 30°] with each other (D) make an angle 2 sin�1 (0.72) with each other 40. A diminished image of an object is to be obtained on a screen 1.0 m from it. This can be achieved by appropriately placing (A) a concave mirror of suitable focal length (B) a convex mirror of suitable focal length (C) a convex lens of focal length less than 0.25 m (D) a concave lens of suitable focal length 41. A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a (A) convergent lens of focal length 3.5 R (B) convergent lens of focal length 3.0 R (C) divergent lens of focal length 3.5 R (D) divergent lens of focal length 3.0 R 42. A hollow double concave lens is made of very thin transparent material. It can be filled with air or either of two liquids L1 and L2 having refractive indices 1 . and .2 respectively ..2 . .1 .1. . the lens will diverge a parallel beam of light if it is filled with (A) air and placed in air (B) air and immersed in L1 (C) L1 and immersed in L2 (D) L2 and immersed in L1 43. A diverging beam of light from a point source Is having divergence angle . , falls symmetrically on a glass slab as shown. The angles of incidence of the two extreme rays are equal. If the thickness of the glass slab is t and the refractive index n, then the divergence angle of the emergent beam is (A) zero (B) . (C) sin 1 1n . . . . . . . (D) 2sin 1 1n . . . . . . . 44. A ray of light passes through four transparent media with refractive indices 1 . , 2 . , 3 . and 4 . as shown in the figure. the surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have (A) 1 2 . . . (B) 2 3 . . . (C) 3 4 . . . (D) 4 1 . . .
OPTICS www.physicsashok.in 16 45. A given ray of light suffers minimum deviation in an equilateral prism p, Additional prism Q and R of identical shape and of the same material as P are now added as shown in the figure. The ray will now suffer P Q R (A) greater deviation (B) no deviation (C) same deviation as before (D) total internal reflection 46. Which one of the following spherical lenses does not exhibit dispersion? The radii of curvature of the surfaces of the lenses are as given in the diagrams. (A) R1 R2 (B) R (C) R R (D) R 47. Two plane mirrors A and B are aligned parallel to each other, as shown in the figure A light ray is incident at an angle 30° at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is (A) 28 (B) 30 (C) 32 (D) 34 48. The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 2 0 cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image. (A) 1.25 cm (B) 2.5 cm (C) 1.05 cm (D) 2 cm 49. A ray of light is incident at the glass-water interface at an angle i, it emerges finally parallel to the surface of water, then the value of g . would be (A) .4 3.sini (B) 1 sin i (C) 4 3 (D) 1 50. A beam of white light is incident on glass air interface from glass to air such that green light just suffers total internal reflection. The colors of the light which will come out to air are (A) Violet, Indigo, Blue (B) All colors except green (C) Yellow, Orange, Red (D) White light
OPTICS www.physicsashok.in 17 51. An equilateral prism is placed on a horizontal surface. A ray pQ is incident onto it. For minimum deviation P Q R S (A) PQ is horizontal (B) QR is horizontal (C) RS is horizontal (D) Any one will be horizontal 52. A source emits sound of frequency 600 Hz inside water. The frequency heard in air will be equal to (velocity of sound in water = 1500 m/s, velocity of sound in air = 300 m/s) (A) 3000 Hz (B) 120 Hz (C) 600 Hz (D) 6000 Hz 53. A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1.5. The distance of virtual image from the surface is (A) 6 cm (B) 4 cm (C) 12 cm (D) 9 cm 54. A convex lens is in contact with concave lens. the magnitude of the ratio of their focal length is 2 3 . Their equivalent focal length is 30 cm. What are their individual focal lengths? (A) �15, 10 (B) �10, 15 (C) 75, 50 (D) �75, 50 55. A container is filled with water .. .1.33. upto a height of 33.25 cm. A concave mirror is placed 15 cm above the water level and the image of an object placed at the bottom is formed 25 cm below the water level. Focal length of the mirror is (A) 15 cm (B) 20 cm (C) �18, 31 cm (D) 10 cm Multiple Choice Question with ONE or MORE THAN ONE correct answer: 56. A convex lens of focal length 40 cm is in contact with a concave lens of focal length 25 cm. The power of the combination is (A) �1.5 dioptres (B) �6.5 dioptres (C) +6.5 diopres (D) +6.67 dioptres 57. A converging lens is used to form an image on a screen. When the upper half of the lens is covered by an opaque screen (A) half the image will disappear (B) complete image will be formed (C) intensity of the image will increase (D) intensity of the image will decrease. 58. A short linear object of length b lies along the axis of a concave mirror of focal length f at a distance u from the pole of the mirror. The size of the image is approximately equal to (A) 1 2 b u f f . . . . . . . (B) 1 2 b f u f . . . . . . . (C)
b u f f . . . . . . . (D) 2 b f u f . . . . . . . 59. A beam of light consisting of red, green and blue colours is incident on a right angled prism, figure. The refractive indices of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively. The prism will
OPTICS www.physicsashok.in 18 45° (A) separate part of the red colour from the green and blue colours (B) separate part of the blue colour from the red and green colours (C) separate all the three colours from one another. (D) not separate even partially any colour from the other two colours. 60. A thin prism P1 with angle 4° and made from glass of refractive index 1.54 is combined with another thin prism P2 made from glass of refractive index 1.72 to produce dispersion without deviation. The angle of the prism P2 is (A) 5.33° (B) 4° (C) 3° (D) 2.6° 61. Two thin convex lenses of focal lengths f1 and f2 are separated by a horizontal distance d (where d . f1 . d . f2 ) and their centres are displaced by a vertical separation . as shown in Figure. Taking the origin of coordinates O, at the center of the first lens the x and y coordinates of the focal point of this lens system, for a parallel beam of rays coming from the left, are given by: (A) 1 2 1 2 x f f , y f f . . . . (B) . . 1 2 1 2 1 2 , f f d x y f f d f f . . . . . . . (C) . . . . 1 2 1 1 1 2 1 2 , f f d f d f d x y f f d f f d . . . . . . . . . . (D) . . 1 2 1 1 2 , 0 f f d f d x y f f d . . . . . . 62. Which of the following form(s) a virtual and erect image for all positions of the object? (A) Convex lens (B) Concave lens (C) Convex mirror (D) Concave mirror.
63. A ray of light travelling in a transparent medium falls on a surface separating the medium from air at an angle of incidence of 45°. The ray just undergoes total internal reflection. If n is the refractive index of the medium with respect to air, select the possible value(s) of n from the following: (A) 1.3 (B) 1.4 (C) 1.5 (D) 1.6 64. A concave mirror is placed on a horizontal table, with its axis directed vertically upwards. Let O be the pole of the mirror and C its centre of curvature. A point object is placed at C. It has a real image, also located at C. If the mirror is now filled with water, the image will be. (A) real, and will remain at C (B) real, and located at a point between C and . (C) virtual, and located at a point between C and O (D) real, and located at a point between C and O
OPTICS www.physicsashok.in 19 Fill in the blanks: 1. A light wave of frequency 5 x 1014 Hz enters a medium of refractive index 1.5, In the medium the velocity of the light wave is .................. and its wavelength is ................ (2 Marks) 2. A convex lens A of focal length 20 cm and a concave lens B of focal length 5 cm are kept along the same axis with a distance d between them. If a parallel beam of light falling on A leaves B as a parallel beam, then d is equal to .......... cm. 3. A monochromatic beam of light of wavelength 6000Å in vacuum enters a medium of refractive index 1.5. In the medium its wavelength is ..., its frequency is ............. (1985) 4. In young�s double-slit experiment, the two slits act as coherent sources of equal amplitude �A� and of wavelength �. �. In another experiment with the same set-up the two slits are sources of equal amplitude �A� and wavelength �. �, but are incoherent. The ratio of the intensity of light at the midpoint of the screen in the first case to that in the second case is ................. (1986) 5. A thin lens of refractive index 1.5 has 7a focal length of 15 cm in air. when the lens is placed in a medium of refractive index 43 , its focal length will become ........... cm. (1987) 6. A point source emits sound equally in all directions in a non-absorbing medium. Two points P and Q are at a distance of 9 meters and 25 meters respectively from the source. The ratio of amplitudes of the waves at P and Q is .................... (1989) 7. A slab of a material of refractive index 2 shown in Figure, has a curved surface APB of radius of curvature 10 cm and a plane surface CD. On the left of APB is air and on the right of CD is water with refractive indices as given in the figure. AB CD P C O n1=1.0 n2=2.0 20 cm 15 cm 3 43 n = An object O is placed at a distance of 15 cm from the pole P as shown. The distance of the final image of O from P, as viewed from the left is ............... (1991) 8. A thin rod of length 3f is placed along the optic axis of a concave mirror of focal length f such that its image which is real elongated, just touches the rod. The magnification is ............. (1991) 9. A ray of light undergoes deviation of 30° when incident on an equilateral prism
of refractive index 2 . The angle made by the ray inside the prism with the base of the prism is ............. (1992) 10. A light of wavelength 6000Å in air, enters a medium with refractive index 1.5. Inside the medium its frequency is ........ Hz and its wavelength is ............... Å. (1997)
OPTICS 11. Two thin lenses, when in contact, produce a combination of power +10 diopters. When they are 0.25 m apart, the power reduces to +6 diopters. The focal length of the lenses are ...... m and .... m. (1997)
12. A ray of light is incident normally on one of the faces of a prism of apex angle 30° and refractive index 2 . The angle of deviation of the ray is .......... degrees. (1997) True / False :
13. Theintensityof light at adistance�r�from theaxisof alongcylindrical sourceisinverselyproportional to �r�. (1981) 14. A convex lens of focal length 1 meter and a concave lens of focal length 0.25 meter are kept 0.75 meter apart. A parallel beam of light first passes through the convex lens, then through the concave lens and moves to a focus 0.5 m away from the concave lens. (1983) 15. A beam of white light passing through a hollow prism give no spectrum. (1983) 16. A parallel beam of white light fall on a combination of a concave and a convex lens, both of the same material. Their focal lengths are 15 cm and 30 cm respectively for the mean wavelength in white light. On the other side of the lens system, one sees coloured patterns with violet colour at the outer edge. (1988) Table Match 17. Match List I and List II and select the correct answer using the codes given below the lists: The arrangement shows different lenses made of substance of refractive index 1.5 and kept in air. R1 = 30 cm, R2 = 60 cm. Match the focal lengths Table I Table II I. R1 R2 A. �120 cm II. R1 R2 B. +40 cm III. R1 R2 C. �40 cm R1
R2
IV. D. +120 cm (A) I-A, II-B, III-D, IV-C (B) I-C, II-A, III-B, IV-D (C) I-D, II-C, III-A, IV-B (D) I-B, II-D, III-C, IV-A www.physicsashok.in 20
OPTICS www.physicsashok.in 21 18. Table I Table II I. An object is placed at focus before A. Magnification is � . a convex mirror II. An object is placed at the centre of B. Magnification is +0.5 curvature before a concave mirror III. An object is placed at focus before C. Magnification is +1 a concave mirror. IV. An object is placed at centre of curvature D. Magnification is � 1 before a convex mirror. (A) I-B, II-D, III-A, IV-E (B) I-A, II-D, III-C, IV-B (C) I-C, II-B, III-A, IV-E (D) I-B, II-E, III-D, IV-C 19. Match the followings: Table I Table II A. Magnification m = +1 (i) Convex mirror B. Magnification 23 m . . (ii) Plane mirror C. Magnification 32 m . . (iii) Concave mirror (A) A . (ii) B . (iii) C . (i) (B) A . (i) B . (ii) C . (iii) (C) A . (ii) B . (i) C . (iii) (D) A . (iii) B . (ii) C . (i) 20. For a concave mirror of focal length 20 cm, match the followings: Table I Table II Objective distance Nature of image A. 10 cm (i) Magnified, inverted and real B. 30 cm (ii) Equal size, inverted and real C. 40 cm (iii) Smaller, inverted and real D. 50 cm (iv) Magnified, erect and virtual (A) A . II, B . I, C . III, D . IV (B) A . IV, B . I, C . II, D . III (C) A . I, B . IV, C . III, D . II (D) A . IV, B . I, C . III, D . II. PASSAGE TYPE QUESTIONS THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focal length decreases. For a clear vision, the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grownup person.
OPTICS www.physicsashok.in 22 A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can ciliary muscles are most strained in this position. For an average grown-up person, minimum distance of object should be around 25 cm. A person suffering for eye defects uses spectacles (eye glass). The function of lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle lens becomes object for eye-lens and whose image is formed on retina. The number of spectacle lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens equal to the numerical value of the power of lens with sign. For example, power of lens required is +3 D (converging lens of focal length 100 3 cm), then number of lens will be +3. For all the calculations required you can use the lens formula and lens maker�s formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens. 1. Minimum focal length of eye-lens of a normal person is (A) 25 cm (B) 2.5 cm (C) 25 9 cm (D) 25 11 cm 2. Maximum focal-length of eye lens of normal person is (A) 25 cm (B) 2.5 cm (C) 25 9 cm (D) 25 11 cm 3. A near-sighted man can clearly see object only upto a distance of 100 cm and not beyond this. The number of the spectacles lens necessary for the remedy of this defect will be (A) + 1 (B) � 1 (C) + 3 (D) � 3 4. A far-sighted man cannot see object only upto a distance of 100 cm from his eyes. The number of the spectacles lens that will make his range of clear vision equal to an average grown up person is (A) + 1 (B) � 1 (C) + 3 (D) � 3 5. A person who can see objects clearly from distance 10 cm to . , then we can say that the person is (A) normal sighted person (B) near-sighted person (C) far-sighted person (D) a person with exceptional eyes having no eye defect. THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Spherical aberration in spherical mirrors is a defect which is due to dependence of focal length ' f ' on angle of incidence '. ' as shown in figure. is given by sec
2f . R . R . C F Principal axis Pole (P) f .. where R is radius of curvature of mirror and . is the angle of incidence. The rays which are closed to principal axis are called paraxial rays and the rays far away from principal axis are called marginal rays. As a result of above dependence different rays are brought to focus at different points and the image of a point object is not a point.
OPTICS www.physicsashok.in 23 6. If p f and m f represent the focal length of paraxial andmarginal rays respectively, then correct relationship is: (A) f p . fm (B) p m f . f (C) p m f . f (D) None 7. If angle of incidence is 60°, then focal length of this rays is: (A) R (B) 2R (C) 2R (D) 0 8. The total deviation suffered by the ray falling on mirror at an angle of incidence equal to 60° is: (A) 180° (B) 90° (C) Can�t be determined (D) None 9. For paraxial rays, focal length approximately is: (A) R (B) 2R (C) 2R (D) None 10. Which of the following statements are correct regarding spherical aberration: (A) It can be completely eliminated (B) it can�t be completely eliminated but is can�t be minimised by allowing either paraxial or marginal rays to hit the mirror (C) It is reduced by taking large aperture mirrors (D) None THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Rainbow is formed during rainy season due to refraction and total internal reflection of rays falling on suspended water droplets. When rays of the sun fall on rain drops, the rain drops disperse the light and deviate the different colours by refraction and total internal reflection to the eye of the observer. A person observing the drops will see different colours of the spectrum at different angles. The rainbow which results from single total internal reflection is called primary rainbow and secondary rainbow is formed due to two total internal reflections suffered by rays falling on water drops. AB CD VRRV Red Red violet violet Rays from sun Secondary rainbow Primary rainbow Figure shows formation of rainbow due to four drops A, B, C and D. The light surffers only one total linternal reflection in drops C and D forming primary rainbow. Secondary rainbow
is formed by drops A and B where light suffers two total linternal reflections. 11. Rainbow is an arc of: (A) Circle (B) Ellipse (C) Parabola (D) Can�t be determined 12. The visibility of the rainbow is due to: (A) All rays (B) Rays undergoing maximum deviation (C) Rays undergoing minimum deviation (D) None
OPTICS www.physicsashok.in 24 13. In primary rainbow, the colour of outer edge is: (A) Blue (B) Violet (C) Red (D) None 14. In secondary rainbow, the colour of inner edge is: (A) Red (B) Violet (C) Indigo (D) None 15. The necessary condition for the observer to see rainbow is: (A) Sun, observer�s eye and the centre of the rainbow arc lie on the same line (B) Sun, observer�s eye and the centre of the rainbow arc lie on the different line (C) From any position provided sun is at the back of the observer (D) None THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE The laws governing the behavior of the rays namely rectilinear propagation, laws of reflection and refraction can be summarised in one fundamental also known as Fermat�s principle. According to this principle a ray of light travels from one point to another such that the time taken is at a stationary value (maximum or minimum). If c is the velocity of light in a vacuum, the velocity in a medium of refractive index n is cn , hence time taken to travel a distance l is nl c . If the light passes through a number of media, the total time taken is 1 nl c . . . .. . . and 1 n dl c . . If refractive index varies continuously. Now, .nl is the total optical path, so that Fermat�s principle states that the path of a ray is such that the optical path in at a stationary value. This principle is obviously in agreement with the fact that the ray are straight lines in a homogenous isotropic medium. It is found that it also agrees with the classical laws of reflection and refraction. 16. If refractive index of a slab varies as . .1. x2 where x is measured from one end, then optical path length of a slab of thickness 1 m is: (A) 43 m (B) 34 m (C) 1 m (D) None 17. The optical path length followed by ray from point A to B given that laws of reflection are obeyed as shown in figure is: A B P (A) Maximum (B) Minimum (C) Constant (D) None 18. The optical path length followed by ray from point A to B given that laws of reflection are obeyed as shown in figure is A B
(A) Maximum (B) Minimum (C) Constant (D) None
OPTICS OPTICS 19. The optical path length followed by ray from point A to B given that laws of refraction are obeyed as shown in figure is A B (A) Maximum (B) Minimum (C) Constant (D) None 20. The optical path length followed by ray from point A to B given that laws of refraction are obeyed as shown in figure is A and B are focii of ellipse A B (A) Maximum (B) Minimum (C) Constant (D) None THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE One hard and stormy night you find yourself lost in the forest when you come upon a small hut. Entering it you see a crooked old woman in the corner hunched over a crystal ball. You are about to make a hasty exit when you hear the howl of wolves outside. Taking another look at the gypsy you decide to take your chances with the wolves, but the door is jammed shut. Resigned to bad situation your approach her slowly, wondering just what is the focal length of that nifty crystal ball. 21. If the crystal ball is 20 cm in diameter with R.I. = 1.5, the gypsy lady is 1.2 m from the central of ball, where is the image of the gypsy in focus as you walk towards her? (A) 6.9 cm from the crystal ball (B) 7.9 cm from the crystal ball (C) 8.9 cm from the crystal ball (D) None 22. The image of old lady is: (A) real, inverted an enlarged (B) erect, virtual and small (C) erect, virtual and magnified (D) real, inverted and diminished 23. The old lady moves the crystal ball closer to her wrinkled old face. At some point you can no longer get an image of her. At what object distance will there be no change of the gypsy formed? (A) 10 cm (B) 5 cm (C) 15 cm (D) None THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE
The table below contains some physical properties of common optical materials. The refractive index of a material is a measure of the amount by which light is bent upon entering the material. The transmittance range is the range of wavelengths over which the material is transparent.
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OPTICS www.physicsashok.in 26 Physical Properties of Optical Materials Material Refractive index for light of 0.589 .m Transmittance range (.m) Useful range for prisms (.m) Chemical resistance Lithium fluoride 1.39 0.12-6 2.7-5.5 Poor Calcium fluoride 1.43 0.12-12 5-9.4 Good Sodium chloride 1.54 0.3-17 8�16 Poor Quartz 1.54 0.20-3.3 0.20-2.7 Excellent Potassium bromide 1.56 0.3-29 15�28 Poor Flint glass* 1.66 0.35�2.2 0.35-2 Excellent Cesium iodide 1.79 0.3�7.0 15-55 Poor *Flint glass is lead oxide doped quartz. 24. According to the table, which material(s) will transmit light at 25 .m � (A) Potassium bromide only (B) Potassium bromide and cesium iodide (C) Lithium fluoride and cesium iodide (D) Lithium fluoride and flint glass 25. A scientist hypothesizes that any material with poor chemical resistance would have a transmittance range wider than 10.m. The properties of which of the following materials contradicts this hypothesis� (A) Lithium fluoride (B) Flint glass (C) Cesium iodide (D) Quartz 26. When light travels from one medium to another, total internal reflection can occur if the first medium has a higher refractive index than the second. Total internal reflection could occur if light were travelling from� (A) Lithium fluoride of flint glass (B) potassium bromide to cesium iodide (C) quartz to potassium bromide (D) flint glass to calcium fluoride 27. Based on the information in the table, how is the transmittance range related to the useful prism range� (A) The transmittance range is always narrower than the useful prism range (B) The transmittance range is narrower than or equal tot he useful prism range (C) The tranmittance range increases as the useful prism range decreases (D) The tranmittance range is wider than and includes within it the useful prism range 28. The addition of lead oxide to pure quartz has the effect of� (A) decreasing the transmittance range and the refractive index (B) decreasing the transmittance range and increasing the refractive index (C) increasing the transmittance range and the useful prism range
(D) increasing the transmittance range and decreasing the useful prism range. THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE A periscope viewing system is to be used to observe the behavior of primates in a large environmentally controlled room on the upper floor of a large research facility. The periscope, like those used on
OPTICS OPTICS submarines, is essentially a large, folded-path, low power telescope (using prisms to fold the light path). A sketch of the preliminary design appears below. Like all Newtonian telescopes, it uses a relatively long focal length objective lens to form a real image in front of the eyepiece lens (of shorter focal length). The observer looks through the eyepiece lens to see the final image, in the same manner that one would use a magnifying glass.
The distance between the lenses is approximately equal to the sum of their focal lengths. The eyepiece,
(A) The focal length of the eyepiece lens is too short. in this design, can be moved forward or back in order to focus on the primates as they move closer to or further away from the objective lens. 29. The total tube length of the three sections is to be 4 m. The objective lens available has a focal length of 3 m. What should the focal length of the eyepiece lens be? (A) 0.75 m (B) 1 m (C) 1.33 m (D) 7 m 30. A visitor seeing the sketch points out an important flaw that will require a design change. what is the flaw? (B) The images of the primates will be inverted (C) The objective lens should be a diverging lens. (D) The prisms cannot be used in this way. 31. A visitor seeing the sketch points out an important flaw that will require a design change. What is the flaw? (A) The focal length of the eyepiece lens is too short. (B) The images of the primates will be inverted (C) The objective lens should be a diverging lens. (D) The prisms cannot be used in this way. 32. What will be the approximate magnification of this periscope? (A) 0.67x (B) 1x (C) 3x (D) 300x 33. The prisms (45�45�90° prisms) turn the light path through 90° by �total internal reflection� from the inside hypotenuse faces of the prisms when the incident angle is 45° as in the sketch. (C) Can one use crown glass with an index of refraction of 1.52 for the prism? (A) yes, because the critical angle for crown glass is 47° (B) yes, because the critical angle for crown glass is 41°. No, because the critical angle for crown glass is exactly 47° (D) No, because the critical angle for crown glass is exactly 41°. 34. Describe the properties of the image that one sees with this preliminary design (A) real, inverted, magnified (B) real, upright, magnified (C) virtual, upright, same size as object. (D) virtual, inverted, magnified 35. The telescope is focused on a primate rather far away on the farside of the large habitat. As the primate moves rather closer to the telescope, what must the observer do to see the primate clearly?
(A) No change, the image remains clear. (B) Move the eyepiece away from the objective. (C) Move the eyepiece closer to the objective. (D) Use an inverting eyepiece because the image flips. www.physicsashok.in 27
OPTICS www.physicsashok.in 28 THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE In the normal human eye, light from an object is refracted by the cornea-lens system at the front of the eye and produces a real image on the retina at the rear of the eye. For a given eye, its lens-to-retina distance is fixed at about 2.5 cm. Most of the focusing of an image is done by the cornea, which has a fixed curvature that is convex with respect to incoming light. The importance of the lens is that its radius of curvature ccan be changed, allowing the lens to fine-tune the focus. The lens is surrounded by the ciliary muscle. Contraction of the muscle decreases tension on the lens. This allows the natural elasticity of the lens to produce an increase in the radius of curvature. when the muscle relaxes, the lens flattens out, decreasing tis radius of curvature. Unfortunately, the lens losses elasticity with age and the ability to alter curvature decreases. The range over which clear vision is possible is bounded by the far point and the near point. In normal vision the far point is infinity and the near point depends on the radius of curvature of the lens. For normal eyes the average near point for reading is 25 cm. AGE, years NEAR POINT, cm 10 7 20 10 30 14 40 22 50 40 60 200 In the myopic (nearsighted) eye, the lens-to-retina length is too long and/or the radius of curvature of the cornea is too great. This causes rays from an object at infinity to focus at a point in front of the retina. The far point is closer than normal .A corrective lens will put a virtual image of a distant object at the position of the actual far point of the eye. In the hyperopic (farsighted) eye, the lens-to-retina length is too short and/or the radius of curvature of the cornea is not great enough. This causes rays from an object at infinity to focus at a point behind the retina. The near point is farther away than normal. A corrective lens will put a virtual image of the close object at the position of the actual near point. The relation among the object (o) and image (i) distances from the eye and the focal length (f) of the lens is given by the lens-distance rule : 1 o .1 i .1 f . When using this equation, all distances are given in centimeters. The power of corrective lenses is usually given in units called diopters. Power, in diopters, is the reciprocal of the focal length in meters : 1 diopter meter P . f . By convection � I.Converging lenses have positive focal lengths, and diverging lenses have negative focal lengths. II. Real images have positive distances from the lens, and virtual images have negative distances from the lens. 36. The lens system of the myopic eye is best described as �
(A) producing too much convergence. (B) producing too little convergence. (C) producing too much divergence. (D) producing too little divergence. 37. An optometrist examined John�s eyes. The farthest object he can clearly focus on with his right eye is 50 cm away. What is the power of the contact lens required to correct the vision in his right eye � (A) �0.50 diopters (B) �2.0 diopters (C) +2.0 diopters (D) +5.0 diopters
OPTICS OPTICS 38. In a mildly hyperopic eye, the focal length of the eye�s natural lens can be corrected by � (A) contracting the muscle and increasing the radius of curvature. (B) contracting the ciliary muscle and decreasing the radius of curvature (C) relaxing the ciliary muscle and increasing the radius of curvature. (D) relaxing the ciliary muscles and decreasing the radius of curvature. 39. Jane must wear a contact lens with a power of +3.00 diopters in one eye to be able to clearly focus on an object 2.5 cm in front of the eye. Based on the vision in this eye, which of the following is the most likely age range for Jane � (A) Less than 40 years old (B) From 40 to 49 years old (C) From 50 to 59 years old (D) 60 years or older (B) RK corrects myopia by increasing the focal length of the eye. (C) RK corrects hyperopia by decreasing the focal length of the eye. 40. George wears eyeglasses that sit 2.0 cm in front of his eyes. His incorrect far point is 50 cm. What is the focal lengths of his eyeglasses � (A) �50 cm (B) +50 cm (C) �48 cm (D) +48 cm 41. In a surgical procedure called radial keratotomy, (RK), a laser is used to flatten the cornea by placing as series of hairline cuts around the perimeter of the cornea. Which statement is most accurate � (A) RK corrects myopia by decreasing the focal length of the eye. (D) RK corrects hyperopia by increasing the focal length of the eye. THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Student are given a variety of lenses and optics equipment, such as lens holders, lighted object sources. Optical benches, meter sticks and tapes, image screens, and several examples of commercial optical equipment, such as microscopes and telescopes. They are to work in an open-ended optics lab in order to learn the general principles of lenses and the optical devices that can be constructed using lenses. 42. A student is given a short focal length converging lens and long focal length converging lens. One lens is placed in a holder. A lighted object is placed 18 cm in front of the lens and it is found that a clear image can be focused on a screen placed 36 cm behind the lens. what is the focal length of this lens? (A) 8 cm (B) 12 cm (C) 27 cm (D) 46 cm 43. What magnification is produced by the above lens when the object is 18 cm in front of the lens and the image is 36 cm behind the lens? (B) 3x (C) 4x (D) 6x 44. (A) 2x is the image and which kind of image is it? A lighted object is placed 6 cm in front of the second lens, which has a focal length of +24 cm. Where (A) 8 cm in front of the lens: a virtual image. (B) 8 cm behind the lens: a real image (C) 16 cm in front of the lens: a real image (D) 16 cm behind the lens; a virtua
l image. 45. The 24 cm focal length lens is used as the objective of a simple refracting telescope and a third converging lens of focal length +8 cm is used as the eyepiece. What is the magnification of this simple refractor? (A) 0.6x (B) 3x (C) 4x (D) 6x 46. A commercial microscope is examined by the student. The objective is marked 20x and the eyepiece is marked 10x. what power objective should replace the above objective so that the microscope�s magnification will be 400x (A) 5x (B) 10x (C) 40x (D) 100x www.physicsashok.in 29
OPTICS www.physicsashok.in 30 47. A lighted object is placed 24 cm in front of a +12 cm focal lengths lens. The image formed by this lens is the object for a second lens of +24 cm focal length. The second lens is placed 72 cm behind the first lens. where is the final image with respect to the second lens? (A) 24 cm in front of # 2 (B) 24 cm behind # 2 (C) 36 cm in front of # 2 (D) 48 cm behind # 2 48. A lens of focal length +24 cm is used to view an object placed 12 cm in front of the lens. The object is 5 cm tall. How tall is the image? (A) 2.5 cm (B) 3.3 cm (C) 7.5 cm (D) 10 cm 49. A diverging lens of focal length �24 cm is now used with the object 12 cm in front of the lens. How tall is the image if the object is 5 cm tall? (A) 2.5 cm (B) 3.3 cm (C) 8 cm (D) 10 cm 50. A near sighted student cannot see objects clearly unless they are as close as 80 cm (his �far-point�). The image that he sees through his new contact lens is a virtual image because he looks through the lens to see the image. what focal length lenses does he need in order to see very distance objects, such as the starts? (A) �20 cm (B) �30 cm (C) �4 cm (D) � 25 cm THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE The phenomenon of refraction has long intrigued scientists and was actually used to corroborate one of the major mysteries of early science: the determination of the speed of light. The refractive index of a transparent material irrelated to a number of the physical properties of light. In terms of velocity, the refractive index represents the ratio of the velocity of light in a vacuum to its velocity in the material. From this ratio, it can be seen that light is retarded when it passes through most types of matter. It is worth noting that prisms break up white light into the seven �colors of the rainbow� because each color has a slightly different velocity in the medium. Snell�s law allows one to follow the behavior of light in terms of its path when moving from a material of one refractive index to another with the same, or different refractive index. It is given by: n1 sin.1 . n2 sin.2 , where �I� refers to the first medium through which the ray passes, �2� refers to the second medium, and the angles refer to the angle of incidence in the first medium . . 1 .and the angle of refraction in the second . . 2 . . A ship went out on a search for a sunken treasure chest. In order to locate the chest, they shone a beam of light down into the water using a high intensity white light source as shown in Figure. The refractive index for sea water is 1.33 while that for air is 1.00. 51. From the information in the passage, how would you expect the speed of light in air to compare with the speed of light in a vacuum (which is given by �c�)? (A) It would be the same (=c) (B) It would be greater than c. (C) It would be less than c. (D) This cannot be determined from the information
given.
OPTICS www.physicsashok.in 31 52. Using the information in the passage, what must the approximate value of 2 . be such that it hit the chest as shown in Figure? (A) 15.2° (B) 30.4° (C) 45.6° (D) 63.4° 53. How does the refractive index in water light compare with that of red light given that violet light travels more slowly in water than red light? (A) nviolet . nred (B) violet red n . n (C) violet red n . n (D) This depends on the relative speeds of the different colors in a vacuum. 54. Total internal reflection first occurs when a beam of light travels from one medium to another medium which has a smaller refractive index at such an angle of incidence that the angle of refraction is 90°. This angle of incidence is called the critical angle. What is the value of the sine of this angle when the ray moves from water towards air? (A) 2 (B) 0.75 (C) 0.50 (D) 0 55. What would happen to the critical angle, in the previous question, if the beam of light was travelling from water to a substance with a greater refractive index than air, but a lower refractive index than water? (A) It would increase (B) It would decrease (C) It would remain the same (D) Total internal reflection would not be possible. 56. Which of the following would you expect to remain constant when light travels from one medium to another and the media differ in their refractive indices? (A) Velocity (B) Frequency (C) Wavelength (D) Intensity. THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE The invention of the compound microscope by Jansen in the late 1500�s truly revolutionized the world of science, particularly the field of cellular and molecular biology. The discovery of the cell as the fundamental unit of living organisms and the insight into the bacterial world are two of the contributions of this instrument to science. It is unseemly that such a relatively simplistic apparantus took generations to be developed. Its main component are two convex lenses: one acts as the main magnifying lens and is referred to as the objective, and another lens called the eyepiece. The two lenses act independently of each other when bending light rays. The actual lens set-up depicted in Figure. Light from the object (O) first passes thought he objective and an enlarged, inverted first image is formed. The eyepiece then magnifies this image. Usually the magnification of the eyepiece is fixed (either x 10 or x 10) and three rotating objective lenses are used : x 10, x 40 and x 60. The most recent development in microscope technology is the electron microscope which uses a beam of electrons instead of light. Photographic film must be used otherwise no image would be formed on the retina. This microscope has a resolution about a hundred times that of the light microscope.
57. Based on the passage, what type of image would have to be produced by the objective magnification? (A) Either virtual or real (B) Virtual (C) Real (D) It depends on the focal length of the lens.
OPTICS www.physicsashok.in 32 58. Where would the first image have to be produced by the objective relative to the eyepiece such that a second, enlarged image would be generated on the same side of the eyepiece as the first image (first image distance = d1)? (A) di . Fe (B) i e d . F (C) 2 e i e . F . d . F (D) 2 i e d . . F 59. Two compound microscopes A and B were compared. Both had objectives and eyepieces with the same magnification but A gave an overall magnification that was greater than that of B. Which of the following is a plausible explanation? (A) The distance between objective and eyepiece in A is greater than the corresponding distance in B. (B) The distance between objective and eyepiece in A is less than the corresponding distance is B. (C) The eyepiece and objective positions were reversed in A. (D) The eyepiece and objective positions were reversed in B. 60. A student attempted to make a compound microscope. However, when she tried to view an object through the apparatus, no image was seen. Which of the following could explain the mishap? I. The object distance = focal length of objective. II. The object distance for eyepiece lens as her eyepiece. III. The student used a diverging lens as her eyepiece. IV. The student used a converging lens as her objective (A) I, II, III and IV (B) I, II, III (C) I, II, IV (D) II, III, IV 61. The magnification of the eyepiece of a compound microscope is x15. The image height is 25 mm and the magnification of the objective is x40. What is the object height? (A) 1.67 mm (B) 0.60 mm (C) 0.38 mm (D) 0.04 mm 62. What is the refractive power of an objective lens with a focal length of 0.50 cm? (A) 0.2 diopters (B) 2.0 diopters (C) 20 diopters (D) 200 THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Magnification by a lens of an object at distance 10 cm from it is �2. Now a second lens is placed exactly at the same position where first was kept, without changing the distance object and lens. The magnification by this second lens is � 3. 63. Now both the lenses are kept in contact at the same place. What will be the new magnification. (A) 13 5 . (B) 12 7 . (C) 6 11 . (D) 57 . 64. What is the focal length of the combination when both lenses are in contact.
(A) 60 17 cm (B) 5 17 cm (C) 12 7 cm (D) 13 9 cm THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE In the case of convex lens, when object is moved from f to 2f, its image is real, inverted and magnified. It moves from f to infinity on other side. 65. Focal length of a convex lens is 10 cm. When the object is moved from 15 cm to 25 cm, the magnitude of linear magnifications. (A) will increase (B) will decrease (C) will first increase then decrease (D) will first decrease than increase.
OPTICS www.physicsashok.in 33 66. Image of object AB shown in figure will be like: 2F A F B (A) F 2F A' B' (B) F 2F A'B' (C) F 2F A' B' (D) F 2F A' B' THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Figure shows a simplified model of the eye that is based on the assumption that all of the refraction of entering light occurs at the cornea. The cornea is a converging lens located at the outer surface of the eye with fixed focal length approximately equal to 2 cm. Parallel light rays coming from a very distant object are refracted by the cornea to produce a focused image on the retina. The retina then transmits electrical impulse along the optic nerve to the brain. cornea retina Two common defects of vision are myopia and hyperopia. Myopia, sometimes referred to as nearsightedness, occurs when the cornea focuses the image of a distance object in front of the retina. Hyperopia, sometimes referred to as farsightedness, occurs when the cornea focuses the image of a nearby object behind the retina. Both of these problems can be corrected by introducing another lens in front of the eye so that the two lens system produces a focused image on the retina. If an object is so far away from the lens system that its distance may be taken as infinite, then the following relationship holds: 1 1 1 c l f f x v . . . , where c f is the focal length of the cornea, l f , is the focal length of the correcting lens, x is the distance from the correcting lens to the cornea, and v is the image distance measured from the cornea. (Note : The index of refraction is 1.0 for air and 1.5 for glass). 67. How far away should the retina be from the cornea for normal vision? (A) 0.5 cm (B) 1.0 cm (C) 2.0 cm (D) 4.0 cm 68. For a distant object, the image produced by the cornea is: (A) real and inverted (B) real and upright (C) virtual and inverted (D) virtual and upright. 69. What kind of lens would be suitable to correct myopia and hyperopia respectively? (Note : Assume that the correcting lens is at the focal point of the cornea so that c x . f .)
(A) Converging, converging (B) Converging, diverging (C) Diverging, diverging (D) Diverging, converging 70. The focal length of a woman�s cornea is 1.8 cm, and she wears a correcting lens with a focal length of .16.5 cm at a distance x = 1.5 cm from her cornea. What is the image distance v measured from the cornea for a distant object? (A) 1.0 cm (B) 1.5 cm (C) 2.0 cm (D) 2.5 cm
OPTICS www.physicsashok.in 34 71. In the case of contact lens, the cornea and the correcting lens are actually touching and act together as a single lens. If the focal length of both the cornea and the contact lens are doubled, then the image distance v for a distant object would: (A) be 1/4 the old value (B) be 1/2 the old value. (C) be the same as the old value (D) be twice the old value. Level # 2 1. In what direction should a beam of light be sent from point A (Figure) contained in a mirror box for it to fall onto point B after being reflected once from all four walls? A B Point A and B are in one plane perpendicular to the walls of the box (i.e., in the plane of the drawing). 2. A concave mirror has the form of a hemisphere with a radius R = 55 cm. A thin layer of an unknown transparent liquid is poured into this mirror, and it was found that the given optical system produces, with the source in a certain position two real image, one of which (formed by direct reflection) coincide with source and the other is at a distance of . = 30 cm from it. Find the refractive index . of the liquid. 3. A point source of light S is placed on the major optical axis of concave mirror at a distance of 60 cm. At what distance from the concave mirror should a flat mirror be placed for the rays to converge again at the point S having been reflected from the concave mirror and then from the flat one?Will the position of the point where the rays meet change if they are first reflected from the flat mirror? The radius of the concave mirror is 80 cm. 4. A pile 4 m high driven into the bottom of a lake is 1 m above the water. Determine the length of the shadow of the pile on the bottom of the lake if the sun rays make an angle of 45° with the water surface. The refractive index of water is 4/3. 5. In figure, a fish water watches a fish through a 3.0 cm thick glass wall of a fish tank. The watcher is in level with the fish; the index of refraction of the glass is 8/5 and that of the water is 4/3. Observer 8.0 cm Wall Water 3.0 cm 6.8 cm (a) To the fish, how far away does the watcher appear to be? (B) To the watcher, how far away does the fish appear to be? 6. A hollow sphere of glass of refractive index . has a small mark on its interior surface which is observed from a point outside the sphere on the side opposite the center. The inner cavity is concentric with external surface and the thickness of the glass is every where equal where equal to the radius of the inner surface. Prove that the mark will appear nearer than it really is, by a distance . .
.3 1. 1 R . . . . , where R is the radius of the inner surface.
OPTICS www.physicsashok.in 35 7. A long rectangular slab of transparent medium of thickness d is placed on a table with length parallel to the x-axis and width parallel to the y-axis. A ray of light is travelling along y-axis at origin. the refractive index . of the medium varies as 1 .x r. 0 .. . . , where .0 and A Y O X medium d r (> 1) are constants. The refractive index of air is 1. (a) Determine the x-coordinate of the point A, where the ray intersects the upper surface of the slab-air boundary. (B) Write down the refractive index of the medium at A. (C) Indicate the subsequent path of the ray in air. 8. A man of height 2.0 m is standing on level road where because of temperature variation the refractive index of air is varying as . . 1. ay , where y is height from road. If a = 2.0 x 10�6 m�1. Then find distant point that he can see on the road. 9. A portion of straight glass rod of diameter 4 cm and refractive index 1.5 is bent into an arc of circle of radius R. A parallel beam of light is incident on R it as shown in the figure. Find the smallest value of R which permits all the light to pass around the arc. 10. A glass sphere has a radius of 5.0 cm and a refractive index of 1.6. A paperweight is constructed by slicing through the sphere on a plane that is 2.0 cm from the centre of the sphere and perpendicular to a radius of the sphere that passes through the 8.0 cm 3.0 cm 5.0 cm Observer center of the circle formed by the intersection of the plane and the sphere. The paperweight is placed on a table and viewed directly above by an observer who is 8.0 cm from the table top as shown in figure. when viewed through the paperweight, how far away does the tabletop appear to the observer? 11. A ray of light is incident on a composite slab at a angle of incidence i as shown in the figure. Find the lateral shift x of the ray when it comes out from the otherside. 12. A prism of apex angle A is made up of a material of refractive index .. The refractive indices of the mediums on the left and right sides are .1 and .2 respectively. A ray of light is incident A .1 .2 . i from the side of medium of refractive index .1 at an angle i and comes out from the other side as shown in the figure. Find the
angle of deviation.
OPTICS www.physicsashok.in 36 13. A hemisphere of radius a/2 and made up of a material of variable refractive index is placed with its base centre O at the origin as shown in the figure. The refractive index of the material of the hemisphere varies as a x a. . . . A ray of light is incident at the point O at an angle . with the normal in the xy plane and comes out through a point P on its curved surface. Find the coordinate of the point P if . . 0 . 14. A ray of light is incident on the sphere of radius R and refractive index . as shown in the figure. The incident ray is parallel to a horizontal diameter and b R u the distance between the incident ray and the horizontal diameter is b. Find the angle of deviation . suffered by the ray. 15. An intense beam parallel to the principal axis is incident on a convex lens. Multiple extra images F1, F2, ...... are formed due to feeble internal reflections, called flare spots as shown in the figure. The radius of curvature of the lens is 30 cm and Principal axis F1 F2 F0 60 cm and the refractive index is 1.5. Find the position of the first flare spot. 16. The image of the object shown in the figure is formed at the bottom of the tray filled with water. From the details given in the figure, calculate the value of h. O 36 cm 1 mh 85 cm ./4 . = 30 cm 17. In the given figure there are two thin lenses of same focal length . arranged with their principal axes inclined at an angle .. The separation between the optical centers of the lenses is 2 . . A point object lies on the principal axis of the O X Y . 2. convex lens at a large distance to the left of convex lens. (a) Find the coordinates of the final image formed by the system of lenses taking O as the origin of coordinate axes, and (B) Draw the ray diagram.
OPTICS www.physicsashok.in 37 18. (a) A prism has refracting angle equal to . /2. It is given that . is the angle of minimum deviation and . is the deviation of the ray entering at grazing incidence. Prove that sin . = sin2 . and cos . = .cos . (b) A ray of light passes through a prism in a principal plane the deviation being equal to angle of incidence which is equal to 2.. It is given that . is the angle of prism. Show that 8 . 1. cos 12 2 . . . . where . is the refractive index of the material of prism. 19. A thin flat glass plate is placed in front of a convex mirror. At what distance b from the plate should a point source of light S be placed so that its image produced by the rays reflected from the front surface of the plate coincides with the image formed by the rays reflected from the mirror? a b S The focal length of the mirror is . = 20 cm and the distance from the plate to the mirror a = 5 cm. How can the coincidence of the images be established by direct observation? 20. A concave mirror forms the real image of a point source lying on the optical axis at a distance of 50 cm from the mirror. The focal length of the mirror is 25 cm. The mirror is cut in two and its halves are drawn 1 cm a distance of 1 cm apart in a direction perpendicular to the optical axis. How will the images formed by the halves of the mirror be arranged? 21. A glass hemisphere of radius 10 cm and . = 1.5 is silvered over tis curved surface. There is an air bubble in the glass 5 cms from the plane surface along the axis. Find the position of the images of this bubble seen by observer looking along the axis into the flat surface of the hemisphere. 22. The height of a candle flame is 5 cm. A lens produces an image of this flame 15 cm high on a screen. Without touching the lens, the candle is moved over a distance of . = 1.5 cm away from the lens, and a sharp image of the flame 10 cm high is obtained again after shifting the screen. Determine the main focal length of the lens. 23. A thin converging lens of focal length . is moved between a candle and a screen. The distance between the candle and the screen is d (> 4 . ). Show that for two different positions of the lens, two different images can be obtained on the screen. If the ratio of dimensions of the image is . , find the value of ( . + 1/ . ). 24. Three convergent thin lenses of focal lengths 4a, a and 4a respectively are placed in order along the axis so that the distance between consecutive lenses is 4a. Prove that this combination simply inverts every small object on the axis without change of magnitude or position. 25. A converging bundle of light rays in the shape in the shape of a cone with the vertex angle of 40° falls on a circular diaphragm of 20 cm diameter. A lens with a focal power of 5 diopters
is fixed in the diaphragm. What will the new cone angle be? 26. A ray of light is incident on the spherical surface of radius of curvature R as shown in the figure. Therefractive index on the right side of spherical surface is .. The medium ont he left side of the spherical surface is air. The distance of the incident ray from the axis of the spherical surface is b. After refraction the ray intersects the axis at a point. F. Find the distance . of the point F from the pole O.
OPTICS www.physicsashok.in 38 27. Consider an arrangement of two equibi convex lenses of focal length in air 10 cm. The refractive index of the glass of which the lenses are between the made is .g = 3/2 and the refractive index of water filling 10 cm 10 cm O AIR AIR .g .g the space two lenses is .w = 4/3. A small object O is placed on the axis at a distance of 10 cm from the first lens in air as shown in the figure. The distance of separation between the two lenses is 10 cm. Find the position and magnification of the final image. 28. A thin convex lens of focal length 1m is cut into three parts A, B and C along the diameter. The thickness of the middle layer C is 1 cm. The middle layer is now removed and the two parts A nad B are put together to form a composite lens. Then the part C is also placed infront of this A A B B CC 1 cm composite lens symmetrically as shown in the figure. A paraxial beam of light is incident along tyhe axis of the part C. Find the distance between the two images formed. 29. An equi biconvex lens of focal length 10 cm in AIR and made up of material of refractive index 3/2 is polished on one side. Another identical lens (not polished) is placed infront of the polished lens at a distance of 10 cm as shown in the figure. The space between 10 cm 10 cm the two lenses is filled with a liquid of refractive index 4/3. An object O is placed infront of the unpolished lens at a distance of 10 cm. Find the final position of the image. 30. Consider an equilateral prism ABC as shown in the figure. A ray of light is incident on the face AB and gets transmitted into the prism. Then total internal reflection takes place at the face BC and the ray comes out of prism through the face AC. The total A B C 60° 60° 60° angle of deviation is 120°. Find the refractive index . of the material of the prism. Level # 3 1. An object is placed 21 cm in front of a concave mirror of radius of curvature 10 cm. A glass slab of thickness 3 cm and refractive index 1.5 is then placed close to the mirror in the space between the object and the mirror. Find the position of the final image formed. (You may take the distance of the near surface of the slab from the mirror to be 1 cm). [IIT 1980] 2. The x-y plane is the boundary between two transparent media. Medium �1 with z . 0 has refractive index 2 and medium �2 with z . 0. 0, has a refractive index 3 . A ray of light in
medium �1 given by the vector k�10 j�3 8 i�A . 6 3 . . . is incident on the plane of separation. Find the unit vector in the direction of the refracted ray in medium � 2. [IIT 1999] 3. An object is placed in front of a convex mirror at a distance of 50 cm. A plane mirror is introduced covering lower half of the convex mirror. If the distance between the object and the plane mirror is 30 cm, it is found that there is no parallux between the images formed by two mirrors. What is the radius of curvature of the convex mirror? [IIT 1973]
OPTICS www.physicsashok.in 39 4. A rectangular block of glass is placed on a printed page lying on a horizontal surface. Find the value of the refractive index of glass for which the letters on the page are not visible from any of the vertical faces of the block. [IIT 1979] 5. A glass lens has focal length 5 cm in air. What will be its focal length in water. (Refractive index of glass is 1.51 and that of water is 1.33). [IIT 1977] 6. A ray of light is travelling form diamond to glass. Calculate the minimum angle of incidence of the ray as the diamond glass interface such that no light is refracted into glass. What will happen if the angle of incidence exceeds the angle? (refractive index of glass is 1.51 and that of diamond is 2.47)[IIT 1977] 7. What is the velocity of light in glass of refractive index 1.5? (Velocity of light in air = 3 x 1010 cm/sec.) [IIT 1976] 8. Photographs of the ground are taken from an aircraft flying at an altitude of 2000 meters by a camera with a lens of focal length 50 cm. the size of the film in the camera is 18 cm x 18 cm. What area of the ground can be photographed by this camera at any one time? [IIT 1976] 9. A rectangular glass block of thickens 10 cm and refractive index 1.5 is placed over a small coin. A beaker filled with water of refractive index 34 to a height of 10 cm and is placed over the glass block. (a) Find the apparent position of the object when it is viewed at near normal incidence. (b) Draw a neat ray diagram. (c) If the eye is slowly moved away from the normal at a certain position the object is found to disappear due to total internal reflection. At which surface does this happen and why? [IIT 1975] 10. A ray of light travelling in air is incident at grazing angle (incident angle = 90°) on a long rectangular slab of a transparent medium of thickness t = 1.0 m (see figure). The point of incidence is the origin A (0, 0). The medium has a variable index of refraction n (y) given by n(y) = [Ky3/2 + 1]½ where K = 1.0 (meter)�3/2 The refractive index of air is 1.0. (a) Obtain a relation between the slope of the trajectory of the ray at a point B (x, y) in the medium and the incident angle at that point. (b) Obtain an equation for the trajectory y (x) of the ray in the medium. (c) Determine the coordinates (x1, y1) of the point P, where the ray intersects the upper surface of the slab-air boundary. (d) Indicate the path of the ray subsequently. [IIT 1995] 11. A quarter cylinder of radius R and refractive index 1.5 is placed on a table. A point object P is kept at a distance of mR from it. Find the value of m for which a ray from P will emerge parallel to the table as shown in the figure. [IIT 1999] 12. A light ray is incident on an irregular shaped slab of refractive index
2 at an angle of 45° with the normal on the incline face as shown in the figure. the ray finally emerges from the curved surface in the medium of the refractive index . = 1.514 and passes through point E. If the radius of curved surface is equal to 0.4 m, find the distance OE correct up to two decimal places. [IIT 2004]
OPTICS OPTICS 13. A point object O is placed at a distance of 12 cm on the axis of a convex lens of focal length 10 cm. On the other side of the lens, a convex mirror is placed at a distance of 10 cm from the lens such that the image formed by the combination coincides with the object itself. What is the focal length of the convex mirror? [IIT 1976]
14. An object of height 4 cm is kept to the left of and on the axis of a converging lens of focal length 10 cm as shown in figure. A plane mirror is placed inclined at 45° to the lens axis 10 cm to the right of the lens (see figure). Find the position and size of the image formed by the lens and mirror combination. trace the rays forming the image. [IIT 1972] 15. An object is placed at 20 cm left of the convex lens of focal length 10 cm. If a concave mirror of focal length 5 cm is placed at 30 cm to the right of the lens find the magnification and the nature of the final image. Draw the ray diagram and locate the position of the final image. [IIT 1974] 16. An object is approaching at thin convex lens of focal length 0.3 m with a speed of 0.01 m/s. Find the magnitudes of the rates of change of position and lateral magnification of image when the object is at a distance of 0.4 m from the lens. [IIT 2004] 17. A thin biconvex lens of refractive index 3 is placed on a horizontal
2 plane mirror as shown in the figure. The space between the lens
4
andthemirroristhenfilledwithwaterofrefractiveindex .
3 It is found that when a point object is placed 15 cm above the lens on its principle axis, the object coincides with its own image. On repeating with another liquid, the object and the image again coincide at a distance 25 cm from the lens. Calculate the refractive index of the liquid. [IIT 2001]
18. A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axes PQ and RS parallel but separated in vertical direction by 0.6 cm as shown. The distance between the lens and mirror is 30 cm. An upright object AB of height 1.2 cm is placed on the optic axis PQ of the lens at a distance of 20 cm from the lens. If A�B� is the image after refraction from the lens and reflection from the mirror, find the distance of A�B� from the pole of the mirror and obtain its magnification. Also locate position of A� and B� with respect to the optic axis RS. [IIT 2000] www.physicsashok.in 40
OPTICS www.physicsashok.in 41 19. A thin equiconvex lens of glass of refractive index 2 . . 3 and of focal length 0.3 m in air is sealed into an opening at one end of a tank filled with water ... .... . 34 . On the opposite side of the lens, a mirror is placed inside the tank on the tank wall perpendicular to the lens axis, as shown in figure. The separation between the lens and the mirror is 0.8 m. A small object is placed outside the tank in front of the lens at a distance of 0.9 m from the lens along its axis. Find the position (relative to the lens) of the image of the object formed by the system. [IIT 1997, May] 20. A thin plano-convex lens of focal length . is split in to two halves: one of the halves is shifted along the optical axis (see figure). The separation between object and image planes is 1.8 m. The magnification of the image formed by one of the half-lenses is 2. Find the focal-length of the lens and separation between the two halves. Draw the ray diagram for image formation. [IIT 1996] 21. A plano convex lens has a thickness of 4 cm. When placed on a horizontal table with the curved surface in contact with it, the apparent depth of the bottom most point of the lens is found to be 3 cm. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of the plane face is found to be 25/8 cm. Find the focal length of the lens. [IIT 1984] 22. The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface. (a) Where should a pin be placed on the optic axis such that its image is formed at the same place? (b) If the concave part is filled with water of refractive index 34, find the distance through which the pin should be moved so that the image of the pin again coincide with the pin. [IIT 1981] 23. Find the focal length of the lens shown in the figure. The radii of curvature of both the surfaces are equal to R. [IIT 2003] 24. The refractive indices of the crown glass for blue and red lights are 1.51 and 1.49 respectively and those of the flint glass are 1.77 and 1.73 respectively. An isosceles prism of angle 6° is made of crown glass. A beam of white light is incident at a small angle of this prism. The other flint glass isosceles prism is combined with the crown glass prism such that there is no deviation of the incident light. Determine the angle of the flint glass prism. Calculate the net dispersion of the combined system. [IIT 2001]
25. A prism of refracting angle 30° is coated with a thin film of transparent material of refractive index 2.2 on face AC of the prism. A light of wavelength 5500A.is incident on face AB such that angle of incidence is 60°, find (a) the angle of emergence, [Given refractive index of the material of the prism is 3 ].
OPTICS www.physicsashok.in 42 (b) the minimum value of thickness of the coated film on the face AC for which the light emerging from the face has maximum intensity. [IIT 2003] 26. A right angle prism (45° � 90° � 45°) of refractive index n has a plate of refractive index n1 (n1 < n) cemented to its diagonal face. The assembly is in air. A ray is incident on AB (see figure), (a) Calculate the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle. (b) Assuming n = 1.352, calculate the angle of incidence at AB for which the refracted ray passes through the diagonal face undeviated. [IIT 1996] 27. A right angled prism is to be made by selecting a proper material and the angles A and B (B . A), as shown in figure. It is desired that a ray of light incident on the face AB emerges parallel to the incident direction after two internal reflections. (a) What should be the minimum refractive index n for this to be possible? (b) For n = 35 is it possible to achieve this with the angle B equal to 30 degrees ? [IIT 1987] 28. Monochromatic light is incident on a plane interface AB between two media of refractive indices n1 and n2 (n2>n1) at an angle of incidence . as shown in the figure. The angle . is infinitesimally greater than the critical angle for the two media so that total internal reflection takes place. Now if a transparent slab DEFG of uniform thickness and of refractive index n3 is introduced on the interface (as shown in the figure), show that for any value of n3 all light will ultimately be reflected back again into medium II. Consider separately the cases. (i) n3 < n1 and (ii) n3 > n1. [IIT 1986] 29. A parallel beam of light travelling in water (refractive index = 34 ) is refracted by a spherical air bubble of radius 2 mm situated in water. Assuming the light rays to be paraxial, (a) find the position of the image due to refraction at the first surface and the position of the final image. (b) draw a ray diagram showing the positions of both the images. [IIT 1988] 30. Light is incident at an angle . on one planar end of a transparent cylindrical rod of refractive index n. Determine the least value of n so that the light entering the rod does not emerge from the curved surface of the rod irrespective of the value of .. [IIT 1992] 31. The radius of curvature of the convex face of a plano convex lens is 12 cm and its refractive index is 1.5. (i) Find the focal length of this lens. (ii) The plane surface of the lens is now silvered. At what distance from the lens will parallel rays incident on the convex face converge. (iii) Sketch the ray diagram to locate the image, when a point object is placed on the axis, 20 cm from the lens (polished). (iv) Calculate the image distance when the object is placed as in (iiii). [IIT 1979] 32. A ray of light is incident at an angle of 60° on one face of prism which has a
n angle of 30° with the incident ray. Show that the emergent ray is perpendicular to the face through which it emerges and calculate the refractive index of the material of the prism. [IIT 1978]
OPTICS www.physicsashok.in 43 33. A pin is placed 10 cm in front of a convex lens of focal length 20 cm., made of a material of refractive index 1.5. The surface of the lens farther away from the pin is silvered and has a radius of curvature are 22 cm. Determine the position of the final image. Is the image real as virtual? [IIT 1978] 34. The refractive index of the material of a prism of refracting angle 45° is 1.6 for a certain monochromatic ray. What should be minimum angle of incidence of this ray on the prism so that no total internal reflection takes place as the ray comes out of the prism. [ I I T 1976] 35. A prism of refractive index n1 and another prism of refractive index n2 are stuck together without a gap as shown in the figure. The angles of the prisms are as shown n1 and n2 depend on . , the wavelength of light, according to 2 4 1 n 1.20 10.8 10 .. . . and 2 4 2 n 1 45 1 80 10 .. . . . . where, . is in nm. (a) Calculate the wavelength . 0 for which rays incident at any angle on the interface BC pass through without bending at that interface. (b) For light of wavelength . 0, find the angle of incidence i on the face AC such that the deviation produced by the combination of prisms is minimum. [IIT 1998] 36. A projector lens has a focal length 10 cm. It throws an image of a 2 cm x 1 cm slide on a screen 5 metre from the lens. Find : (a) the size of the picture on the screen and (b) ratio of illuminations of the slide and of the picture on the screen. [IIT 1975] 37. A ray of light incident normally on one of the faces of a right angled isosceles prism is found to be totally reflected as shown in the figure. What is the minimum value of the refractive index of the material of the prism? When the prism is immersed in water, trace the path of the emergent rays for the same incident ray, indicating the values of all the angles. ... .... . . 34 . [IIT 1973]
OPTICS OPTICS Answer Key
Assertion & Reasion
Que. 1 2 3 4 5 6 7 8 9 10 Ans. A A A A B E C A C A Que. 11 12 13 14 15 16 Ans. B D D B B D Assertion & Reasion Level # 1
Objective Type Que. 1 2 3 4 5 6 7 8 9 10 Ans. B C A A C C B D D C Que. 11 12 13 14 15 16 17 18 19 20 Ans. D A B C A D B CD C C Que. 21 22 23 24 25 26 27 28 29 30 Ans. B B B A D C A A A B Que. 31 32 33 34 35 36 37 38 39 40 Ans. C C C A D A A C D C Que. 41 42 43 44 45 46 47 48 49 50 Ans. A D B D C C B B B C Que. 51 52 53 54 55 56 57 58 59 60 Ans. B C A A B A BD D A C Que. 61 62 63 64 Ans. C BC CD D
Fill in the Blanks / True�False / Match Table
1. 2 x 108 m/s, 0.4 x 10�6 m 2. d = +15 cm 3. 4000Å, 5 x 1014 Hz 4. 2 25
5. 60 cm 6. 7. 35 cm 8. 1.3 9
9. Zero 10. 5 x 1014 Hz, 4000Å 11. 0.125 m, 0.5 m 12 . 15° Que. 13 14 15 16 17 18 19 20 Ans. T F T T D A C B
www.physicsashok.in 44
OPTICS www.physicsashok.in 45 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. D B B C D B D D B A A C C A A Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans. A B A B C A D B B A D D B B B Que. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 Ans. C C B D B A B C C C B B A A B Que. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans. C D D B C A B C B A B C A A B Que. 61 62 63 64 65 66 67 68 69 70 71 Ans. D D C A B C C A D C D Passage Type Level # 2. 2. 1.6 3. 90 cm., Yes. 4. 2.88 m 5. (a) 13.3 cm (b) 14.975 cm 7. (a) . . .. . . .. ... ... . . . . . . . 2 20 20A 1 rx r 1 d (b) 21 2 2020 0 A 1 rd .. .. . .. .. . ... ... . . . . . . . . (c) Ray will become parallel to y-axis. 8. �2 Km. 9. R . 12 cm . 10. 7.42 cm 11. sin i sin i t 1 cosi sin i x t 1 cos i 2 2 2 2 2 1 1
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. . 12. .. .. . .. .. . . . .. .. . . i . A . sin. sin A sin2 i cosAsin i 2 1 1 1 13. ... ... 15 8, a 8a 14. . . .. . . .. . . . . . . . . 22 2 1 2 R1 b Rb R 2sin b 15. The first spot is at 12 cm on left side from the optical centre. 16. �20 cm 17. ..
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. . , 0 cos 12cos 1 19. b = 15 cm. 20. At a distance of 15 cm. from the mirror 2 cm. from each other. 21. First Image at a distance of 3.33 cm from flat surface and the second at infinity. 22. 9 cm 23. 2 2 d 2 . . .. . . .. . . . 25. 81°40� 26. . . . . . .. . . . . . .. . . . . . . . 22 22 2 R1 b Rb R 1 1 27. 25 cm from the second lens on the right side magnification m = �2. 28. 0.5 cm 29. 6 cm back side of unpolished lens. 30. 3. . 7
OPTICS www.physicsashok.in 46 Level # 3 1. �7.67 cm from the mirror. 2. 5 2 k�5 j � 4 i�3 . . 3. 25 cm 4. 2 5. 18.84 cm 6. . = sin�1 0.6115 = 38° No ray is refracted into glass. 7. 2 x 1010 cm per second. 8. 720 m x 720 m 9. (a) 14.16 cm below the water surface. (b) No glass water interface. 10. (a) cot i dx Slop dy . (b) 4 2 4x K y ... ... . (c) (4m, 1m) (d) Ray emerges parallel to the positive x-axis. 11. 3m . 4 12. OE = 6.06 m 13. 25 cm. 14. At a perpendicular distance of 20 cms from the lens axis 8 cm is size oriented parallel to lens axis. 15. At the position of the object magnification = �1. 16. 0.09 m/s, 0.3 per second. 17. 1.6 18. Distance of A�B� from pole of mirror a15 cm, magnification = �1.5. Distance of A� above RS is 0.3 cm, Distance of B� below RS is 1.5 cm. 19. 0.9 m from the lens (0.1 m behind the mirror) 20. (a) . = 0.4 m, d = 0.6 m. 21. 75 cm 22. (a) 15 cm (b) 1.15 cm towards the lens 23. 3 1 3 R. . . . 24. 4°, 0.04° 25. (a) 0 (b) 1250Å 26. (a) .. . .. . .. . .. . . . . 1 21 1 n2 n n 2 sin 1 (b) . . 73. 2 sin 1 1.352 27. (a) sin B 1 (b) No. 29. (a) Image due to first surface at a distance of 6 mm before the first surface final image at a distance of 1mm before the first surface. 30. 2 31. (a) 24 cm (b) 12 cm (d) 80 cm. 32. . . 3 .
33. 17 cm infront of lens, Real. 34. sin.1.0.176. . 10.1. 35. (a) .0 . 600 nm (b) i . sin.1 ¾ 36. (a) 100 cm x 50 cm (b) 2401 picture slide . .. 37. . . 2 , Angle of refraction in water r = sin.1 ¾ . �X�X�X�X�
ELECTROSTATICS
DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125
ELECTROSTATICS www.physicsashok.in 1 ELECTROSTATICS ELECTRIC CHARGE Charge is the propertyassociatedwithmatter due towhichis produces and experiences electrical andmagnetic effects. The study of electrical effects of charge at rest is called electrostatics. The strength ofparticle�s electric interactionwithobjects around it depends on its electric charge,whichcan be either positive or negative.An object withequal amounts of two kinds of charge is electricallyneutral,whereas onewith an imbalance is electricallycharged. In the table given below, if a body in the first column is rubbed against a bodyin the second column, the body in first columnwill acquire positive charge,while that in the second columnwillacquire negative charge. TABLE S.No. FirstColumn SecondColumn 1. Glass rod Silk rod 2. Flannes or cat skin Ebonite rod 3. Woollen cloth Amber 4. Woollen cloth Rubber shoes 5. Woollen cloth Plastic objects Electric Charge : Electric charge canbewritten as newhere n is a positive or negative integer and e is a constant ofnature called the elementarycharge (approximately1.60 × 10�19C). Electric charge is conserved, the (algebraic) net charge of any isolated systemcannot be changed. Regarding charge following points areworthnothing : (a) Like charges repel each other and unlike charges attract each other. (b) Charge is a scalar and can be of two types; positive or negative. (c) Charge is quantized, i.e., the charge on anybodywill be some integralmultiple of e, i.e., q =± ne. where n = 1, 2, 3,........... Charge on any body can never be 1 e 3 . . .. .. , 1.5e etc. (d) The electrostatic unit of charge is stat-coulomb and electromagnetic unit is ab-coulomb inCGS system. But in SI systemthe unit of charge is coulomb, 1 coulomb 1 ab coulomb 3 109 stat coulomb. 10 . . . . . NOTE : Recently, it has been discovered that elementary particles such as proton or neutron are composed of quarks having charge . . .. .. ± 1 e 3 and . . .. .. ± 2 e 3 . However, as quarks do not exist in free state, the quantum of charge is still e.
Example-1 : Howmanyelectrons are there in one coulomb of negative charge ? Sol. : The negative charge is due to presence of excess electrons, since they carry negative charge. Because an electron has a charge whosemagnitude is e = 1.6 × 10�19 C, the number of electrons 19 n q 1.0 e 1.6 10. . . . n = 6.25 × 1018
ELECTROSTATICS www.physicsashok.in 2 COULOMB�S LAW The force acting between two point charges is directly proportional to the product of the charges and it is inverselyproportional to the square of distance between them.Mathematically. 1 2 2 F q q r . 1 2 2 F kq q r . 9 2 2 0 k 1 9.0 10 Nm /C 4 . . . .. where .0 = permittivityof free space .0 = 8.854 × 10�12 C2/N�m2 RegardingCoulomb�s lawfollowing points areworthnoting : (a) When two charges exert forces simultaneously on a third charge, the total force acting on that charge is the vector sumof forces that the two chargeswould exert individually.This important property, called the principle of superposition of forces, holds for anynumber of charges. Thus, Fnet . F1 . F2 . F3 . ...... Fn . . . . . (b) If some dielectric (K) is placed in the space between the charges, the net force acting on eachcharge is altered because charges are induced in themolecules of intervenningmedium. q1 q2 Fe Fe r In vacuum Thus, 1 2 e 2 0 F 1 . q q 4 r . .. (invacuum) 1 2 e 2 0 F' 1 . q q 4 k r . .. (Inmedium) or 1 2 e 2 F' 1 . q q 4 r . .. Here 0. . . K is called permittivityof themedium (c) The coulomb�s lawexpresses the force betweentwo point charges at rest. Inapply
ingit to the case of extended bodies of finite size care should be taken in assuming the whole charge of a body to be concentrated at its centre as that is true onlyfor sphercal charged body, that is too for external point. (d) �Acharged paritcle under the action of coulombian forces onlycan never be stable.� This statement is called Earnshow�s theorem. IMPORTANT FEATURE 1. Suppose the position vectors of two charges q1 and q2 are 1 r . and 2 r . then electric force on charges q1 due to charge q2 is 1 2 . . 12 3 1 2 0 1 2 F 1 . q q r r 4 r r . . .. . . . . . . Similarly, electric force on q2 due to charges q1 is 1 2 21 3 2 1 0 2 1 F 1 . q q (r r ) 4 |r r | . . .. . . .. .. ... ..
ELECTROSTATICS www.physicsashok.in 3 Here q1 and q2 are to be subsitiuted with sign. . r1. x1i . y1 j. z1k . . . and . 2 2 2 2 r . x i . y j. z k . . . where 1 1 1 (x , y ,z ) and 2 2 2 (x , y , z ) are the co-ordinates of charges q1 and q2. 2. Suppose two charges q1 and q2 are placed in vacuumat a distance r0 and the electric force between themis 1 2 0 2 0 0 F 1 . q q 4 r . .. Now, the same charges are placed in a dielecricmediumof dielectric constant K at distance r (<r0) such that the electric force between them 1 2 2 0 F 1 . q q 4 k r . .. remains the same or F = F0 Then, 1 2 1 2 2 2 0 0 0 1 . q q 1 q q 4 k r 4 r . .. .. or 0 r . k r Thus, we see that for the electric force between two charges at distance r in a dielectricmediumis equivalent to a distance 0 r . k r in vacuum. Example-2 :What is the smallest electric force between two charges placed at a distance of 1.0 m? Sol. 1 2 e 2 0 F 1 . q q 4 r . .. For Fe to bemimimumq1 q2 should bemimimum. 19 1 2 q . q . e . 1.6 .10. C Substituting in above relationwe have 9 19 19 e 2 (F ) (9.0 10 )(1.6 10 )(1.6 10 ) (1.0). . . . . .
. 2.304 .10.28N Example-3 :Three point charges +q, �q and +q are placed at the vertices P, Q and R of an equilateral triangle as shown in fig. If 22 0 1 q F , 4 r . .. where r is the side of the triangle, +q y P x r r �q r +q the force on charge at P due to charges at Q and R is : Q R (A) F along positive x-direction (B) F along negative x-direction (C) 2 F along positive x-direction (D) 2 F along negative x-direction. Sol. Refer to Fig. The charge at Q exerts an attractive force F on charge at P along PQ. The charge at R exerts a repulsive force on charge at P along PS ofmagnitude F. The angle between these two forces is 120º. From parallelogramlaw, the magnitude of the resultant force is 2 2 2 2 r F = F + F + 2F cos 120º +q y x r �q r +q Q R 120º P Fr S FF = 2F2 � F2 = F2 or Fr = F.As shown in the figure, the direction of the resultant force is along the negative x-direction. Hence the correct choice is (b). Example-4 : Three charges q1 = 1µc, q2 = �2µc and q3 = 3µc are placed on the vertices of an equilateral triangle of side1.0m.Find the net electric force acting on charge q1.
ELECTROSTATICS www.physicsashok.in 4 Sol. Let us assume a co-ordinate axeswith q1 at origan as shown infigure. The co-ordinates of q1, q2 and q3 in this co-ordinate systemare (0,0,0), (1m, 0,0) and (0.5m, 0.87m, 0) respectively. q3 q1 q2 x y Now, 1 F . = force on q1 due to charge q2. 1 2 3 1 2 0 1 2 1 . q q (r r ) 4 |r r | . .. . .. .. .. .. 9 6 6 3 (9.0 10 )(1.0 10 )(3.0 10 ) (1.0) . . . . . . .[(0 . 0.5).i . ?(0 . 0.87).j. (0 . 0)k.] . (1.8 .10.2 i)N and F2 . . force on q1 due to charge q3 1 3 3 1 3 0 1 3 1 . q q (r r ) 4 |r r | . .. . .. .. .. .. 9 6 6 3 (9.0 10 )(1.0 10 )(3.0 10 ) (1.0) . . . . . . .[(0 . 0.5).i . (0 . 0.87).j. (0. 0)k.] . (.1.35.i . 2.349.j) .10.2N Therefore net force on q1 is F . F1 . F2 . . . . (0.45.i . 2.349.j) .10.2N ELECTRIC FIELD Electirc field is the region around an electric charge (or a group of electric charges) inwhich the electric force can be experienced. If an electric charge is placed in such region, it experiences either an attractive or a repulsive force. Electric field at a point can be defined in terms of either a vector functionE.. called electric field strength or a
scalar functionVcalled electric potential. The electric field can also be visualised graphicallyinterms of lines of force. Electric Field Intensity : The intensity of electric field at anypoint is defined as the force acting on a unit positive charge placed at that point. If the electrostatic force experienced by a small test charge q0 is e F . , then intensity of electric field e0 E Fq . .. . The electric field is a vector quantityand its directionis the same as the direction of the force e F . on a positive test charge. The SI unit of electric field isN/C. Example-5 : An electric field of 105N/C points duewest at a certain spot.What are themagnitude and direction of the force that acts on a charge of + 2µC and �5µC at this spot ?
ELECTROSTATICS www.physicsashok.in 5 Sol. Force on + 2µC = qE = (2 × 10�6) (105) = 0.2 N (due west) Force on � 5µC = (5 × 10�6) (105) = 0.5 N (due east) Example-6 : An inclined planemaking an angle 30º with the horizontal is placed in a uniformhorizontal electric field . of 100 V/m(see figure). Particle of mass 1 kg and charge 0.01 C is allowed to slide down 1m e= 100 V/m q 30º fromrest froma height of 1 m. If the coefficient of friction is 0.2, find the time it will take the particle to reach the bottom. Sol. The different forces on the particle are shown in figure. Fromfigure, N = mg cos 30º + q. cos 60º Friction f = µN = µ mg cos 30º + µ q. cos 60º Now the total force F acting along the inclined plane is F = mg sin 30º � µ N � q..cos 30º q 30º mg sin 30º q cos 30º e qe µN q sin 30º emg cos 30º N 30 º 30º or F = mg sin 30º � mg cos 30º � µq. cos 60º � q..cos 30º Thus acceleration is or a F g sin 30º µg cos30º µq cos 60º q cos30º m m m . . . . . . . or a 9.8 0.5 0.2 9.8 . 3 / 2. 0.2 0.01 100 0.5 0.01 100 3 1 1 2 . . . . . . . . . . . . or = 4.9 � 0.98 × 1.732 � 0.10 � 0.551.732 or = 4.9 � 1.697 � 0.10 � 0.866 = 2.237 Now, distance travelled in time t is s 0 1 at2 2 = + or t 2 2 a. . . . . . . . [As s = 1/sin30 = 2] or 4 1.345 sec. 2.237 . . . . .. .. Example-7 : In space horizontalElectric field (E= (mg)/q) exist as shown in figure and a massm attached at the end of a light rod. Ifmassmis released fromthe positionshown infigure find the angular velocityof the rodwhenit passes through
m m +qq q=45º mg E q = the bottommost position (A) gl (B) 2g l (C) 3g l (D) 5g l Sol. According towork energytheorm: w = .T WE + Wg = 12 mv2 � 0 ...(1) WE = qE l sin ., Wg = mg (l � l cos .) m m +qq q=45º l mg E q = +q l sinq qE l cosq l � l cosq qE l sin . + mg (l � l cos.) = 12 mv2 fromeqn (1) mg l sin. +mg l �mg l cos. = 12 mv2 { E mg q . = }
ELECTROSTATICS www.physicsashok.in 6 g g g 1 mv2 2 2 2 l + l - l = . v = 2gl v 2g 2g w = = = l l l l . w = 2g l Ans. TABLE : Electric field Intensity of Various System S.No. System Electric Field Intensity 1. Isolated charge 2 0 E 1 . q 4 r . .. q r p Isolated charge E 2. Aring of charge . .2 2 3/ 2 0 E 1 . qx 4 R x . .. . 3. Adisc of charge 2 2 0 E 1 x 2 x R . . . . . . . . . . . 4. Infinite sheet of charge 0 E 2. . . 5. Infinitelylongline ofcharge 0 E 2 r . . .. r
ELECTROSTATICS www.physicsashok.in 7 6. Finite line of charge . . 0 E sin sin 4 x . . . . . . .. . . | | 0 E cos cos 4 x . . . . . .. 7. Charged spherical shell (a) Inside 0 . r . R, E = 0 (b) Outside r . R 2 0 E q 4 r . .. 8. Solid sphere of charge (a) Inside 0 . r . R 0 E r 3. . . (b) Outside r . R 2 0 E q 4 r . .. where q = charge on sphere Electric Lines of Force : Faraday gave a newapproach for representationof electric field in the formof electric lines of force. Electric lines offorce are graphical representationof electric field. �Anelectric line of force is animaginarylineor curve drawn through a region of space so that its tangent at any point is in the direction of the electric field vector at that point.� EP EQ Q P Thismodelofelectric field has the following characteristics : (i) Electric lines of force are originated frompositive charge and terminated into negative charge. (ii) The number of electric lines of force originates froma point charge q is q/.0. Electric lines of forcemay be fraction. (iii) The number oflines per unit area that pass througha surface perpendicular to the electric field lines isproportional
to the strength of field inthat region.
ELECTROSTATICS ELECTROSTATICS (iv) No electric lines of force cross each other. If two electric lines of force cross each other, it means electric field has two directions at the point of cross. This is not physically possible. E1 E2 Lineofforce Lineofforce (v) Electric lines of force for two equal positive point charges are said to have rotational symmetryabout the axis joining the charges. (vi) Electric lines offorce for point positive charge and a nearby negative point charge that are equal in magnitude are said to have rotational symmetry about an axis passing through both charges in the plane of the page. (vii) Electric lines of force due to infinitely large sheet ofpositive charge is normalto the sheet. (viii) No electrostatic lines of force are present inside a conductor.Also electric lines of force are perpendicular to the surface of conductor. For example if a conducting sphere is placed in a region where uniform electric field is present, then induced charges are developed on the sphere. (ix) Ifachargedparticleisreleasedfromrestinregionwhereonlyuniformelectricfieldispresent,thencharged particle move along an electric line offorce. But if charged particle has initialvelocity, thenthe charged particle mayormaynor followtheelectriclinesofforce.
(x) Electriclinesofforceinsidetheparallelplatecapacitorisuniform.Itshows that field inside the parallel plate capacitor is uniform. But at the edge of plates, electric lines of force are curved. It shows electric lines of force at the edge of plates is non-uniform. This is knownas fringing effect. If the size of plates are infinitelylarge, then fringing effect can be neglected. (xi) If a metallic plate is introduced between plates of a charged capacitor, then electric lines of force can be discontinuous. www.physicsashok.in 8
ELECTROSTATICS www.physicsashok.in 9 (xii) If a dielectric plate is introduced between plates of a charged capacitor, then number of lines of forces in dielectric is lesser than that in case of vacuumspace. (xiii) Electrostatics electric lines of force can never be closed loops, as a line can never start and end on the same charge.Also if a line offorce is a closed curve,work done round a closed pathwillnot be zero and electric field willnot remain conservative. (xiv) Lines of force have tendency to contract longitudinally like a stretched elastic string producing attraction between opposite charges and repel each other laterally resulting in, repulsion between similar charges and edge-effect (curving of lines of force near the edges ofa charged conductor). ELECTRIC POTENTIAL AND ELECTRIC POTENTIAL DIFFERENCE : Electric Potential : �Electric potential at anypoint in a electric field is equalto the ratio of thework done in bringing a test charge frominfinityto that point, to the value of test charge.� Suppose,Wbe thework required in bringing a test charge q0 frominfinity to a point b against the repulsive force F acting on it, then potential at the point b is b b 0 W V q.. . Since,Wand q0 both are scalar quantities, the potential is also a scalar quantity. Electric Potential Difference : The potentialdifference between two points in an electric field is equal to the ratio ofwork done inmoving a test charge fromone point to the other, to the value oftest charge. SupposeWwork be done inbringing a small test charge q0 fromthe point a to a point b against the repulsive force acting on it, then potential difference between the points is a b b a 0 V V Wq. . . Obviously, potentialdifference is also a scalar quantity. IMPORTANT FEATURES 1. Following three formulae are veryusefulin the problems related to work done inelectric field. (Wa � b)electric force = q0 (Va � Vb) (Wa � b)external force = q0 (Vb � Va) = �(Wa � b)electric force (W. � a)external force = q0Va 2. Electric potential due to a point charge q : Fromthe definition of potential, 0 0 0 0 1 . qq
U 4 r V q q .. . .
ELECTROSTATICS www.physicsashok.in 10 or 0 V 1 . q 4 r . .. Here, r is the distance fromthe point charge q to be point at which the potential is evaluated. If q is positive, the potential that it produces is positive at allpoints; if q is negative, it produces a potential that is negative everywhere. In either case, Vis equal to zero at r = .. 3. Electric potential due to a systemof charges : Just as the electric field due to a collection of point charges is the vector sumof the fields produced by each charge, the electric potential due to a collection of point charges is the scalar sumof the potentials due to each charge. i 0 i i V 1 q 4 r . .. . 4. In the equation i 0 i i V 1 q 4 r . .. . , if the whole charge is at equaldistance r0 fromthe point where Vis to be evaluated, thenwe canwrite, net 0 0 V 1 . q 4 r . .. where qnet is the algebraic sumof all the charges ofwhich the systemismade. Example-8 : The electric potential at pointAis 20Vand at Bis �40V. Find thework done by a external force and electrostatic force inmoving an electron slowly fromBtoA. Sol. Here, the test charge is an electron, i.e., q0 = �1.6 × 10�19 C VA = 20V and VB = �40 V Work done by external force (WB � A)external force = q0(VA � VB) = (�1.6 ×10�19)[(20) � (�40)] = �9.6 × 10�18 J Work done by electric force (WB � A)electric force = �(WB � A)external force = � (�9.6 × 10�18 J) = 9.6 × 10�18 J Example-9 : Find thework done bysome external force inmoving a charge q = 2µCfrominfinityto a point where electric potential is 104V. Sol. using the relation, (W. � a)external force = q Va We have (W. � a)external force = (2 × 10�6) (104) = 2 × 10�2 J
Example-10 : Three point charges q1 = 1µC, q2 = �2µC and q3 = 3µC are placed at (1m, 0, 0), (0, 2m, 0) and (0, 0, 3m) respectively. Find the electric potential at origin. Sol. The net electric potential at origin is, 1 2 3 0 1 2 3 1 q q q V 4 r r r . . . . . . . .. . . Substituting thevalues,we have
ELECTROSTATICS www.physicsashok.in 11 V .9.0 109 . 1 2 3 10 6 1.0 2.0 3.0 . . .. . . .. . . . . V = 9.0 × 103 volt TABLE : Electric Potential of Various Systems S.No. FirstColumn SecondColumn 1. Isolated charge 0 V q 4 r . .. q p r 2. Aring of charge 2 2 0 V q q 4 R x . .. . 3. Adisc of charge 2 2 0 V R x x 2. . . . . . . . . 4. Infinite sheet of charge Not defined 5. Infinitelylongline ofcharge Not defined r 6. Finite line of charge 0 V ln sec tan 4 sec tan . . . . . .. . . . p x ++++++++
ELECTROSTATICS www.physicsashok.in 12 7. Charged spherical shell (a) Inside 0 . r . R 0 V q 4 R . .. (b) Outside r > R 0 V q 4 r . .. 8. Solid sphere of charge (a) Inside 0 . r . R 2 22 0 V R 3 r 6 R . . . . . . . . . . (b) Outside r . R 0 V q 4 r . .. Example-11 : In a regular polygon of n sides, each corner is at a distance r fromthe centre. Identicalcharges are placed at (n � 1) corners.At the centre, the intensityis E and the potential isV. The ratioV/E hasmagnitude. (A) r n (B) r(n � 1) (C)(n � 1)/r (D) r(n � 1)/n Sol. 2 0 E q 4. r . . and 0 v (n 1)q 4. r . . . . 0 2 0 (n 1)q v 4 r q (n 1)r E 4 r . . .. . . . .
Example-12 : The figure shows a nonconducting ring which has positive and negative charge non uniformly distributed on it such that the total charge is zero.Which of the following statements is true ? (A) The potential at all the points on the axis will be zero. axis O + + + ++++++ + ������ � ��� �� (B) The electric field at all the points on the axiswill be zero. (C) The direction of electric field at all points on the axis will be along the axis. (D) If the ring is placed inside a uniformexternal electric field then net torque and force acting on the ring would be zero. Sol. 2 2 0 V q 0 4 R x . . . .. . ( . .q = 0) There for the potential at all the points on the axis will be zero. O ++ + ++++++ + ������ � ��� �� xR2 + x2 Hence (A) is correct. Example-13 : Two concentric rings, one of radius �a�and the other of radius �b�have the charges +q and �(2/5)�3/2 q respectively as shown in the figure. Find the ratio b a q =+q A q = �(2/5) q B �3/2 z=a b/a if a charge particle placed on the axis at z = a is in equilibrium.
ELECTROSTATICS www.physicsashok.in 13 Sol. At equilibrium F = 0 . QE = 0 . E = 0 A B 3 3 2 2 2 2 2 2 0 0 q z q z 0 4 (z a ) 4 (z b ) . . . . . . . . . A B 3 3 2 2 2 2 2 2 0 0 q z q z 4 (z a ) 4 (z b ) . . . . . . . . 32 3 3 2 2 2 2 2 2 0 02 qa qa 5 4 (a a ) 4 (a b ) . . . .. .. . . . . . . . . . . . .3 3 2 2 2 2 3 2 2 2 a b 2a a 5 . . . . . . . . . . . . . . 3 3 2 2 2 2 3 2 2 a b 52a 2 . . . . . . . . . 2 2 2 a b 5 2a 2 . .
b a q =+q A q = �(2/5) q B �3/2 z=a 2a2 + 2b2 = 10a2 . 2b2 = 8a2 22 b 4 a . . b 2 a . Ans. Example-14 : Acircular ring of radiusRwith uniformpositive charge density. per unit lengthis located in the yz planewith its centre at the originO.Aparticle ofmassmand positive charge q is projected fromthe point P(R 3 , O,O) on the positive x-axis directly towardsO,with an initial kinetic energy 0 q 4.. . (A) The particle crossesOand goes to infinity. (B) The particle returns to P (C) The particlewill just reachO. (D) The particle crosses Oand goes to �R 3 . Sol. According tomachenical energyconser Ui + Ti = Uf + Tf ...(i) 1 i 0 U q q 4. (x) . . Where q1 is the charge on ringh q + ++++++ +++++O R l R 3 R2 + 3R2 = x . q1 = (2.R). i 0 U q(2 R ) 4 (2R) . .
.
.
. i 0 U q 4. . . i 0 T q 4. . . 1 f 0 U q q 4. R . . f 0 0 U q(2 R ) q 4 R 2 . . . . . . . . 2 f T 1 mv 2 = where v is a velocityofcharge putting the volues in eqn (1)
ELECTROSTATICS www.physicsashok.in 14 2 0 0 0 q q q 1 mv 4 4 2 2 . . . . . . . . . 2 0 0 q q 1 mv 2 2 2 . . . . . . . 1 mv2 0 2 . . v . 0 The particlewill just reach centre of ring. Example-15 : Find the electric field at centre of semicircular ring shown in figure. ����� � � + +++++ Y R X �q q Sol. . 2 2 0 0 0 E . E . E . 2 E But 0 0 E sin 2 R 4 . . . . . But 0 K 4 l . . . , 2 2q R . . . ����� � � + ++ +++ 45º 45º 45º �q +q E0 E0 O X E Y 45º
. 2 E 4Kq .R . directed negative x-axis. Example-16 : Asimple pendulumof length l and bobmassmis hanging in front ofa large nonconducting sheet having surface charge density.. Ifsuddenlya charge +q is given to the bob&it is released fromthe position shown in figure. Find the maximum + + ++ + ++ + + l s angle throughwhich the string is deflected fromvertical. Sol. The electric field intensitydue to nonconducting sheet is 0 E 2. . . ...(1) According towork energy theorem Wnet = .k = 0 Since initial and finalvelocity ofbob is zero Therefore, WE + Wg = 0 ...(2) WE = qE . sin . Wg = �mg ( . � . cos.) Putting the values in eqn (2) + + ++ + ++ + + cosq s mgv=0 qE qsinq qE . sin. �mg. (1�cos.) = 0 qE . sin. =mg. (1�cos.) qE 1 cos mg sin . . . . . . . 2 sin2 qE 2
mg 2sin cos 2 2 . . . . qE tan mg 2 . . . tan 1 qE 2 mg . . . . . . . . . 2 tan 1 qE mg . . . . . . . . . Putting the value ofE fromeqn (1) 1 0 2 tan q 2 mg . . . . . . . . . . . . 1 02 tan q 2 mg . . . . . . . . . . . Ans.
ELECTROSTATICS www.physicsashok.in 15 Example-17 : A particle ofmassmand negative charge q is thrown in a gravityfree spacewith speed u fromthe pointAon the large non conducting charged sheet with surface charge density ., as shown in figure. Find themaximumdistance fromAon sheet where the particle can strike. + + ++ + ++ + u A Sol. Electric field intensitydue to nonconducting sheet 0 E 2. . . ...(1) where ux = u sin. uy = u cos. x a qE m = + + ++ + ++ + u A ucos usin X Y 2 x x x u T 1 a T 2 = + where x = 0 2 x x 0 u T 1 a T 2 = + . 0 .u sin .T 1 qE T2 2 m . . . .u sin .T 1 qE T2 2 m . . . T 2umsin qE
.
. y = uyT y= (u cos.)T y u cos 2u msin qE . . . . . y u2msin 2 qE . . y 2 0 2u m sin 2 q . . . . fromeqn (1) . 2 0 y 2u msin 2 q . . . . 2 0 max y 2u m q.. . At ymax, sin2. = 1 Example-18 : The figure shows three infinite non-conducting plates of charge perpendicular to the plane of the paper with charge per unit area +., +2. and �., Find the ratio of the net electric field at that pointAto that at point B. + + � + + � + + � + + � + + � + + � + + � A B 2.5m 2.5m 5 m 5 m +s +2s �s Sol. The electric field intensity at point Adue to plate x, y and z EA = Ex + Ey + Ez 0 0 0 2 2 2 2 . . . . . . . .
EA = 0 ...(1) A B +s +2s �s x y z 2s 2Î 0 s 2Î 0 s Î 0 At point B B 0 0 0 E 2 2 2 2 . . . . . . . . . B 0 E 4 2 . . . ...(2) Fromeqn (1) and (2) AB E 0 E =
ELECTROSTATICS www.physicsashok.in 16 Example-18 : A metallic solid sphere is placed in a uniformelectric field. The lines of force follow the path(s) shown in figure as : (A) 1 (B) 2 1234 432 1 (C) 3 (D) 4 Sol. Electric field lines never enter a metallic conductor (E = 0, inside a conductor) and they fall normally on the surface of a metallic conductor (because whole surface is at same potential and lines are perpendicular to equipotential surface). Example-19 : Anon-conducting solid sphere of radius Ris uniformly charged. Themagnitude of electric field due to the sphere at a distance r from its centre : (A) increases as r increases for r < R (B) decreases as r increases for 0 < r < . (C) decreases as r increases for R < r < . (D) is discontinuous at r = R (E) both (a) and (c) are correct Sol. 0 E r 3.. . r < R E .r r < R 2 0 E Q 4.. r . r > R 2 E 1 r . Hence, both (a) and (c) are correct. Example-20 :Aparticle ofmassmand charge �qmoves along a diameter of a uniformlycharged sphere of radius Rand carrying a total charge+Q. Find the frequencyofS.H.M. of the particle if the amplitude does not exceed R. Sol. Electric field intensitydue to nonconducting sphere 0 E x 3. . . ...(1) (where x < R) Where . is volume charge density. The force onnegative charge is opposite direction of electric field. �qE = ma Fromeqn (1) 0
q x ma 3. . . . . 0 a q x 3 m . . . . ...(2) xR �q For S.H.M. a = �.2x ...(3) Fromeqn (2) and (3) 2 0 q 3 m . . . . . 0 q 3 m . . . . 0 2 f q 3m. . . . {. . . 2.f } 0 1 q f 2 3m. . . . . 3 0 1 qQ f 2 4 mR . . .. 3 Q 4 / 3 R . . . . . . . . . . Example-21 : Apositive chargeQis uniformlydistributed throughout the volume of a dielectric sphere of radius R.Apointmass having charge +q andmassmis fired towards the centre of the spherewith velocityv froma point at distance r (r >R) fromthe centre of the sphere. Find theminimumvelocityvso that it canpenetrateR/
ELECTROSTATICS www.physicsashok.in 17 2 distance of the sphere. Neglect any resistance other than electric interaction. Charge on the smallmass remains constant throughout themotion. Sol. Frommachenicalenergyconservation U i + T i = Uf + Tf qV i + ½mv2 = qVf + 0 ...(1) i 0 V Q 4. r . . , 2 2 f 3 0 1 Q(3R r ) V 2 4. R . . . where r R2 = R O vmin 2R r +q 2 2 f 3 0 Q 3R R1 4 V 2 4. R . . . . . . . . . . . 2 2 . f 3 0 Q 12R R V 32. R . . . f 0 V Q(11) 32. R
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. . f 0 V 11Q 32. R . . Putting the values ofVi andVf ineqn. (1) 2 0 0 qQ 1 mv 11qQ 4. r 2 32. R . . . . . 2 0 0 1 mv 11qQ qQ 2 32. R 4. r . . . . 2 0 0 mv 11qQ qQ 16. R 2. r . . . . . 2 0 mv qQ 11 1 2. 8R r . . . . . .. .. 2 0 v qQ 11 R 2m. R 8 r . . . . . .. .. . v2 2kqQ 1 3 R mR 8 r . . . . . .. .. v2 2kqQ r R 3 mR r 8 . . . . . . . . . . 12kqQ r R 3 2 v mR r 8 . . . .. . . . . .. . . .. Example-22 : The diagramshows a small bead ofmassmcarrying charge q. The beamcan freelymove on the smooth fixed ring placed on a smooth horizontal plane. In the same plane a charge +Qhas alos been fixed as shown. The potential at the point P due C4a a+Q B xg P to +Q isV. The velocitywithwhich the bead should projected fromthe point P so that it can complete a circle should be greater than
(A) 6qV m (B) qV m (C) 3qV m (D) none Sol. According to conservationprincipalofmechanical energy. Ui + Ti = Uf + Tf 20 0 qV 1 mv qQ 0 2 4. a . . . . ...(1) where v0 is velocityof at point �p� the potential at the point p due to +Qis C4a a+Q B xg P v0 0v Q 4. (4a) . . fromeqn (1)
ELECTROSTATICS www.physicsashok.in 18 20 qv 1 mv 4qV 2 . . 201 mv 3qV 2 . 20v 6qV m . . 0 v 6qV m = Ans. Example-23 : Two spherical, nonconducting, and verythin shells of uniformlydistributed positive chargeQand radius d are located a distance 10d fromeach other.A positive point charge q is placed inside one of the shells at a distance d/2 from d/2 10 d d Q Q the center, on the line connecting the centers of the two shells, as shown in the figure. What is the net force on the charge q ? (A) 2 0 qQ 361.. d to the left (B) 2 0 qQ 361.. d to the right (C) 2 0 362qQ 361.. d to the left (D) 2 0 360qQ 361.. d to the right Sol. Electric force on charge q due to sphreAis zero. But electric force due to sphere B on charge q is 2 0 F qQ towards left 4 19 d 2 .. . . . .. .. Q Q B q F 19 d 2
A d/2 2 0 qQ towards left 361.. d . Example-24 : The diagramshows three infinitely long uniformline charges placed on the X, Yand Z axis. The work done in moving a unit positive charge from(1, 1, 1) to (0, 1, 1) is equal to Y X Z l 3l 2l (A) (. ln 2) /2..0 (B) (. ln 2)/..0 (C) (3. ln 2) /2..0 (D) None Sol. Here r = 1+ x2 E1 = The magintude of electric field due to wire along y-axis = 2 03 2 1. x . .. directed paerpendicular to y-axis E2 = The magnitude of electric field due to wire 1+ x2 E along z-axis = 2 1 x l+ directed perpendicular to z-axis. The electric field due to wire along x-axis is directed perpendicular to x-axis. . The net component of electric field along x-axis is x 2 2 0 0 E 3 cos cos 2 1 x 2 1 x . . . . . . . . .. .. where . = angle made by E1 with x-axis and . = angle made by E2 with x-axis But 2 cos cos x 1 x . . . . . . x . 2 . 0E 4 x 2 1 x . . . ..
ELECTROSTATICS www.physicsashok.in 19 . �dV = Ex dx or . . x 0 2 x 1 0 dV 4 x dx 2 1 x .. . . . . . . .. 21 v x 0 2 v 0 x 1 dV 4 x dx 2 1 x .. . . . . . . .. x 0 1 2 2 0 x 1 V V 4 x dx 2 1 x .. . . . . . .. Put 1 + x2 = z or 2x dz dx = or dz = 2 xdx x 0 1 2 0 x 1 V V 2 x dz z 2x .. . . . . . .. x 0 0 x 1 dz z .. . . . ..
x 0 x 1 0 lnz .. . .. .. . .. 01 0 . 2 ln(1.x) .. .. . .. 01 0 . ..ln1.ln2.. . ..2 1 0 v . v . ln 2 . .. Example-25 : A particle ofmass 1 kg & charge 1 µC 3 is projected towards a non conducint fixed spherical shell having the same charge uniformly V from 1 mm distributed on its surface. Find the minimuminitial velocity of projection required if the particle just grazes the shall. (A) 2 m/ s 3 (B) 2 2 m/ s 3 (C) 2 m/ s 3 (D) none of these Sol. Apply conservation principle of angular momentum, m vd = mv0 r . 0 v vd v r 2 . . d r 0.5 mm 2 . . . . ... .. Applying mechanical energy conservation principle. Ui + Ti = Uf + Tf 2 2 20 0 1 q 1 0 mv mv 2 4 r 2 . . . ..
or 2 2 20 0 1 q 1 mv mv 2 4 r 2 . . ..
ELECTROSTATICS www.physicsashok.in 20 or 2 2 2 0 1 mv q 1 m v 2 4 r 2 2 . . . . .. .. .. or 2 2 2 0 1 mv q mv 2 8 4 r . . .. or 2 2 0 1 3 q mv 2 4 4 r . . .. or 2 2 0 1 q mv 2 3 r . .. or 2 2 0 8q v 3 4 mr . . .. 9 12 3 8 9 10 1 10 9 8 3 1 1 10 3 . . . . . . . . . . . v 8 2 2 m/ s 3 3 = = ELECTRIC POTENTIAL ENERGY Two like charges repel each other while the two unlike charges attract each othe
r. So, when the charges are moved awayfromeach other or they are brought near each other, either somework is obtained or somework is done.Thiswork is stored in the systemofcharges in formof electric potential energy. �The electric potential energyof a systemof different point charges is equal to thework done inbringing those charges frominfinityto formthe system.� It is represented byU. The electric potentialenergyof a systemof two point charges q1 and q2 invacuumat a separation r is given by, 1 2 0 U 1 q q J 4 r . .. Electric Potential Energy of a System of Charges : The electric potential energy of a systemof charges is given by i j i j 0 ij 1 q q U J 4 . r . . .. This sumextends over allpairs of charges.We do not let i=j, because that would be an interaction ofa charge with itself andwe include onlytermswith i < j to make sure that we count each pair only once. For example, electric potential energyof four point charges q1, q2, q3 and q4 would be given by, 4 3 4 2 4 1 3 2 3 1 2 1 0 43 42 41 32 31 21 1 q q q q q q q q q q q q U 4 r r r r r r . . . . . . . . . . .. . . Here, all the charges are to be substitutedwith sign. NOTE : Total number of pairs formed by n point charges are n.n - 1. . 2 Example-26 : Four charges q1 = 1µC, q2 = 2µC, q3 = �3µC and q4 = 4µC are kept on the vertices of a square of side 1m. Find the electric potential energy of this systemof charges. q4 q3 q1 q2 1 m 1 m Sol. In this problem,
ELECTROSTATICS www.physicsashok.in 21 r41 = r43 = r32 = r21 = 1m and 2 2 r42 . r31 . (1) . (1) . 2 m Substituting the proper valueswith signin the relation givenabove in the theory,we get U = (9.0 × 109) (10�6) (10�6) (4)( 3) (4)(1) (4)(1) ( 3)(2) ( 3)(1) (2)(1) 1 2 1 1 2 1 . . . . . . . . . . . . . . . .9.0 10 3 . 12 52 . . . .. . . .. .. = �7.62 × 10�2 J NOTE : Here negative sign of U implies that positive work has been done by electrostatic force in assembling these charges at respective distances from infinity. Example-27 : Consider the configurationof a systemof four charges each of value +q. Find thework done byexternal agent in changing the +q +q +q +q a a fig (i) fig (ii) +q +q +q +q a configuration of the systemfromfigure (i) to fig (ii). Sol. Ui =Electricalpotential energyinsquare arrangement 2 2 i 0 0 U 4q 2q 4 a 4 2a . . . . . . Uf = Electric potential energyof circular arrangement of charge q a2 a a q q q 2 2 f 0 0 U 4q 2q 4 2a 4 (2a) . . . . . . ext w . .U . . 2 f i 0 q U U 3 2 4 a
. . . . .
. . Example-28 : Aconemade of insulatingmaterialhas a total chargeQspread uniformly over its sloping surface.Calculate the energyrequired to take a test charge q frominfinity to apexAof cone. The slant length is L. B A AB = L Sol. ext f i w . .U . U . U A q(V V ) . . . wext A A . q(V . 0) . qV ...(1) HereVA = Electric potential at pointA. The electric charge on the considered ring dq Q2 rdx RL . . . . . Electric potentialdue to consideredring is 0 dV dq 4 x . . . . 0 dV Q2 rdx RL4 x . . . . . . sin r R x L . . . Ar dx q x
ELECTROSTATICS www.physicsashok.in 22 . r Rx L . . 0 2 Q R x dx dV L RL4 x . . .. .. . . . . . . L A 2 0 0 V dv Q dx 2 L . . . . . . 0 Q L 2 = p Î . ext A 0 w q V qQ 2 L . . . . fromeqn (1) RELATION BETWEEN ELECTRIC FIELD AND POTENTIAL In case of cartesian co-ordinates x y z E . E i� . E �j. E k� .. Here, x E Vx . . . . . . (partialderivative ofVw.r.t. x) y E Vy . . . . . . (partialderivative ofVw.r.t. y) z E Vz . . . . . . (partial derivative ofVw.r.t. z) . E V �i V �j V k�
x y z . . . . . . . . . . . . . . . . .. This is alsowritten as, E . .grad V . ..V .. .. Example-29 : The electric potential in a region is represented as, V= 2x + 3y� z, Obtain expression for electric field strength. Sol. E V �i V �j V k� x y z . . . . . . . . . . . . . . . . .. Here, . . V 2x 3y z 2 x x . . . . . . . . V .2x 3y z. 3 y y . . . . . . . . V .2x 3y z. 1 z z . . . . . . . . . . E . .2 i� . 3�j. k� .. In polar co-ordinates r E Vr . . . . and E 1 . V r . . . . .. For example, electric potential due to a point charge q at distance r is 0 V 1 . q 4 r . .. . E. = 0 V 0 . . . . . . . .. . . and r 2 0 E 1 . q 4 r . ..
ELECTROSTATICS www.physicsashok.in 23 Thus, electric field E dV dr . . or E = � (slope of v�r graph) NOTE : � If electric field .. E is known, then electric potential can be determined from the relation as given below: . d .. . dV = -E r or . . d b .. . b a a V -V = - E r Here, dr . dx �i . dy �j. dz k� . � In uniform electric field V = Ed Example-30 : FindVab in an electric field E .2 i� 3�j 4 k� . NC . . . .. where . . a r . i� . 2�j. k� m . and . . b r . 2�i . �j. 2k� m . Sol. Here, the given field is uniform(constant). So using, dV . .E . dr .. . or a ab a b b V . V . V . .. E . dr .. .. . . . . . .1, 2, 1. 2, 1, 2 2�i 3�j 4k� . dx �i dy�j dz k� . . . .. . . . . . . . . .1, 2,1. 2,1, 2 2dx 3dy 4dz . . . .. . . . .. . .1, 2, 1. 2,1, 2 2x 3y 4z . . . . . . = � 1 V Example-31 : Auniformelectric field having strengthE.. is existing in x�yplane as shown in figure. Find the p.d. between originO &A(d, d, 0)
(A) Ed(cos. + sin.) (B) �Ed(sin. � cos.) E A(d,d,0) o x y z (C) 2 Ed (D) none of these Sol. Here E . Ecos. �i . Esin. �j .. . OA . d �i . d �j . .... . ..v . E . .. .. ..v . Ed cos. . Ed sin. . Ed(cos. . sin. ) A 0 . . (V .V ) . Ed(cos. . sin.) 0 A . V . V . Ed(cos. . sin. ) Example-32 : Uniformelectric field ofmagnitude 100V/min space is directed along the line y= 3 + x. Find the potential difference between pointA(3, 1)&B (1, 3) (A) 100 V (B) 200 2 V (C) 200 V (D) 0 Sol. . y = 3 + x . tan. = 1
ELECTROSTATICS www.physicsashok.in 24 . . = 45º E .100cos. �i .100sin. �j .. E 100 �i 100 �j 2 2 . . .. .r . AB . i� . 3�j. 3�i . �j . .2�i . 2�j ... .... .V . .E . .r .. ... 100 � 100 � V i j r 2 2 . . . . .. . . . . . . ... V 100 �i 100 �j . 2�i 2�j. 2 2 . . . . .. . . . . . . . . .V . .100 2 .100 2 . A B V - V = 0 Example-33 : A, B, C, D, P andQare points in a uniformelectric field. The potentials a these points are V (A) = 2 volt. V(P) =V(B) =V(D) = 5 volt. V(C) = 8 volt. The electric field at P is 0.2m AB P CQD 0.2m (A) 10 Vm�1 along PQ (B) 15 2 Vm�1 along PA (C) 5 Vm�1 along PC (D) 5Vm�1 along PA Sol. A D x E V V 2 5 30 15 v /m 0.2 0.2 2 . . . . . . . . Also A B y E V V 2 5 30 15 v /m AB 0.2 2 . . . . . . . . E = - 15�i - 15�j .. Py x 15 15 x�A 45º
| E |= (- 15)2 + (- 15)2 = 15 2 v /m .. PA E .15 2 Hence (B) is correct. Example-34 : Variation of electrostatic potential along x-direction iswhosn in the graph. The correct statement about electric field is (A) xcomponent at point Bismaximum (B) x component at pointAis towards positive x-axis x A B C v (C) x component at point C is along negative x-axis (D) x component at point C is along positive x-axis Sol. The negative slope v � x graph give x � component of electric field. In the given graph, slope at C is negative. Hence, x � component of electric field is positive. EQUIPOTENTIAL SURFACE Equipotential surface is an imaginarysurface joining the points of same potential inan electric field. So,we can say that the potential difference between anytwo points on anequipotential surface is zero.The electric lines of force at each point ofan equipotential surface are normal to the surface. Fig. shows the electric lines of force due to a point charge +q. +q This spherical surfacewillbe the equipotential surface and the electrical lines of force emanating fromthe point chargewill be radial and normal to the spherical surface.
ELECTROSTATICS www.physicsashok.in 25 Regarding equipotentialsurface, following points areworth noting : (a) Equipotential surfacemaybe planar, solid etc.But equipotential surface never be a point. (b) Equipotential surface is single valued. So, equipotential surfaces never cross each other. (c) Electric field is always perpendicular to equipotential surface. (d) electric lines of force cross equipotentialsurface perpendicularly. (e) Work done to move a point charge q between two points on equipotential surface is zero. (f) The surface ofa conductor in equilibriumis equipotential surface. (g) equipotential surface due to isolated point charge is spherical. (h) Equipotential surface are planar in uniformelectric field. (i) Equipotential surface due to line charge is cylindrical. (j) Equipotential surface due to an electric dipole is shown inthe figure. �q +q Example-35 : The electric field in a region is givenby : E . .4axy z .�i . .2ax2 z .�j. .ax2y / z .k�, where a is a positive constant.The equation of anequipotential surfacewill be of the form (A) z = constant/[x3y2] (B) z= constant/[xy2] (C) z = constant/[x4y2] (D) none Sol. At equipotential surface potential is constant. Therefore E v r = - d d .. dv . . E . dr .. . . E .d r . constant .. . r . x i� . y�j. z k� . . dr . dx�i . dy�j. dzk� . . E . (dx �i . dy�j. dz k�) .. = Constant
ELECTROSTATICS www.physicsashok.in 26 .(4axy z )i� . (2ax2 z)�j. (ax2y / z)k� . . (dx�i . dy�j. dzk�) . . . = constant 2 4axy z dx 2ax2 z dy ax ydz z . . . . . = constant. 2 4ax y z 2ax2y z 2ax2y z 2 . . = constant 6 ax2 y z = constant 2 z constant 6ax y = .. 4 2 z constant x y . Example-36 : The equation of an equipotential line an electric field is y= 2x, thenthe electric field vector at (1, 2) may be (A) 4�i +3�j (B) 4�i +8�j (C) 8i� + 4�j (D) �8�i + 4�j Sol. Electric field is perpendicular to equipotential line . y = 2 x or dy 2 dx = . m1 = 2 . m1m2 = � 1 . 2 m 12 = - . tan 12 . . . where . is anglewith x-axis. In option (D), tan 4 1 8 2 . . . . . ELECTRIC DIPOLE Electric dipole is a systeminwhich two equal and opposite point charges are placed at a smalldistance. The product of any of the charges and distance between two charge is called electric dipole 2l �q P +q moment p. It is directed fromnegative charge to positive charge (fig.). The line joining the two charges is called axis of dipole. Let charges of an electric dipole are �q and +q and are separated by a small distance 2l. The dipolemoment of such a dipole is given by p = q × 2l = 2ql Electric Potential and Field due to an Electric Dipole Let an electric dipole is placed along y-axis and its centre is at origin, thene
lectric potential at pointA(x, y, z) due to this dipole. +q�q z y l l x2 + z2 + (y+l)2 x2 + z2 + (y�l)2 A(x, y, z) x
ELECTROSTATICS www.physicsashok.in 27 2 2 2 2 2 2 0 1 q q V 4 x (y ) z x (y ) z . . . . . . .. .. . . l . . . l . .. x E Vx . . . . , y E Vy . . . . and z E Vz . . . . Special Cases : (i) On the axis of dipole (axial position) : x = 0, z = 0 (a) 2 2 0 V 1 p 4 (y ) . .. . l 2 2 0 V 1 p 4 (r ) . .. . l If y = r or axis 2 0 V 1 P 4 r . .. If r > > l (b) Ex = 0 = Ez and y 2 2 2 0 E E 1 2py 4 (y ) . . .. . l (alongp . ) 2 2 2 0 1 2pr 4 (r ) .
.. . l If y = r axis 3 0 E 1 2p 4 r . .. (ii) On the perpendicular bisector of dipole (equatorical position) : Say along x-axis (it may be along z-axis also) y = 0, z = 0 (a) bisec tor V 0 . . (b) Ex = 0, Ez = 0 and y 2 2 3/ 2 0 E 1 p 4 (x ) . . .. . l (opposite to p . ) Here, negative sign implies that the electric field is along negative y-direction or antiparalleltop . . Further at a distance r fromthe centre of dipole (x= r), then 2 2 3/ 2 0 E 1 . p 4 (r ) . .. . l or bisec tor 3 0 E 1 p 4 r . . .. if r >> l (iii) In polar co-ordinates (r, .) : (a) 2 0 V 1 p cos 4 r . . .. (b) r E Vr . . . . 3 0 1 . 2pcos 4 r . . .. E 1 . V r . . . . .. 3 0 1 psin
4 r . . .. E Er E P r O A(r, ) . 2 2 r E E E. . . . 2 3 0 E p 1 3cos 4 r . . . .. (c) In Figure, tan tan2 . . .
ELECTROSTATICS www.physicsashok.in 28 Example-37 : 4 charges are placed each at a distance �a� fromorigin. The dipolemoment of configurationis (A) 2qa�j (B) 3qa�j 3q�2q �2q qy x (C) 2aq[i� + �j] (D) none Sol. Here 1 P . 2q . 2 a 1 P = 2 2 qa 2 P . q . 2a . 2 qa 3 P . q . 2a . 2 qa 3q �2q �2q q y x P1 45º 45º 45º P3 P2 2a2a 2a . x 1 2 3 P = P cos 45 - P cos 45 - P cos 45 x P = 2 2qa cos 45 - 2qa cos 45 - 2qa cos 45 x P = 0 y 1 2 3 P = P sin 45 + P sin 45 - P sin 45 y 2 2qa 2qa 2qa P 2qa 2 2 2 = + - = . P . 2 q a�j .. Example-38 : Two point charges +3 µC and �3 µC are placed at a small distance 2 × 10�3 mfromeach other. Find: (a) electric field and potential at a distance 0.6mfromdipole in its equatorialposition. (b) electric field and potential at the same distance fromdipole as in (a) if the dipole is rotated through 90º. Sol. Dipolemoment p = q × 2l = (3 × 10�6) (2 × 10�3) p = 6 × 10�9 C-m (a) Electric field in equatorialposition, 3 0 E 1 . p 4 r . .. . . 9
9 3 E 9 10 6 10 (0.6). . . . . . . . . . E = 250 N/C Electric potential,V= 0 (b) On rotating the dipole through 90º, the same point nowwill be in axial position. So, electric field 3 0 E 1 . 2p 500 N/C 4 r . . .. and electric potential 2 0 V 1 p 4 r . .. . . 9 9 2 V 9 10 6 10 (0.6). . . . = 150 V
ELECTROSTATICS www.physicsashok.in 29 Example-39 : Ashort electric dipole is situated at the origin of coordinate axis with its axis along x-axis and equator along y-axis. It is found that the magnitudes of the electric intensity and electric potential due to the dipole are equal at a point distance r = 5 m fromorigin find the position vector of the point. Sol. Let P be such a point at distance r and angle . fromequator. Now |EP| = |VP| or P 2 P 3 2 K 1 3sin K sin r r . . . . or 1 3sin2 sin 5 . . . . Squaring both nets y x rP q 1 + 3 sin2. = 5 sin2. or sin 12 . . or ..= 45º Positive vector of r . point P is r 5 (�i �j) 2 = + . Electric Dipole in Uniform Electric Field : (i) Torque :When a dipole is placed in a uniformelectric field as shown in Fig. the net force on it, F . ..qE . ..q. E.. . 0 .. .. while the torque, �q +q P qE qE l E . = qE × 2l sin . . . pEsin . . or . . p . E . . ..
Fromthis expression it is clear that torque acting on a dipole ismaximum(= pE)when dipole is perpendicular to the field andminimum(=0) when dipole is parallel or antiparallel to the field. NOTE :If the electric field is not uniform the dipole will experience both a resultant force and a torque so its motion will be combined translatory and rotatory (if .. 0 or 180º) (ii) Work :Work done in rotating a dipole in a uniformfield througha small angle d.willbe, dW= .d..= pE sin .d. So,work done in rotating a dipole fromangular position .1 to .2with respect to field, 2 1 1 2 W pEsin d pE [cos cos ] . . . . . . . . . . So, if a dipole is rotated fromfield direction, i.e., .1 = 0 to position ., i.e., .2 = .. W= pE[1 � cos.]
ELECTROSTATICS www.physicsashok.in 30 �q +q P= 0 E Wmin = 0 (a) +q �q P = 180º E (b) Wmax = 2pE (iii) Potential energy : In case of a dipole (in a uniformfield), potential energy of dipole is defined aswork done in rotating a dipole froma directionperpendicular to the field to the given direction. i.e., U = (W. � W90º) U = pE (1 � cos .) � pE (1 � cos 90º) U . .pE cos. . .p . E . .. (iv) AngularSHM:Adipolewhen placed in a uniformelectric field, align itself parallel to the field. If it is given a small angular displacement . about its equilibriumposition, the restoring torque produced in itwillbe, . = �pE sin . = � pE. (. sin . . .) or 22 d I pE dt. . . . or 2 2 2 ddt. . .. . with 2 pEI . . This is standard equation of angular SHMwith period T 2. .. . . . . . . . So, dipolewill execute angular SHMwith time period T 2 I PE . . Interaction of Two Dipoles : If a dipole is placed in the field ofother dipole, then depending onthe positions of dipoles relative to eachother, force, torque and potential energyare different. TABLE : Dipole-Dipole Interaction S.No. Relative position of Dipole Force F Torque . PotentialEnergyU 1. r
p1 F F p2 1 2 0 4 1 6p p (along r) 4 r . .. zero 1 2 3 0 1 2p p 4 r . .. 2. p r 1 p2F F 1 2 4 0 1 3p p 4.. r (along r) zero 1 2 3 0 1 p p 4.. r 3. r F p F p 2 1 1 2 4 0 1 3p p 4 r . .. on 1 2 1 3 0 p , 1 2p p 4.. r zero (perpendicular to r) on 1 2 2 3 0 p , 1 p p 4.. r Example-40 : Two point charges +3.2 ×10�19 C and �3.2 × 10�19 C are placed at a distance 2.4 × 10�10m, from each other. This dipole is placed in a uniformelectric field of 4.0 × 105V/m. (a) Find the necessarytorque required to rotate the dipole through 180º fromits equilibriumposition. (b) What is thework done inrotating the dipole through 180º ?
ELECTROSTATICS www.physicsashok.in 31 (c) What is the potential energy of dipole in this position ? Sol. Electric dipolemoment p = q × 2l Here, q = 3.2 × 10�19 C 2l = 2.4 × 10�10 m . p = (3.2 × 10�19) (2.4 × 10�10) p = 7.68 × 10�29 C-m (a) necessary torque . = pE sin . ..= 7.68 × 10�29 × 4 × 105 × sin 180º ..= 0 ( . sin 180º = 0) (b) Work done inrotating the dipole through 180º is given by W= pE (cos.1 � cos.2) Here, .1 = 0, .2 = 180º, p = 7.68 × 10�29 C-m and E = 4.0 × 105 V/m . W = 7.68 × 10�29 × 4 × 105 × (cos 0º � cos 180º) W = 7.68 × 10�29 × 4 × 105 × (1 + 1) W = 6.14 × 10�23 J (c) The potential energyof dipole in this positionis given by U. = � pE cos . where ..is the angle between the axis of dipole and electric field. In equilibriumposition, ..=0 . U0 = � pE or U0 = � (7.68 × 10�29) (4.0 × 105) U0 = �3.07 × 10�23 J Example-41 : Awheelhavingmassmhas charges +q and �q on diametrically opposite points. It remains in equilibriumon a rough inclined plane in the presence of uniform vertical electric field E = +q �q E (A) mg q (B) mg 2q (C) mg tan 2q q (D) none Sol. The torque of electric force about centre is balanced bytorque due to friction about the centre. r f= PE sin. . . = PE sin. But f=mg sin. (for equilibrium) . mgr sin. =PE sin. +q -q mgsin 90- 90- E f P E mgr mgr P q 2r
. .
.
. P = q × 2r (dipolemoment) E mg 2q .
ELECTROSTATICS www.physicsashok.in 32 Example-42 : Anonconducting ring ofmassmand radius Ris charged as shown. The charged densityi.e. charge per unit length is .. It is then placed on a rough nonconducting horizontalsurface plane.At time t = 0, a uniformelectric field E = E0i .. is swithced on and the ring start rollingwithout sliding. y � x � � � +++ + Determine the frictionforce (magnitude and direction) acting on the ring, whenit startsmoving. Sol. Where f is friction surface. f = macm ...(1) Let d. element on ring then dq . . d. dq . .Rd. { d dR . . . . } y � x � � � +++ + p f rough non-conducting surface d dP E . . . ... .. d. . 2RdqEsin .. .. . {. dp = (2Rdq)} d. . 2RE .Rsin.d. / 2 2 E 0 . 2R E . sin d . . . . . 2 E . . 2R .E 2R2 .E � fR = mR2. �dq f dqq q dq E d p p- q 2R.E � f = mR.
2R.E � macm = mR. (fromeqn (1)) for pure rolling, acm = R. . 2R.E = macm + mR. 2R.E = 2 macm macm = .RE cm 0 f . ma . .RE Example-43 : In the figure shown S is a large nonconducting sheet of uniformcharge density ..Arod Rof length l andmass �m�is parallel to the sheet and hinged at itsmid point. The linear charge densities on the upper and lower half of the rod are shown in the x l - l R Ss figure. Find the angular acceleration of the rod just after it is released. Sol. Electric field due to non-conducting sheet is 0 . 2 . . d. = x E dq × 2 = 2x Edq / 2 2 /2 0 0 d 2x E dx 2E x2 . . . . . . . . . . . . . . . . . 2E 2 E 2 8 4 . . . . . . . . E 2 I 4 . . . . Edq Ss Edq �dq +dq xx or m 2 E 2 12 4 . . . . .
ELECTROSTATICS www.physicsashok.in 33 . 2 2 E 12 3E 4 m m . . . . . . . . . 0 3 2 m . .. . . ELECTRIC FLUX Electric flux throughan elementaryarea dS is defined as the scalar product of area and field, i.e., d.E . E . dS .. . d.E = EdS cos . n E i.e., E dS . . . E . dS .. . It represents the total lines offorce passing through the given area. Regarding electric flux following points areworthnoting : (a) Electric fluxwillbemaximumwhen cos. = 1, i.e., . = 0, i.e., electric field is normal to the area with (d.E)max = EdS. En (d ) = EdS E max (b) Electric fluxwillbeminimumwhen cos. = 0, i.e., . = 90º, i.e., field is parallel to the surfacewith (d.E)min = 0. E n (d ) = 0 E min (c) For a closed body outward flux is taken to be positive [Fig. (a)] while inward negative [Fig. (b)] n (a) Positive flux (b) Negative flux GAUSS�S LAW This lawgives a relation between the net electric flux through a closed surface and the charge enclosed by the surface. According to this law, �the net electric flux through any closed surface is equal to the net charge inside the surface divided by .0.� In symbols it can bewritten as, in E S 0 . . E . n� dS . q. . .. . ...(i)
but this formofGauss�s lawis applicable only under following two conditions : (i) The electric field at every point onthe surface is either perpendicular or tangential. (ii) Magnitude ofelectric field at everypoint where it is perpendicular to the surface has a constant value (sayE). Here, S is the areawhere electric field is perpendicular to the surface. Applications of Gauss�s Law : To calculate Ewe choose animaginaryclosed surface (calledGaussian surface) inwhichEqs. (i) or (ii) can be applied easily. Inmost of the caseswewill use Eq. (ii). Let us discuss fewsimple cases.
ELECTROSTATICS www.physicsashok.in 34 (i) Electric field due to a point charge : FromEq. (ii), in 0 ES . q. Here, S = are of sphere = 4.r2 and qin = charge enclosing theGaussian surface = q q r E . . 2 . 0 E 4.r . q . . 2 0 E 1 q 4 r . .. It is nothing but Coulomb�s law. (ii) Electric field due to a linear charge distribution : Consider a long line chargewith a linear charge density(charge per unit length) ..We have to calculate the electric field at a point, a distance r fromthe line charge.We construct aGaussian surface, a cylinder of any arbitrary length l of radius r and its axis coinciding with the axis of the l r E ++++ + + + + E line charge. Hence, we can apply theGauss�s lawas, in 0 ES . q. E Curved Surface E Plane Surface Here, S = area of curved surface = (2.rl) and qin = net charge enclosing this cylinder = .l . 0 E(2 ) . .. . . l l . 0 E 2 r . .
.. i.e., E 1r . (iii) Electric field due to a plane sheet of charge : FromGauss�s Law in 0 ES . q. . 0 0 0 E(2S ) ( ) (S ) . . . + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + E E S0 . 0 E 2. . . (iv) Electric field near a charged conducting surface : This is similar to the previous one the onlydifference is that this time charges are on bothsides.Hence, applying in 0 ES . q.
ELECTROSTATICS www.physicsashok.in 35 Here, S = 2S0 and qin = (.) (2S0) + E S0 E + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + . + 0 0 0 E(2S ) ( ) (2S ) . . . . 0 E . . . Example-44 : Which of the following statements is/are correct ? (A) Electric field calculated byGauss lawis the field due to only those charges, which are enclosed inside the Gaussian surface. (B) Gauss law is applicable onlywhen there is symmetrical distribution of charge. (C) Electric flux through a closed surface is equal to total flux due to all the charges enclosed within that surface only. (D) None of these Sol. Since, electric field at a point is equal to electric flux passing per unit area, therefore, . E. dS .. . . is the net flux emanating from a closed surface. Though net flux through the closed surface depends upon the charges enclosed in that surface only but electric field E at a point depends not onlyupon charges enclosed but it depends upon charges lying outside the surface also. Hence (A) is wrong. Gauss lawis applicable to a closed surface. The surfacemay have any shape. It means, it is a general law. Hence (B) is wrong. Gauss law is 0E. dS q . . . . . .. . . It means, net flux through a closed surface depends upon .q. But it is equal to net charge enclosed within the surface only. Hence (C) is correct. Obviously (D) is wrong.
Example-45 : In a region of uniformelectric field E, a hemispherical body is placed in such away that field is parallel to its base (as shown in figure). E OC The flux linked with the curved surface is : (A) zero (B) �.R2E (C) .R2E (D) R2 E 2 . Sol. The flux linked with the body is zero as it does not enclose any charge. . = .cs + .pb = 0 As field is parallel to base, the flux linked with base .pb = E × .R2 cos 90º = 0 E n . .paralled to base = 0 .curved surface = 0 Example-46 : If a point charge q is placed at the centre of a cubewhat is the flux linked (a) with the cube (b) with each face of the cube ? Sol. (a)According to Gauss�s Law, in total 0 0 . . q . q . . (b) Nowas cube is symmetrical bodywith 6-faces and the point charge is at its centre, so electric flux linkedwith each facewill be total 0 ´ q 6 6 . . . . .
ELECTROSTATICS www.physicsashok.in 36 NOTE : If charge is not at the centre of cube (but any where inside it), total flux will not change but the flux linked with different faces will be different. Example-47 : If a point charge q is placed at one corner of a cube, what is the flux linkedwith the cube ? Sol. By placing three cubes at three sides of given cube and four cubes above, the chargewill be in the centre. So, the fluxwith each facewillbe one-eight of the flux (q/.0). . Flux throughthe cube 0 q 8 . . Example-48 : Calculate flux through the cube and flux through the each surface of cube when q is placed at one of its corner. Sol. To cover charged particle completlywe have to use 7 extra cubes around the charged particle. So, Flux Through cube 0 ( ) q 8 . . . Flux through 0 ABCD q 3 24 . . . . Flux through 0 ABFE q 3 24 . . . . D C AE F B HG q Flux through 0 ADEH q 3 24 . . . . Flux through EFGH = 0
Flux through FGCB = 0 Flux through GCDH = 0 Example-49 : The electric field in a region is given by 0 E = E x i .. . l . Find the charge contained inside a cubical volume bounded by the surfaces x = 0, x = a, y = 0, y = a, z = 0 and z = a. Take E0 = 5 × 103N/C, l = 2cm and a = 1cm. Sol. 0 E . E x �i .. . Totalflux 0 . q . 0 EA . q . whereAis area of surface 0 2 0 E a . a . q . . . 3 0 0 E a q . . . 3 0 0 E a q . . . . q . .3 12 2 3 2 5 10 8.85 10 1 10 2 10. . . . . . . . . . q 5 103 8.85 10 4 2 . . . . . . q.2.2.10.12 C
ELECTROSTATICS www.physicsashok.in 37 ELECTRIC FIELD AND POTENTIAL DUE TO CHARGED SPHERICAL SHELL OR SOLID CONDUCTING SPHERE Electric Field The allpoints inside the charged spherical conductor or hollowsphericalshell, electric field E . 0, .. as there is no charge inside such a sphere.We can construct a Gaussian surface (a sphere) of radius r > R. FromGauss�s Law C 0 E . ds . q . . .. . 2 0 E(4.r ) . q . . 2 0 E 1 q 4 r . .. r R Gaussian surface E + q +++ + + + + + ++ + Hence, the electric field at anyexternalpoint is the same as if the total charge is concentracted at centre. At the surface of sphere r = R, . 2 0 E 1 q 4 R . .. Thus, we canwrite, Einside = 0 surface 2 0 E 1 q 4 R . .. outside 2 0 E 1 q 4 r .
.. The variation ofelectric field (E) with the distance fromthe E O R r 1 qR2 = E r2 1 centre (r) is as shown in fig. NOTE : � At the surface graph is discontinuous. � 2 surface 2 0 0 0 E = 1 q = q/4pR = s 4pe R e e Potential : As we have seen, outside 2 0 E 1 q 4 r . .. . outside 2 0 dV 1 q dr 4 r . . . . . . . . .. E dV dr . . . . .. .. . V r 0 outside 2 0 dV q dr 4 . r . . .. . . (V. = 0) . 0 V 1 q or V 1 4 r r . . .. Thus, at externalpoints the potentialat anypoint isthe samewhenthewhole charge is assumedto the concentrated at the centre.At the surface of the sphere, r = R. . 0 V 1 q 4 R . .. At some internalpoint electric field is zero everywhere, therefore, the potential is same at allpointswhich is equal to the potential at surface. Thus,we canwrite,
inside surface 0 V V 1 q 4 R . . ..
ELECTROSTATICS www.physicsashok.in 38 and outside 0 V 1 q 4 r . .. V O R r 1 qR = V r 1R The potential (V) varieswith the distance fromthe centre (r) as shown in Fig. Example-50 : The diameter of hollowmetallic sphere is 60 cmand charge on sphere is 500 µC. Find the electric field and potential at a distance 10 cmfromcentre of sphere. Sol. The point at 10 cmdistance fromcentre of spherewillbe inside the sphere. Inside hollowconducting sphere, electric field is zero everywhere and potential is uniform(same as at the surface). Hence, intensityof electric field E = 0 and potential 0 V 1 q 4 R . .. Here, q = 500 µC = 500 × 10�6 C and R = 30 cm= 0.30 m . . . . . 6 9 500 10 V 9.0 10 0.30 . . . . . V = 1.5 × 107 V ELECTRIC FIELD AND POTENTIAL DUE TO A UNIFORMALY NON-CONDUCTING SPHERE Electric field : Let positive charge q is uniformlydistributed throughout the volume of a solid sphere of radius R.We have to find the intensityof electric field due to this charged sphere at point P distance r fromcentreO. ApplyingGauss�s law in 0 ES . q. ...(i) Here, S = 4.r2 and 3 in q ( ) 4 r 3 . . . . . .. .. Here, 3 . = charge per unit volume = q 4/3.R
Substituting these values inEq. (i),we have 3 0 E 1 . q . r 4 R . .. or E . r At the centre r = 0, so, E = 0 At surface r = R, so, 2 0 E 1 q 4 R . .. To find the electric field outside the charged sphere,we use a sphericalGaussian surface ofradius r(>R).This surface encloses the entire charged sphere, so qin = q, and Gauss�s law gives. 2 0 E(4.r ) . q . or 2 0 E 1 q 4 r . .. or 2 E 1 r . Thus, for a uniformly charged solid sphere,we have the following formulae formagnitude of electric field. inside 3 0 E 1 . q . r 4 R . .. surface 2 0 E 1 . q 4 R . .. outside 2 0 E 1 . q 4 r . .. E O R r 1 qR E r E r2 2 1
ELECTROSTATICS www.physicsashok.in 39 The variation of electric field (E)with the distance fromthe centre of the sphere (r) is shownin fig. Potential : The field intensityoutside the sphere is, outside 2 0 E 1 . q 4 r . .. outside outside dV E dr . . . V r outside 2 0 dV 1 . q dr . . 4 r . . .. . . . 0 V 1 q as V 0 4 r . . . .. or V 1r . At r = R, 0 V 1 q 4 R . .. i.e., at the surface of the sphere potential is S 0 V 1 q . 4 R . .. The electric intensityinside the sphere, inside 3 0 E 1 . q . r 4 R . .. inside inside dV E dr . . .
S V r V inside 3 R 0 dV 1 q r dr 4 R . . .. . . . 2 r S 3 0 R V V 1 q r 4 R 2 . . . . . . . .. . . Substituting S 0 V 1 . q , 4 R . .. we get22 0 V 1 . q 3 1 r 4 R 2 2 R . . . . . . .. . . At the centre r = 0 and C S, 0 V 3 1 q 3 V 2 4 R 2 . . . . . . . .. . i.e., potential at the centre is 1.5 times the potential at surface. Thus, for a uniformlycharged solid spherewe have the following formulae for potential. outside 0 V 1 q 4 r . .. surface 0 V 1 q 4 R . .. and 2 inside 2 0 V 1 q 3 1 r 4 R 2 2 R . . . . . . .. . . 1 q
R 1 qR 32 V The variationof potential (V)withdistance formthe centre (r) is as shown in figure.
ELECTROSTATICS www.physicsashok.in 40 Example-51 : The radius of a solidmetallic non-conducting sphere is 60 cmand charge on the sphere is 500 µC. Find the electric field and potential at a distance 10 cmfromcentre of sphere. Sol. The point 10 cmfromcentre of spherewill be inside the sphere. Hence, inside 3 0 E 1 . qr 4 R . .. Here, q = 500 × 10�6 C, r = 10 cm = 0.10 m R = 60 cm= 0.6 m . . . 6 9 6 inside 3 E 9.0 10 500 10 0.10 2 10 N/C (0.60) . . . . . . . . and 2 inside 2 0 V 1 q 3 1 r 4 R 2 2 R . . . . . . .. . . . . 6 2 9 2 9 10 (500 10 ) 3 1 (0.10) (0.60) 2 2 (0.60) . . . . . . . . . . . . 7.5 106 3 1 2 72 . . . . . . .. .. 7.5 106 107 72 . . . = 1.1 × 107 V Example-52 : Asolid non conducting sphere ofradiusRhas a non-uniformcharge distribution ofvolume charge density, 0 r , R . . . where .0 is a constant and r is the distance fromthe centre of the sphere. Showthat: (a) the total charge on the sphere isQ = ..0R3 and (b) the electric field inside the sphere has amagnitude given by, 2 4 KQr E . R .
Sol. (a) dv = 4.r2dr 0 2 dq dV r 4 r dr R . . . . .0 3 4 r dr R . .. R 0 3 0 4 Q r dr R . . .. r dr R 4 0 3 0 4 R Q Q R R 4 . . . . .. .. (b) r 0 3 in 0 4 q r dr R . . .. 4 4 0 0 4 r 4 r R 4 4R . . . .. .. in 2 0 E q 4 r . . . 4 0 2 0 4 r E 4R 4 r . . . ...
But Q = ..0R3
ELECTROSTATICS www.physicsashok.in 41 0 3 Q R .. . and 0 K 1 4 . . . . 2 4 E KQr R . Example-53 : Three concentric metallic shellsA, B and C of radii a, b and c (a < b < c) have surface charge densities ., �. and . respectively. (i) Find the potential of three shellsA, B and C. (ii) If the shellsAand C are at the same potential, obtain the relation between the radii a, b and c. Sol. The three shells are shown in figure. (i) Potential ofA= (Potential ofAdue to + ..onA) + (Potential ofAdue to � ..on B) + (Potential ofAdue to + ..on C) 4 a2 4 b2 4 c2 K a b c . . . . . . . . . . . . . . . . . 0 . a . b . c .. c b C B A a +s +s - s Potential of B = (Potential due to + ..onA) + (Potential due to � ..on B) + (Potential due to + ..on C) 4 a2 4 b2 4 c2 K a b c . . . . . . . . . . . . . . . 2 0 a b c b . . . . . . . . . . .
Potential of C = (Potential due to + ..onA) + (Potential due to � ..on B) + (Potential due to + ..on C) 4 a2 4 b2 4 c2 K a b c . . . . . . . . . . . . . . . 2 2 0 a b c b c . . . . . . . . . . . (ii) Given that VA =VC or . . 2 2 0 0 a b c a b c b c . . . . . . . . . . . . . . . . or a2 b2 a b c c b c . . . . . Solving we get, c = (a + b). Example-54 : An electric field converges at the originwhosemagnitude is givenby the expression E = 100N/C, where r is the distancemeasured fromthe origin. (A) total charge contained in any sphericalvolumewith its centre at origin is negative. (B) total charge contained at anysphericalvolume, irrespective ofthe location of its centre, is negative. (C) total charge contained in a spherical volume of radius 3 cm with its centre at origin has magnitude 3×10�13 C. (D) total charge contained in a spherical volume of radius 3 cm with its centre at origin has magnitude 3×10�9 C.
ELECTROSTATICS www.physicsashok.in 42 Sol. Since, electric lines of forces are terminated into the sphere. Hence, total charge contained by should the sphere should be negative. So, options (A) and (B) are correct E E E E . 2 0 E q 4 r . .. Here r = 3 × 10�2C . q = 3 × 10�13 m . Hence, option (C) is correct. Example-55 : Figure shows a section through two long thinconcentric cylinders of radii a&bwith a < b. The cylinders have equal and opposite charges per unit length .. Find the electric field at a distance r fromthe axis for b a (A) r < a (B) a < r < b (C) r > b Sol. a l b - l (A) r < a C 0 0 E . ds . . . .. . r b Ea 0 E . 2 r . O. . . E = 0 (B) a < r < b C 0 q E . ds . . . .. . r b a 0 E . 2 r . .. . . . 0 E 2 r . . .
. (C) r > b C 0 E ds 0 . . . . . . .. . . . . . E . 2.r. . 0 a b r E . 0 THINKINGTYPEPROBLEMS 1. The electric charge ofmacroscopic bodies is actually a surplus or deficit of electrons.Why not protons? 2. What are insulators and conductors? 3. Acharged rod attracts bits of dry cork dust which, after touching the rod, often jump away fromit violently. Explain. 4. Atruck carrying explosives has ametal chaing touching the ground.Why? 5. Howis the Colulomb force between two charges affected bythe presence of a third charge ? 6. Aperson standing on an insulating stool touches a charged insulated conductor. Is the conductor completely discharged ? 7. Electric lines of fore never cross.Why ?
ELECTROSTATICS ELECTROSTATICS 8. The electric field inside a hollow charged conductor is zero. Is this true or false ? 9. The electric field inside a hollow charged sphericalconductor is the same at allpoints and is equalto the field at the surface. Is this true or false ? 10. Theelectricpotentialinsideahollowchargedsphericalconductor isthesameat allpointsandisequalto the potential of the surface. Is this true or false ? 11. AchargeAsituatedoutsideanunchargedhollowconductorexperiencesaforceifanother chargeBisplaced insidetheconductor, but Bdoesnot experienceanyforce.Why?Doesitnotviolatethethirdlawofmotion? 12. If only one charge is available, can it be used to obtain a charge many times greater than it in magnitude? 13. In the previous question, does it make a difference which face of B is touched in order to remove the free charge (include positive charge) from B? 14. Can a small spherical body of radius 1 cm have a static charge of 1 C? 15. A metal leaf is attached to the internal wall of an electrometer insulated from the earth. The rod and housing of the electrometer are connected by a wire, and then a certain charge is imparted to the housing. Will the leaves of the electrometer be deflected ? What will happen to the leaves if the wire is removed and the rod is then charged ? 16. Abroad metalplate is connected to the earth through a galvanometer. Apositively charged ball flies along a straight line above the plate at a distance much less than the linear dimensions of the plate. Draw an approximate diagram showing how the current flowing through the galvanometer depends on time. 17. The work done in carrying a point charge from one point to another in an electrostatic field depends on the path along which it is taken. Is this true of false? Briefly explain your answer. 18. The figure shows lines of constant potential in a region in which on electric field is present. The values of the ptentials are written in brackets. Of the points A, B and C, the magnitude of the electric field is greatest at the point...... Give reasons for your answer in brief. 19. Can two balls with like charges be attracted to each other? 20. The housing of the electrometer of question no.15 is given a charge. Will the leaves of the electrometer be deflected in this case? Will the deflection change if the rod is earthed ?There is no connecion between the rod and the housing. 21. By touching different points of a metal bucket with a test ball B connected by a wire to an earthed electrometer it can be observed that deflection ofthe leaves of the electrometer is the same for any position of the ball. But if the wire is removed and the charge is transferred by the ball to the ball of the electrometer, the deflection depends on which surface of the bucket (external or internal) is touched before that. Why ?
22. The gravitationalfield strength is zero inside a sphericalshellof matter. The electric field strength is zero not only inside anisolated charged spherical coonductor but inside an isolated conductor of anyshape. Is the gravitational fieldinside,say, acubicalshellofmatter zero ?Ifnot, inwhatrespect istheanalogynot complete?
23. Cantwo equipotentialsurfacesintersect? 24. An isolated, conducting spherical shell of radius R carries a negative charge Q. What will happen if a small, positivelychargedmetalobject isplacedinsideandconnectedbyawire?Assume that thepositivechargeis (a) less than, (b) equal to and (c) greater than Q. 25. AnunchargedmetalspheresuspendedbyasilkthreadisplacedinauniformexternalelectricfieldE.What is the magnitude of the electric field for points inside the sphere ? Will your answer change if the sphere carries a charge? 26. Is charge uniformlydistributed over the surface of an insulated conductor of anyshape ? If not, what is the rule for the distribution of charge over a conductor? 27. Areequipotentialsurfaceswhichariseduetoapointchargeandwhosepotentialsdiffer byaconstant amount (say 1 volt) evenly spaced in radius ? 28. Two point charges Q and 4Q are fixed at a distance of 12cm from each other. Sketch the lines of force and www.physicsashok.in 43
ELECTROSTATICS www.physicsashok.in 44 locate the neutralpoint, if any. 29. Acharge +Qis fixed at a distance d in front of an infinite metal plate. Draw the lines of force indicating the direction clearly. 30. Indicate the surface distribution of charge on a squaremetal plate byusing dots in such away that the greater the surface densityof charge, the farther awaythe dots are fromthe plate (shown in fig.). 31. Ametal plate ismoved with a constant acceleration a parallel to its plane. Does any charge develop on its surfaceswhich are perpendicular to itsmotion? 32. Is an electric field of the type shown by the electric lines in the figure physicallypossible? 33. Zero work is donewhen a charged particle is transferred fromone equipotential surface to another. Is this true or false ? 34. ApotentialdifferenceVis set up betweena filament emitting electrons in avaccumtube and a thinmetallic ring. The electron beampasses past the ring through its central regionwithout spreading. The kinetic energyof the electrons inthe beamincreaseswhile the batteryproducing the potentialdifferenceVperforms nowork, since no current flows through the circuit. Howcan his fact be reconciledwith the principle of conservation of energy? 35. There is an electric field near the surface of a conductor carrying direct current. Is this true or false? 36. Ordinary rubber is an insulator. But the special rubber tyres of aircrafts are made conducting.Why is this necessary? 37. Asmall sphere is charged to a potentialof 50Vand a big hollowsphere is charged to a potentialof 100V. How can youmake electricityflowfromthe smaller sphere to the bigger one? 38. An electric dipole is placed in a non-uniformelectric field. Is there a net force on it? 39. Apoint charge is placed at the centre of a spherical gaussion surface. Does electric flux e . change (a) if the sphere is replaced by a cube of the same volume, (b) if a second charge is placed near, and outside, the original sphere, and (c) if a second charge is placed inside the gaussion surface? 40. Aspherical rubber balloon carries a charge that is uniformlydistributed over its surface.As the balloon is blown up, how does E vary for points (a) inside the balloon, (b) on the surface of the balloon, and (c) outside the balloon? THINKINGPROBLEMS SOLUTION 1. Because protons are tightlybound in the nucleus.They cnnot be removed fromthere easily. 2. Conductors arematerials inwhichthere are a fewfree electrons per atomof thematter. Insulators arematerials inwhich the electrons are not as free as in conductors. 3. Acharged rod first attracts the dust by producing unlike charge at the near end and like charge at the far end. When the cork dust touches the rod, however, it acquires like charge and so is repelled strongly bythe charge on the rod. 4. To conduct away the charge produced by friction. This charge, unless conducte
d away to the earth, may produce sparks causing the explosives in the truck to catch fire and explode. 5. The force on the charge due to another does not depend on the presence of other charges, that is, electrical forces are physically independent. 6. No, theman acts as a conducting body. He shares chargewith the charged body. 7. Because if theycross, the electric field at the point ofintersectionwillhave two directions simultaneously,which is physicallyimpossible. 8. It is true. The electric field inside a hollowcharged conductor is zero. This follows fromthe conditionthat all points on conductor have the same potential.
ELECTROSTATICS www.physicsashok.in 45 9. It is false. The field inside is zero but on the surface, the field is finite. 10. It is true. 11. This is so because there is an electric field overAbut there is no electric field over B as it lies inside a hollow conductor.No, this is not in violationof the third lawofmotion. In fact, force arises betweenAand the hollow sphere, the charge B is simplyan internal part of the sphere. 12. Yes, by repeating the induction process. Place an insulated conductor B close to the charged body.A. yinduction, an equal unlike charge is induced at the near end ofBand equal like charge at the far end.Touch the body B with your finger tip in the presence ofA.The like charge at the far end will flowto the earth. But the unlike charge at the near endwill remain bound due to the force of attraction. Nowremove the bodyB to a distant place. The unlike charge, equal inmagnitude to the inducing charge,will spread over the surface ofB as there is no force of attraction to keep it on one side. Next, deliver this charge to a big body by conduction, that is, by touching this bodywithB. Repeat the process after completely dischargingB. To obtain a similar charge in largemagnitude, first deliver unlike charge to a bodyC and thenrepeat the induction process betweenC and the body inwhich like charge is to be stored. 13. No. The free chargewillgo to the earth via the finger, irrespective ofwhich part ofB is touched, because the free positive charge inB exists at a higher potential due to the positive charge onA. It is a fact that positive charge always flows fromhigher to lower potential, irrespective ofthe path. 14. E 9 109 1 9 1013 NC 1 0.01 . . . . . .The field is everylarge.Air cannot sustain sucha strong electric field, so a small bodycannot have a charge of 1 coulomb. 15. The housing and the rod connected togetherwillhave the same potential, so the leaveswillnot be deflected. After thewire is removed and the rod is charged, both the leaves will be deflected because of the potential difference between the rod and the housing. 16. As the charge approaches the plate, electrostatic induction causes the induced positive charge to pass into the earth,while the induced negative charge accumulates on the upper surface of the plate.Apositive current pulse passes through the galvanometer. No current is produced when the chargemoves above the plate.An opposite current is producedwhen the chargemoves away from the plate. 17. It is false. thework done in carrying a point charge fromone point to another does not depend on the path alongwhich it is carried because electric field is conservative. 18. It is greatest at the pointB. Since the electric field is the rate of fallof potential, the stronger the field, the closer the equipotentialsurfaces. In the figure, the equipotential surfaces are closest in the neighbourhood ofB, so the field at Bis the greatest. 19. Theycan, if the charge of one ball ismuch greater than that of the other.The forces of attraction caused by the induced chargesmay exceed the forces of repulsion. 20. When the housing of the electrometer is given some charge, the potentialof the rod cannot be the same as that
of the interior of the housing because the rod lies partlyoutside.On account of thepotentialdifference between the rod and the housing, the leaveswill be deflected. When the rod is earthed, the potentialdifference between the rod and the housing is increased further, so the deflectionwillbe greater. 21. The electrometermeasures the potentialdiference between the bodyand the housing of the electrometer. Since the surface of the bucket is equipotential, the leaves showthe same deflectionwherever the testing ballBmaytouch the surface of the bucket. But when the ballB is disconnected fromA, the ballBcollects charge by conduction.When it delivers this charge toA, the potentialof the leaves (and also of the rod) is raised above the potential of the housing, so there is deflection of the leaves.When B touches the inner
ELECTROSTATICS www.physicsashok.in 46 surface, it collects no charge because there is no potential difference between it and the inner surface of the bucket. Unless there is a potentialdifference, there can be no flowof charge fromone body to another. So no deflection takes placewhen the ballB touches the inner surface and then touchesA. But when it touches the outer surface, it collects a certain amount of charge, so there is deflection. 22. No, the field inside a cubical shellofmatter is not zero. The analozyis not complete in respect of distribution of mass and charge.Mass is uniformlydistributed but charge is not distributed uniformly. It ismore concentrated at the edges and corners. 23. No, two equipotentialsurfaces can never intersect because if they intersect at a point, the electric field at that point can have two directions simultaneously,whichis physicallyimpossible. 24. For the sake of simplicity, let us consider themetal bodyto be a sphere placed at the centre of the shell. VA (potentialof the surface of the shell) 0 1 Q q 4 R . . . .. where R= radius of the shell and q is the charge of themetal body. VB (potential of themetal body) 0 1 q Q 4 r R .. . .. . . . . . .... . . B A 0 V V q 1 1 4 r R . ... . .. . .. .... . . B A V . V So chargewillflowfromthemetalbodyto the shell. SinceVB =VA at equilibrium, qmust be reduced to zero, i.e., the entire charge on themetalbodymust flowto the outer shell. V, the common potential 0 q Q 4 R . . .. . Thus, (a) when q <Q,Vis negative, (b) when q =Q,V= 0, (c) when q > Q, Vis positive. 25. The field inside themetal sphere at all points is zero because r . is infinityformetals and 0 r E , E . . So E = 0. No, the answer does not change if the sphere carries a charge.
26. Charge is not distributed uniformlyover the surface ofa conductor ofanyshape. Charge is distributed according to the curvature of the surface.The greater the curvature, the greater the surface density of charge. 27. No. . . 2 1 1 2 0 1 2 0 1 2 q 1 1 q r r V V 4 r r 4 r r .. .. . . . .. . ... .. .. .. .. or . . 0 1 1 1 q r 4 r r r . . .. . . where 2 1 .r . r .r Obviously, . r is not constant. So the equipotential surfaces differing by 1Vare not equispaced. 28. The lines of force are shown in the figure. Here, all the lines near the edge of the figurewill appear to radiate uniformlyfromthe point P, the centre ofgravity of the charges.At the neutral point N, the total field is zero. Let it be at a distance x from4Q. Then . .2 2 0 0 1 4Q 1 Q 4 x 4 12 x . .. .. . or x = 8 cm 29. The lines of force are shown in the figure. Explanation Since it is ametalplate, its surface is an equipotential surface. So lines of forcemust terminate normallyon it. Note : One everyimportant result follows fromthismap of electric flued due to a point charge and the induced charges on ametalplate. If a charge �Qis placed as far behind the plate as +Q
ELECTROSTATICS www.physicsashok.in 47 is infront and the plate is removed,AB(positionof the plate) is still an equipotential surface. So for interaction between +Qand the plate or for electric fields between them,wemayuse the formalismof replacing the plate by �Q. This is called the electrical image of +Q. 30. The distributionof charge is shownin the figure. Explanation Since the surface density of charge is proportional to curvature and corners have the greatest curvature and a plane surface has zero curvature, dots are equidistant fromthe straight portion and far away fromthe corners. 31. Since the plate is accelerated, the electrons inside are also acceleratedwith the same acceleration.Aforcemay be applied on an electron, onlywhen an electric field is created. So theremust develop charges on the faces that are perpendicular to the direction ofmotion, so that there develops an electric field inside themetal plate. The front facewill be charged positively, and the rear one negatively.The intensityof the electric field is given by eE = ma. Since E = ./.0, the surface density of charge developed is given by . = .0E = .0ma/e 32. No. The concentration of lines at the botomindicates a field which is stronger at the botoomthan on top. Imagine a rectangular pathwith its two sides perpendicular to the electric lines.Nowmove a charge along this path. Some networkwillbe performed.But inanelectric field,work done is essentiallyzero as it is a conservative field. Hence, an electric field of the type represented is physically impossible. 33. False,Work is always done byanexternal agent when a charged particle is transferred fromone equipotential surface to another. Thework done per unit charge = potential difference. 34. There is an electric field between the filament and the ring.As the electron is emitted fromthe filament, it experiences a force towards the ring and so it is accelerated. It electricalpotential energy,which it acquires on emerging fromthe filament, is converted into its kinetic energy. Thus, though the batterydoes not supply any energy, the electron can gain velocityat the cost of its ownpotential energy. 35. True. There is an electric field inside a current-carrying conductor.This is equal to V/l, whereVis the voltage across the conductor and l is the length of the conductor. Now, consider a closed path abcda as shown in the figure.Work done byan external agent along abcda is zero. . inside outside inside E .ab.0.E .cd.0.0 . E . Eoutside Hence, there is a field near the surface of the conductor. 36. To conduct awayelectricityproduced byfriction. 37. Place the smaller one inside the bigger one.As potential inside a hollowconductor is the same as that of its surface, the potential inside the hollow conductor is 100V. The potential of the smaller conductor is now 100+50 =150V. Connecte the two conductors byawire. Chargewill flowfromthe smaller one to the bigger one as the smaller one is at 150Vand the bigger one at 100V. 38. Yes, there is a net force on the dipole given by F p E .
. .
.l 39. Total flux E 0 . . E ..s. 1 q . . .. . , byGauss�s law. (a) No, the fluxwill not change as it depends onlyon the total charge inside the surface and not on the extent and shape of the surface. (b)No, it remains the same as the total fluxis determined solely bythe charge inside the surface and not the charge outside. (c) Yes, flux will change as the total charge inside the surface has changed. 40. (a) For points inside the balloon,E = 0, (b)E decreases as the surface densityof charge decreases . . 0 E ../ . , (c) E remains constant because 2 0 E . q / 4.. r , where r is the distance of the point fromthe centre of the balloon.
ELECTROSTATICS www.physicsashok.in 48 Reasioning Type Questions THE NEXT QUESTIONS REFER TO THE FOLLOWING INSTRUCTIONS A statement of assertion (A) is given and a Corresponding statement of reason (R) is given just below it of the statements, mark the correct answer as � (A) If both A and R are true and R is the correct explanation of A. (B) If both A and R are true but R is not the correct explanation of A. (C) If A is true but R is false. (D) If both A and R are false. (E) If A is false but R is true. 1. Assertion (A) : If there exists colomb attraction between two bodies, both of them may not be charged. Reason (R) : Due to induction effects a charged body can attract a neutral body. 2. Assertion (A) : A small metal ball is suspended in a uniform electric field with an insulated thread. If high energy X-ray beam falls on the ball, the ball will be deflected in the electric field. Reason (R) : X-rays emit photoelectrons and metal becomes negatively charged. 3. Assertion (A) : Electric current will not flow between two charged bodies when connected if their charges are same. Reason (R) : Current is rate of flow of charge. 4. Assertion (A) : The surface densities of two spherical conductors of radii r1 and r2 are equal. Then the electric field intensities near their surfaces are also equal. Reason (R) : Surface charged density = charge/area. 5. Assertion (A) : When charges are shared between two bodies, there occurs no loss sof charge, but there does occur a loss of energy. Reason (R) : In case of sharing of charges, the energy of conservation fails. 6. Assertion (A) : Two adjacent conductors, carrying the same positive charge have a potential difference between them. Reason (R) : The potential of a conductor depends upon the charge given to it. 7. Assertion (A) : Dielectric breakdown occurs under the influence of an intense light beam. Reason (R) : Electromagnetic radiations exert pressure. 8. Assertion (A) : The tyres of aircrafts are slightly conducting. Reason (R) : If a conductor is connected to ground, the extra charge induced on conductor will flow to ground. 9. Assertion (A) : Metallic shield in the form of a hollow shell, can be built to block an electric field. Reason (R) : In a hollow spherical shell, the electric field inside it is zero at every point. 10. Assertion (A) : A bird perches on a high power line and nothing happens to the bird. Reason (R) : The level of bird is very high from the ground. 11. Ametal sphere is suspended from a nylon thread. Initially, the metal sphere is uncharged . When a positively charged glass rod is brought close to the metal sphere, the sphere is drawn toward the rod. But if the sphere touches the rod, it suddenly flies away from the rod. Explain why the sphere is first attracted, then repelled. Level # 1. Objective Type Question 1. Two copper spheres of same radii one hollow and other solid are charged to th
e same potential then (A) both will hold same charge (B) solid will hold more charge (C) hollow will hold more charge (D) hollow cannot be charged 2. Through the exact centre of a hydrogenmolecule, an . -particle passes rapidly, moving on a line perpendicular to the internuclear axis. The distance between the two hydrogen nuclei is b The maximum force experienced by the . -particle is (A) 2 2 0 4 3 3.. e b (B) 2 2 0 8 3..e b (C) 2 2 0 8 3 3. . e b (D) 2 2 0 4 3..e b 3. A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at x = +1 cm and C be the point on the y-axis at y = +1 cm. Then the potential at the points A, B and C satisfy (A) VA > VB (B) VA < VC (C) VA < VB (D) VA > VC
ELECTROSTATICS www.physicsashok.in 49 4. On an equilateral triangle of side 1 m there are three point charges placed at its corners of 1C, 2C and 3C. The work required to move these charges to the corners of a smaller equilateral triangle of side 0.5 m is (A) 19.8 x 1010 J (B) 39.6 x 1010 J (C) 9.9 x 1010 J (D) 4.45 x 1010 J 5. The potential field depends on x and y coordinates as V = (x2 � y2). Correspondence electric field lines in x.y plane as shown in figure. (A) (B) (C) (D) 6. Two circular rings A and B, each of radius a = 130 cm are placed coaxially with their axes horizontal in a uniform electric field E = 105 NC�1 directed vertically upwards as shown in Figure. Distance between centers of these rings A and B is h = 40 cm, ring A has a positive charge q1 .10.C while ring B has a negative charge of magnitude 2 q . 20.C . A particle of mass m = 100 gm and carrying a positive charge q . 10.C is released from rest at the centre of the ring A. Calculate its velocity when it has moved a distance of 40 cm. (A) v . 6 2ms.1 (B) v . 4 2ms.1 (C) v . 7ms.1 (D) v . 32ms.1 7. Three identical spheres each having a charge q and radius R, are kept in such a way that each touches the other two. the magnitude of the electric force on any sphere due to other two is (A) . . 2 0 1 3 4 4 q.. R . . .. .. (B) . . 2 0 1 2 4 4 q.. R . . .. .. (C) . . 2 0 1 2 4 4 R.. q . . . . . . (D) . . 2 0 1 3 2 4
q.. R . . .. .. 8. A system consists of a uniformly charged sphere of radius R and a surrounding medium filled by a charge with the volume density . .. r , where . is a positive constant and r is the distance from the centre of the sphere. Find the charge of the sphere for which the electric field intensity E outside is independent of r. (A) 0 . 2 . (B) 0 2 . . (C) 2.. R2 (D) None of the above 9. Three charges each of +q, are placed at the vertices of an equilateral triangle. The charge needed at the centre of the triangle for the charges to be in equilibrium is (A) 3.q (B) . 3q (C) 3q (D) . 3q 10. Hollow spherical conductor with a charge of 500.C is acted upon by a force 562.5 N. What is electric intensity at its surface? (A) zero (B) 1.125 x 106 N/C (C) 2.25 x 106 N/C (D) 4.5 x 106 N/C 11. A hemisphere of radius R is charged uniformly with surface density of charge . What will be the potential at centre? (A) 0 2 . R . (B) 0 4.. (C) 0 2.. (D) 0 43 . R .
ELECTROSTATICS www.physicsashok.in 50 12. A circular cavity is made in a conductor. A positive charge q is placed A B at the centre (A) The electric field at A and B are equal (B) The electric charge density at A = the electric charge density at B (C) Potential at A and B are equal (D) All the above. 13. An isolated metallic object is charged in vacuum to potential v0, its electrostatic energy being W0. It is then disconnected from the source of potential, its charge being left unchanged and is immersed in a large volume of dielectric, with dielectric constant k. The electrostatic energy will be. (A) kW0 (B) 0 Wk (C) 0 2Wk (D) W0. 14. A charge +q is fixed at each of the points x = x0, x = 3x0, x = 5x0 .... ad inf. on the x-axis, and a charge � q is fixed at each of the points x = 2x0, x = 4x0, x = 6x0 .... ad inf. Here x0 is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be Q/( 4...r ). Then, the potential at the origin due to the above system of charges is. (A) Zero (B) 8 x log 2 q ..0 0 e (C) . (D) 0 0 e4 xq log 2 .. 15. Apositively charged thinmetal ring of radius R is fixed in the xy-plane with its centre at the originO. Anegatively charged particle P is released from rest at the point (0, 0, z0) where z0 > 0. Then the motion of P is. (A) Periodic, for all values of z0 satisfying 0 < z0 < . (B) Simple harmonic, for all values of z0 satisfying 0 < z0 . R (C) Such that P crosses O and continues to move along the negative z-axis towards z = � . (D) None of these 16. For an infinite line of charge having charge density . lying along x-axis, the work done in moving charge from C to A arc CA is. (A) log 2 2q e ..0 . (B) log 2 4q e ..0 .
(C) log 2 4q e ..0 . (D) 2log 1 2q e ..0 . 17. A particle A has charge +q and particle B has charge + 4q with each of them having the same mass m. When allowed to fall from rest through same electrical potential difference, the ratio of their speeds vA : vB will be (A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4 18. Three point charges q1, q2 and q3 are taken such that when q1 and q2 are placed close together to form a single point charge, the force on q3 at distance L from this combination is a repulsion of 2 units inmagnitude. when q2 and q3 are so combined the force on q1 at distance L is an attractive force of magnitude 4 units. Also q3 and q1 when combined exert an attractive force on q2 of magnitude 18 units at same distance L. The algebraic ratio of charges q1, q2 and q3 is. (A) 1 : 2 : 3 (B) 2 : � 3 : 4 (C) 4 : � 3 : 1 (D) 4 : � 3 : 2 19. An electric potential is given by V = k(xy), where k is a constant. A particle of charge q0 is first taken from (0, 0) to (0, a) to (a, a), then directly from (0, 0) to (a, a) and lastly from (0, 0) to (a, 0) to (a, a). If W1, W2 and W3 be the work done for the individual paths respectively then (A) W1 = W2 = W3 = � q0ka2 (B) W1 = W3 = � 2q0ka2 (C) W1 = W3 > W2 (D) W1 > W2 > W3
ELECTROSTATICS www.physicsashok.in 51 20. Four charges 2C, � 3C, � 4C and 5C respectively are placed at all the corners of a square. Which of the following statements is true for the point of intersection of the diagonals ? (A) Electric field is zero but electric potential is non-zero (B) Electric field is non-zero but electric potential is zero (C) Both electric field and electric potential are zero (D) Neither electric field nor electric potential is zero 21. Two metallic identical spheres A and B carrying equal positive charge + q are a certain distance apart. The force of repulsion between them is F. A third uncharged sphere of the same size is brought in contact with sphere. A and removed. It is then brought in contact with sphere B and removed. What is the new force of repulsion between A and B ? (A) F (B) 8 3F (C) 2F (D) 4F 22. An electron of mass me, initially at rest, moves through a certain distance in a uniform electric field in time t1. A proton of mass mp, also initially at rest, takes time t2 to move through an equal distance in this uniform electric field. Neglect the effect of gravity, the ratio t2 /t1 is nearly equal to (A) 1 (B) (mp /me)1/2 (C) (me /mp)1/2 (D) 1836 23. Eight dipoles of charges of magnitude are placed inside a cube. The total electric flux coming out of the cube will be (A) 08e . (B) 016 e . (C) 0 e . (D) Zero 24. A particle of mass m and charge q is released from rest in a uniform electric field E. The kinetic energy attained by the particle after moving a distance x is. (A) qEx2 (B) qEx2 (C) qEx (D) q2Ex 25. Two point charges +q and �q are held fixed at (�d, 0) and (d, 0) respectively of a x-y coordinate system. Then (A) The electric field E at all points on the x-axis has the same direction. (B) Work has to be done in bringing a test charge from . to the origin [1995] (C) Electric field at all points on y-axis is along x-axis. (D) The dipole moment is 2qd along the x-axis. 26. Three charges Q, +q and +q are placed at the vertices of a ring-angled isosceles triangle as shown. The net electrostatic a Q +q +q energy of the configuration is zero if Q is equal to [2000] (A) 1 2
.q
. (B) 2 2 2 . q . (C) �2q (D) +q 27. Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketches as in [2001] (A) (B) (C) (D) 28. Two equal point charges are fixed at x = -a and x = +a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to [2002] (A) x (B) x2 (D) x3 (D) 1 x
ELECTROSTATICS www.physicsashok.in 52 29. A metallic shell has a point charge �q� kept inside its cavity. Which one of the following diagrams correctly represents the electric lines of forces? [2003] (A) (B) (C) (D) 30. Six charges of equal magnitude, 3 positive and 3 negative are to be placed on PQRSTU corners of a regular hexagon, such that field at the centre is double that of what is would have been if only P Q R T S U O one +ve charge is placed at R. [2004] (A) +, +, +, �, �, � (B) �, +, +, +, �, � (C) �, +, +, �, +, � (D) +, �, +, �, +, � 31. A Gaussian surface in the figure is shown by dotted lime. q1 �q1 q2 The electric field on the surface will be [2004] (A) due to q1 and q2 only (B) due to q2 only (C) Zero (D) due to all 32. Three infinitely long charge sheets are placed as shown in figure. The electric field at point P is [2005] (A) 0 2 k� . . (B) 0 4 k� . . Z Z=a Z=�a Z=�2a P x �2. �.. (C) 0 2 k� . . . (D) 0 4 k� . . . 33. Two equal negative charges �q are fixed at points (0, �a) on y-axis. A positive charge Q is released from rest at the point (2a, 0) on the x-axis. The charge Q will [1984] (A) execute simple harmonic motion about the origin. (B) move to the origin remain at rest
(C) move to infinity (D) execute oscillatory but not simple harmonic motion 34. A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to: [1987] (A) 2. Q (B) 4. Q (C) 4. Q (D) 2. Q 35. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of �3Q, the new potential difference between the same two surfaces is: [1989] (A) V (B) 2V (C) 4V (D) �2V 36. A metallic solid sphere is placed in a uniform electric field. The lines of force follow the path(s) shown in Figure as [1996] (A) a (B) 2 (C) 3 (D) 4
ELECTROSTATICS www.physicsashok.in 53 37. A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the electric field due to the sphere at a distance r from its center [1998] (A) increases as r increases, for r < R (B) decreases as r increases, for 0 < r < . (C) is discontinuous at r = R. (D) None of these 38. An ellipsoidal cavity is carved within a perfect conductor Figure. a positive charge q is placed at the centre of the cavity. The points A A B q and B are on the cavity surface as shown in the figure. Then [1999] (A) electric field near A in the cavity = electric field near B in the cavity (B) charge density at A = charge density at B (C) potential at A = potential at B (D) total electric field flux through the surface of the cavity is 0 q . . More Than One Choice Questions: 39. Three concentric spherical metallic shells A, B and C of radii a, b and c (a < b < c) have charge densities of . , .. , and . respectively. The potentials of A, B and C are: (a) . . 0 1 A V a b c . . . . . (b) 2 0 1 B V a b c c . . . . . . . . . . . (c) 2 2 0 1 C V a b c c c . . . . . . . . . . . (d) . . 0 1 A B C V V V a b c . . . . . . . 40. When positively charged spheer is brought near a metallic sphere, it is observed that a force of attraction exists between two. It means:
(a) metallic sphere is necessarily negatively charged. (b) metallic sphere may be electrically neutral. (c) metallic sphere may be negatively charged. (d) mothing can be said about charge of metallic sphere. 41. A conducting sphere of radius R has a charge. Then: (a) the charge is uniformly distributed over its surface, if there is no external electric field. (b) distribution of charge over its surface will be non-uniform, if an external electric field exists in the space. (c) electric field strength inside the sphere will be equal to zero only when no exxternal electric field exists. (d) potential at every point of the sphere must be same. 42. A small sphere of amss m and having charge q is suspended by a light thread, then: (a) tension in the thread may reduce to zero if anohter charged sphere is placed vertically below it. (b) tension in the thread may increase to twice of its original value if another charged sphere is placed vertically below it. (c) tension in the thread is greater than mg if another charged sphere is held in the same horizontal line in which first sphere stays in equilibrium. 43. Two point charge: Q and �Q/4 are separated by a distance x. Then: (a) potential is zero at a point on the axis which is x/3 on the right side of the charge �Q/4. (b) potential is zero at a point on the axis which is x/5 on the left side of the charge �Q/4. (c) electric field is zero at a point on the axis which is at a distance x on the right side of the charge �Q/4. (d) there exist two points on the axis, where electric field is zero. 44. Two equal and oppositely charged particles are kept some distance apart from each other. A spherical surface having radius equal to separation between the particles and concentric with their midpoint is considered. Then: (a) electric field is normal to the surface at two points only. (b) electric field is zero at no point. (c) electric potential is zero at every point of one circle only. (d) net electric flux through the surafce is zero.
ELECTROSTATICS www.physicsashok.in 54 45. Two identical charges +Q are kept fixed some distane apart. A small particle P with charge q is placed midway between them. If P is given a small displacement . , it will undergo simple harmonic motion if: (a) q is positive and . is along the line joining the charges. (b) q is positive and . is perpendicular to the line joining the charges. (c) q is negative and . is perpendicular to the line joining the charges. (d) q is positive and . is along the line joining the charges. 46. Select the correct statement(s): (a) Charge cannot exist without mass but mass can exist without charge. (b) Charge is conserved but mass is not. (c) Charge is independent of state of rest or motion. (d) None of these 47. Which of the following quantities do not depend on the choice of zero potential or zero potential energy? (a) Potential at a point. (b) Potential difference between two points. (c) Potential energy of a system of two charges. (d) Change in potential energy of system of two charges. Fill in the blanks: 1. Figure shows line of constant potential in a region in which an electric field is present. The value of the potential are written in brackets. Of the points A, B and C, the magnitude of the electric A B C (50 V) (40 V) (30 V) (20 V) (10 V) field is greatest at the point ............. [1984] 2. Two small balls having equal positive charges Q (coulomb0 on each are suspended by two insulating strings of equal length L (metre) from a hook fixed to a stand. The whole set up is taken in a satellite into space where there is no gravity (state of weightlessness). The angle between the two string is ............ and the tension in each string is ......... newtons. [1988] 3. A point charge q moves from point P to point S along the path PQRS (figure) in a uniform electric field E pointing parallel to the positive direction of the X-axis. The coordinates of the points X Y P Q R S E P, Q, R and S are (a, b, O), (2a, O, O) (a, �b, O) and (O, O, O) respectively. The work done by the field in the above process is given by the expression ..................... [1989] 4. The electric potential V at any point x, y, z (all in metres) in space is given by V = 4x2 volts. The electric field at the point (1m, 0.2 m) is ............... V/m. [1992] 5. Five point charges, each of value +q coul, are placed on five vertices of a regular hexagon of side L metres. the magnitude of the force on the point charge of �q coul. Placed at he centre of the hexagon 1
2 3 5 4 q q q q q �qL is .......... newton. [1992] 6 True / False : 6. The work done in carrying a point charge from one point to another in an electrostatic field depends on the path along which the point charge is carried. [1981] 7. Two identical metallic spheres of exactly equal masses are taken. One is given a positive charge Q coulombs and the other an equal negative charge. Their masses after charging are different. [1983] 8. A small metal ball is suspended in a uniform electric field with the help of an insulated thread. If high energy X-ray beam falls on the ball, the ball will be deflected in the direction of the field. [1983]
ELECTROSTATICS www.physicsashok.in 55 9. A ring of radius R carries a uniformly distributed charge +Q. A point charge �q is placed in the axis of the ring at a distance 2R from the center of the ring and released from rest. The particle executes a simple harmonic motion along the axis of the ring. [1988] Table Match 10. Match List I and List II and select the correct answer using the codes given below the lists: The electric fields due to various charge distribution are (q is the total charge on the body, . is the surface charge density, . is the linear charge density) List I List II I. At a distance x from the centre of a A. 2 2 0 1 2 E x x R . . . . . . . . . . . uniformly charged ring of radius R. The point is on the line passing through the centre of the ring and perpendicular to the plane of the ring II. At a distance x from the centre of a uniformly B. 0 2 E x . . . . charged ring of radius R. The point is on the line passing through the centre of the ring and perpendicular to the plane of the disc. III. At a distance x from an infinite sheet of C. 0 2 E . . . uniform distribution of charge IV. At a distance x from an infinite line of charge D. 2 2 3 2 0 1 4 E qx . R x . . . . . . . . (A) I-A, II-C, III-B, IV-D (B) I-D, II-A, III-C, IV-B (C) I-C, II-B, III-D, IV-A (D) I-B, II-D, III-A, IV-C Passage Type Questions THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE E is the electric field created by q1. V is the voltage at a given point in the field E. Assume that the
electric field created by q2 is negligible compared to E. k is Coulomb�s Law constant. m1 si the mass of q1, and m2 is the mass of q2. 1. E can best be described as � (A) constant (B) decreasing as r increases (C) increasing as r increases (D) increasing as q2 increases 2. What is the work done on q2 when it is moved at constant velocity along the distance d? (A) Zero (B) 1 Vq (C) 2 Vq (D) 2 Edq 3. Which of the following represents the work done on q2 when moved from its present position to a distance r from q1? (A) 1 2 12 kq q r (B) 1 2 kq q r (C) 1 2 2kq q r (D) 2 1 2 kq q r 4. If q1 and q2 have opposite charges, then when q2 is moved from its present location to a distance r from q1, the force on q2 due to q1 � (A) decreases by a factor of 4 (B) remains the same (C) increases by a factor of 2 (D) increases by a factor of 4 d r 2r q1 q2
ELECTROSTATICS www.physicsashok.in 56 5. If q1 is positive then when q2 is moved from its present location to a distance r from q1, the magnitude of the voltage experienced by q2 due to E� (A) decreases by a factor of 4 (B) remains the same (C) increases by a factor of 2 (D) increases by a factor of 4 6. The strength of the electric field E at r is � (A) half the field strength at 2r (B) the same as the field strength at 2r (C) twice the field strength at 2r (D) four times the field strength at 2r 7. If q1 and q2 are both positively charged and q2 is released, what is the maximum velocity that can be achieved by q2? (A) 1 2 2 kq q m r (B) 1 2 2 2 kq q m r (C) 1 2 2 2 2kq q m r (D) 1 2 2 2kq q m r 8. Electrostatic precipitators use electric forces to remove pollutant particles from smoke, in particular in the smokestacks of coal-burning power plants. One form of precipitator consists of a vertical hollowmetal cylinder with a thin wire, insulated from the cylinder, running along its axis. A large potential difference is established between the wire and the outer cylinder, with the wire at lower potential. This sets up a strong radial electric field directed inward, producing a region of ionized air near the wire. Smoke enters the precipitator at the bottom, ash and dust in the smoke pick up electrons, and the charged pollutants are accelerated toward the outer cylinder wall by the electric field. Suppose the radius of the central wire is 80.0 .m, the radius of the cylinder is 12.0 cm, and the potential difference between the wire and the cylinder is 60.0 kV. Assume that the wire and cylinder are both very long in comparison to the cylinder radius. (a) What is the electric-field magnitude midway between the wire and the cylinder wall? (b) What magnitude of charge must a 30.0- .g ash particle have if the electric field computed in part (a) is to exert a force ten times the particle�s weight? 9. A Geiger counter detects radiation such as alpha particles by using the fact that the radiation ionizes the air along its path. The device consists of a thin wire on the axis of a hollow metal cylinder and insulated from it.
A large potential difference is established between the wire and the outer cylinder, with the wire at higher potential; this sets up a strong radial electric field directed outward. When ionizing radiation enters the device, it ionizes a few air molecules. The free electrons produced ar accelerated by the electric field toward the wire and, on the way there, ionize many more air molecules. Thus a current pulse is produced that can be detected by appropriate electronic circuitry and converted to an audiable �click.� Suppose the radius of the central wire is 50.0.m and the radius of the hollow cylinder is 2.00 cm. What potential difference between the wire and the cylinder is required to produce an electric field of 6.00 x 104 N/C at a distance of 1.50 cm from the wire? (Assume that the wire and cylinder are both very long in comparison to the cylinder radius.
ELECTROSTATICS www.physicsashok.in 57 Level # 2 1. As shown a solid spherical region having a spherical cavity whose diameter R is equal to the radius of spherical region P C Q R that has a total charge Q. Find the potential at a point P, which is at a distance �x� from C. 2. A spherical water drop of radius a has a charge Q spread uniformly over its face. The drop is split into two identical spherical droplets (each of which has charge Q/2 spread uniformly over its face), which are kept very far from one another. (a) Compute the change in the electrostatic potential energy caused by the splitting. (b) Repeat the above calculation for the case in which the charge is uniformly distributed in the drop volume, before and after the splitting. 3. An electric dipole is placed at distance x from centre O on the axis of a charged ring of radius R and charge Q uniformly distributed over it. (a) Find the net force acting on the dipole. (b) What is the work done in rotating the dipole through 180°. (c) If the dipole is slightly rotated about its equilibrium position, find the time period of oscillation. Assume that the dipole is linearly restrained. 4. Particle 1 located far from particle 2 and possessing the kinetic energy T0 and mass m1 strikes particle 2 of mass m2 through the aiming parameter ., the arm of the momentum vector relative to particle 2 as; in the figure. Each particle carries a charge +q. 1 2 r P P0 . Find the smallest distance between the particles when m1 < < m2. 5. An electric field line emerges from a positive point charge +q1 at an angle . to the straight line connecting it to a negative point charge q2. At what angle . will the field + . . enter the charge � q q1 q2 2 ? 6. Two point charges each carrying a positive charge of 5e (e being the magnitude of the electronic charge) are separated by a distance 2d. An electron describes a circular path due to the attraction of the charges in a plane bise- A B d 5e 5e O d P R cting perpendicularly the line joining the two point charges. If the radius of the circular path described by the electron is R, determine the orbital speed of the electron. 7. Five thousand lines of forces enter a certain volume of space, and three thousand lines emerge from it.
What is the total charge in coulombs with in the volume ? 8. Determine the strength E of the electric field at the centre of a hemisphere produced by charges uniformly distributed with a density . over the surface of this hemisphere. 9. Two small identical balls lying on a horizontal plane are connected by a weight-less-spring. One ball (ball 2) is fixed and the other (ball 1 is free. The balls are charged identically as a result of which the spring length + � 12 increases . = 2 times. Determine the change in frequency. O 10. A semi-circular ring of mass M and radius R with linear charge density . hinged at its centre is placed in a uniform electric field as shown in the figure. (i) Find the net force acting on the ring. (ii) It the ring is slightly rotated about O and released, find its time period of oscillation (iii) What is the work done by an external agency to rotate it through an angle ..
ELECTROSTATICS www.physicsashok.in 58 11. Determine the force F of interaction between two hemispheres of radius R touching each other along the equator if one hemisphere is uniformly charged with a surface density, . 1 and the other with a surface density . 2. 12. Suppose in an insulating medium, having di-electric constant k = 1, volume density of positive charge varies with y-coordinate to law . = ay. A particle of mass m having positive charge q is placed in the medium at point A(0, y0) and projected with velocity v = v0 . i and assuming o x y v A 0 electric field strength to be zero at y = 0, calculate slope of trajectory of the particle as a function of y. 13. A metal ball of radius r and density . is charged by direct contact from the Earth�s surface till it acquires its maximum value. What should be this charge Qmax on the ball and the charge Q on the Earth (assuming it to be uniform sphere of mass m and radius R) such that it may be launched from the Earth�s surface with zero launch velocity. 14. A particle of mass m having negative charge q moves along an ellipse around a fixed positive charge Q so that its maximum and minimum distances from fixed charge are equal to r1 and r2 respectively. Calculate angular momentum L of this particle. 15. Three concentric, conducting spherical shells A, B and C have radii a = 10 cm, b = 20 cm and c = 30 cm respectively. The innermost shell A is earthed and charge q2 = 4 .C and q3 = 3 .C are given to shells B and C respectively. Calculate charge q1 induced on shell A and energy stored in the system. 16. The system consists of a hemispherical dielectric with volume charge density � . �. Find the potential difference between points A and B. . BA 17. A space is filled up with volume density of charge r 3 0 e . . . . . where .0 and . are positive constants, r is the distance from centre of the system. Find the magnitude of the electric field strength vector as a function of r. 18. Two identical balls are suspended from the same point by two threads. The balls are given equal charges and immersed in kerosene. Determine the density of the material of the balls if the threads deflect equally in vacuum and kerosene. The density of kerosene .0 = 0.8 g/cm3 and its relative permittivity .0 = 2. 19. A charge �q� is placed on the surface of an originally uncharged soap bubble of radius R0. Due to the mutual repulsion of the charged surface, the radius is increased to a somewhat large value R. Show that q = 1/ 2 20 0 2 0 0
2. PR R(R RR R ) 3 32 ... ... . . . . in which P is the atmospheric pressure. Level # 3 1. Two fixed charges �2Q and Q are located at the points with coordinates (�3a, 0) and (+3a, 0) respectively in the x-y plane. (a) Show that all points in the x-y plane where the electric potential due to the two charges is zero, lie on a circle. Find its radius and the location of its centre. (b) Given the expression V(x) at a general point on the x-axis and sketch the function V(x) on the whole xaxis. (c) If a particle of charge +q starts form rest at the centre of the circle, show by a short quantitative argument that the particle eventually crosses the circle. Find its speed when it does so. [IIT 91] 2. A blank to be filled appears in each of the following statements. Write in your answer book the subquestion number and write down against it your answer corresponding to each blank. In your answer, the sequence of the sub-questions should be the same as given in the question paper. (i) If .0 and .0 are respectively, the electric permittivity and magnetic permeability of free space, . and . the corresponding quantities in a medium, the index of refraction of the medium in terms of the above parameters is .....
ELECTROSTATICS www.physicsashok.in 59 (ii) The electric potential V at any point x, y, z (all in meters) in space is given by V = 4x2 volts. The electric field at the point (1m, 0, 2m) is ...... V/m. (iii) Five point charges, each of value +qC, are placed on five vertices of a regular hexagon of side L metre. The magnitude q q q q q of the force on the point charge of value � qC placed at the centre of the hexagon is ..... newton. [IIT 92] 3. A circular ring of radius R with uniform positive charge density . per unit length is located in the y-z plane with its centre at the origin O. A particle of mass m and positive charge q is projected from the point P(R 3, 0, 0) on the positive x-axis directly towards O, with an initial speed v. Find the smallest (non-zero) value of the speed v such that the particle does not return to P. 4. Consider the classical-model of an atom such that a nucleus of charge +e is uniformly distributed within a sphere of radius 2 Å. An electron of charge (� e) at a radial distance 1 Å moves inside the sphere. Find the force attracting the electron to the centre of the sphere. Calculate the frequency with which the electron would oscillate about the centre of the sphere. [REE 95 5. A charge 10�9 coulomb is located at origin in free space and another charge Q at (2, 0, 0). If the x-component of the electric field at (3, 1, 1) is zero calculate the value of Q. Is the y-component zero at (3, 1, 1) ? [REE 95] 6. A radioactive source in the form of ametal sphere of radius 10�2m, emits beta particles at the rate of 5 × 1010 particles per s. The source is electrically insulated. How long will it take for its potential to be raised by 2 volt assuming that 40% of the emitted beta particles escape the source ? [REE 97] 7. A non-conducting disc of radius a and uniform positive surface charge density . is placed on the ground, with its axis vertical. A particle of mass m and positive charge q is dropped, along the axis of the disc, from a height H with zero initial velocity. The particle has q/m = 4.0g/.. [IIT 99] (a) Find the value of H if the particle just reaches the disc. (b) Sketch the potential energy of the particle as a function of its height and find its equilibrium position. 8. A particle of charge q and mass m moves rectilinearly under the action of an electric field E = A � Bx where B is a + ve constant and x is a distance from the point where the particle was initially at rest. Calculate (a) Distance travelled by the particle till it comes to rest and (b) Acceleration at that moment. 9. Four point charges + 8 .C, � 1 .C, � 1 .C and + 8 .C are fixed at the points 27 / 2m, . 3 / 2 m, . 3/ 2m and . 27 / 2 m respectively on the y-axis. A particle of mass 6 × 10�4 kg and of charge + 0.1 .C moves along the � x direction. Its speed at x = + . is v0. Find the least value of v0 for which the particle will cross the origin. Find also the kinetic energy of the particle at the origin. Assume that space is gravity free. Given
1/(4..0 ) = 9 × 109 Nm2/C2. [IIT 2000] 10. Eight point charges are placed at the corners of a cube of edge a as shown in figure. Find the work done in disassembling this system of charges. [JEE 2003]
ELECTROSTATICS www.physicsashok.in 60 Answer Key Assertion-Reason Que. 1 2 3 4 5 6 7 8 9 10 Ans. A C D B C B B A A C Level # 1 Q. 1 2 3 4 5 6 7 8 9 10 Ans. A C A C A A A C A B Q. 11 12 13 14 15 16 17 18 19 20 Ans. A A B D A A B D A B Q. 21 22 23 24 25 26 27 28 29 30 Ans. B B D C C B C B C C Q. 31 32 33 34 35 36 37 38 39 40 Ans. D C D B A D A C BC BC Q. 41 42 43 44 45 46 47 Ans. ABD ABC ABC ABCD AC ABC BD Fill in the blanks, True-False & Match the column 1. B 2. 180°, 22 0 1 4 4Q .. L . 3. �qEa 4. �8 5. 2 0 1 4 q q .. L. 6. F 7. T 8. T 9. F 10. B Que. 1 2 3 4 5 6 7 Ans. B A A D C D D Passage Type 8. (a) 1.37 x 105 volt/m (b) 2.15 x 10�11 CLevel # 2 1. . . .. . . .. . . .. . 0 2 R2 4x2 1 x2 7 V Q 2. (a) The fractional change = � ... ..
. . . 22 / 3 1 1 (b) The fractional change in this case is also the same,. 3. (a) 2 2 2 2 5/2 0 aqQ R 2x 2 (R x ) . . . . .. . . . . (b) 2 2 3/ 2 0 aqQx .. (R . x ) 4. 2 2 0 02 min (1 1 (2 T / q ) 2Tr q . . . . . . , 4 0 1.. . .
ELECTROSTATICS www.physicsashok.in 61 5. sin / 2 qq 2 sin 21 . . ... ... . 6. 2 2 3 / 2 2 0 (R d ) V Re 5.. . . 7. 17.7 nc 8. 4.0 . 9. n 3n . 2 times 10. (i) ..RE (ii) 2 N 2 E . . (iii) 4.ER2 11. 1 2 02 2F R . . . . . 12. (y y ) 3m v a q 303 200 . . 13. 2 0 max Q 4 r Rg 3 .. . . , 3 0Q 4 r gR 3 .. . . 14. 2 (r r ) mr r Qq0 1 2 1 2 .. . 15. q1 = � 3 .C, Energy = 9.45 J 16. 0 3
02 3 V R 2 2 R .. . . . . 17. . r3 . 0 2 0 1 e 3 r. . .. .. 18. 1.6 gm/cm3 Level # 3 1. (a) radius = 4a, centre dt (5a, 0) (b) .. . .. . . . .. . . | x a | 2 | x 3a | 1 4V Q0 (c) At x = 9a where V = 0, the charged particle eventually crosses the circle, 8 ma V 9q ..0 . 2. (i) .0.0 .. (ii) � 8 V/m (iii) 2 9 2 L 9 .10 q (N) 3. 2 m V q.0 . . 4. 9 × 1014 Hz 5. Q = � 4.27 × 10�10 C, 0 4 11 11 E 2 10 0 9 y . ..
.
. .
. 6. 700 .F 7. (a) H = 34a and H = 0 (b) 3 a .0 . 8. x = 0, x = B2A , a = n . qA 9. Vmin = 3 m/s, K.E at origin = 2.5 × 10�4 J 10. W = 5.824 . .. . . .. . .. a . q 4 1 2 0 �X�X�X�X�
CURRENT ELECTRICITY
DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125
ELECTRIC CURRENT www.physicsashok.in 1 THEORY OF ELECTRICCURRENT CURRENT ELECTRICITY The time rate of flow of charge through any cross-section is called current. Therefore, if through a crosssection, .q charge passes in time .t, the average electric current through that area av i qt . . . and instantaneous current t 0 i lim q dq . . t dt . . . . . Regarding electric current following points areworth noting : (a) Current is assumed to be a fundamentalquantityin physicswith unit ampere and dimension [A].TheCGS unit of current is emu of current and is called biot (Bi), i.e., 1A = 1C = (1/10)emu of charge s s 1A = 1 Bi 10 (b) The current is same for allcross-sections of a conductor of non-uniformcross-section. Similar towater flow, charge flows fasterwhere the conductor is smaller in cross-section and slowerwhere the conductor is larger in cross-section, so that charge rate remains unchanged. (c) Though conventionallya direction is associatedwith current (opposite to the motion of electrons) it is not a vector as the directionmerely represents the sense of charge flow and not a true direction. Further current does not obey i1 i2 i = i1 + i2 the lawof parallelogramof vectors, i.e., iftwo currents i1 and i2 reach a point we always have i = i1 + i2 whatever be the angle between i1 and i2. (d) By convention, the direction of current is taken to be that inwhich positive chargemoves and opposite to the direction of flowof negative charge. (e) As charge is conserved and current is the rate of flowof charge, the charge entering at one end per second of a conductor is equal to the charge leaving the other end per second. (f) Current indifferent situations is due to motion of different charge carries. Current inconductors and vacuum tubes isdue tomotion ofelectrons, inelectrolytes due tomotion ofbothpositive and negative ions, indischarge tube due tomotion of positive ions and negative electrons and in semiconductors due tomotion of electrons and holes. Example 1. The current in awire varieswith time according to the relation i (3.0A) 2.0 A t s . . . . .. .. (a) How many coulombs of charge pass a cross-section of the wire in the time int
erval between t = 0 and t = 4.0 s? (b)What constant current would transport the same charge in the same time interval ? Sol. (a) i dq dt . . q 4 0 0 . dq . . idt . 4 0 q . . (3. 2t)dt. . 2 40 q . ..3t . t .. . 12 .16 . 28 C
ELECTRIC CURRENT www.physicsashok.in 2 (b) i q 28 7 A t 4 . . . Types of Electric Current According to itsmagnitude and direction current is usuallydivided into two types : (i) Direct current : If themagnitude and direction of current does not varywith time, it is said to be direct current (DC). Cell, battery or DC dynamo are its sources. (ii) Alternating current : If a current is periodic (with constant amplitude) and has half cycle positive and half negative, it is said to be alternating current (AC).ACdynamo is the source of it. NOTE : It is worthy to note here that rectifier converts AC into DC while inverter DC into AC. Rectifier Inverter DC AC Current Density Current densityJ . at at a point is defined as a vector havingmagnitude equalto current per unit area surrounding that point and normal to the directionof charge flowand directioninwhich current passes through that point. So, if at point P current .I passes normally through area .S as shown in Fig., current densityJ . at P will be given by S 0 J lim i n . . S . . . . . + � dS i P n J i.e., J di n dS . . . Regarding current densityfollowing points areworth noting : (a) If the cross-sectional area is not normal to the current, the cross-sectional area normal to the current in accordancewith Fig., will be dS . dS cos . . J di dScos . . i J dS i.e., di = JdS cos . dS cos
or di . J . dS . . i.e., i . . J . dS . . (b) In case of conductors asV= iRand by definitions, E V and R V L S . . . So, .EL. i LS . . i.e., J i 1 E S . . . or J . . E . .. . 1 . . . . . . . . .
ELECTRIC CURRENT www.physicsashok.in 3 (c) In case of uniformcharge flowthrough a cross-section normal to it as i = nqvS So, J i n (nqv) n S . . . . . or charge J . nq v . v (.) . . . [(.)charge = nq] Drift Velocity The average uniformvelocity acquired by free electrons inside ametalby the application of an electric field which is responsible for current through it is called drift velocity. It is represented by�vd�. The current flowing throughconductor d d i neAv or v i neA . . where n= number ofmoving electrons per unit volume A= area of cross-section NOTE : Some books have taken average drift velocity as half of initial and final drift velocities of an electron giving vd = (eE./2m). This is wrong as vd is the average of drift velocities of large number of electrons at same instant and as for each electron v .. d = a f with .. / . . . .. a= e E m constant , . . . . v v . . / . .. d a = e E m . Example 2. An n-type silicon sample ofwidth 4 × 10�3m, thickness 25 × 10�5 mand length 6 × 10�2 mcarries a current of 4.8mAwhen the voltage is applied across the length of the sample.What is the current density ? If the free electron density is 1022 m�3, then find howmuch time does it take for the electrons to travel the full length of the sample ?Given that charge on an electron e = 1.6 × 10�19C. Sol. Bydefinition, . . J i i S b d . . . [as s = (b × d)] So, . . . . 3 3 2 5 3 J 4.8 10 4.8 10 A/m 25 10 4 10
.
. .
.
. . .
. . . and as in case of electric conduction inmetals J = nevd or d v J ne . So, 3 d 22 19 4.8 10 v 3 m/ s 10 1.6 10. . . . . . Hence, time taken by electron to travel the length L (= 6 × 10�2 m) of the conductor 2 d L 6 10 t 0.02 s. v 3 . . . . . Example 3. The area of cross-section, length and density of a piece of a metal of atomicweight 60 are 10�6m2, 1mand5 × 103 kg/m3 respectively. Find the number of free electrons per unit volume ifeveryatomcontributes one free electron.Also find the drift velocityof electrons in themetalwhen a current of 16Apasses through it. Given thatAvogadro�s number NA = 6 × 1023/mol and charge on an electron e = 1.6 × 10�19C.
ELECTRIC CURRENT www.physicsashok.in 4 Sol. As according toAvogadro�s hypothesis, A A A N m so n N N m N d as d m N M V VM M V . . . . . . . .. .. . . . . . 23 3 28 3 3 6 10 5 10 n 5 10 /m 60 10. . . . . . . Nowas each atomcontributes one electron, the number of electrons per unit volume ne = 1 × n = 1 × 5 × 1028 = 5 × 1028/m3 Further as here, 6 2 6 J i 16 16 10 A/m A 10. . . . . and as J = nevd . . . . 6 3 d 28 19 v J 16 10 2 10 m/ s ne 5 10 1.6 10 . . . . . . . . . . NOTE : � From this example it is clear that : An electron will take 1/(2 × 10�3), i.e., 500 s = (8.3 min.) to travel 1 m length of wire if it can. � If resistivity of metal . is taken to 2 × 10�8 ohm�m, the electric field inside the metal E = .J = (2 × 10�8) × (16 × 106) = 0.32 V/m and not zero as in electrostatics. RESISTANCE AND OHM�S LAW For somematerials, especiallymetals, at a given temperature, J . is nearlydirectly proportional to E.. and the ratio ofthemagnitudesE and J is constant.This ratio is called the resistivity(.) and this relationship is called the Ohm�s law.Thus, resistivity EJ . . SI units ofresistivity are .�m(ohm�metre).The reciprocalof resistivityis conductivity(.).
Thus, . . 1. Amaterial that obeysOhm�s lawreasonablywell is called an ohmic conductor or a linear conductor.Materials which show substantial departures fromOhm�slaw arecalled non-ohmic or non-linear. Resistance Suppose a conductingwire has a uniformcross-sectional areaAand length l as shown in Fig. Let Vbe the potentialdifference between the ends of the wire. If themagnitudes of the current densityJ . and the electric field E.. are JE A + � Vl i uniformthroughout the conductor, the total current i is given by i= JAand the potential differenceVbetween the ends isV= El. . V E i JA A . l . . l Here, A . l is constant for ohmicmaterials. This is called the resistance R.
ELECTRIC CURRENT www.physicsashok.in 5 Thus, R Vi . The resistanceR of a particular conductor is related to the resistivity . ofitsmaterial by R A . . l The equation, V = iR is often calledOhm�s law. NOTE : The equation R= Vi defines resistance R for any conductor, whether or not it obeys ohm�s law, but only when R is some constant we can correctly call this relationship Ohm�s law. Thus, for ohmic conductors V�i group is a straight line possing through origin. The slope of this line is equal to the resistance of the conductor. V i . R = V = tan i . Reciprocal of resistance is called conductance (G), i.e. G = 1 = i r V SI unit of G is ohm�1 which is called mho. Example 4. Two copper wires of the same length have got different diameters, whichwire has : (a) greater resistance and (b) greater specific resistance ? Sol. (a) For a givenwire,R A . . l i.e. R 1A . So, the thinner wirewill have greater resistance. (b) Specific resistance (.) is amaterial property. It does not depend on l orA. So, both thewireswillhave same specific resistance. Example 5. Awire has a resistance R.What will be its resistance if it is stretched to double its length ? Sol. Let Vbe the volume ofwire, then V =Al A . Vl Substituting this in R A . . l ,we have 2 R V . . l So, for given volume andmaterial, (i.e., Vand . are constant)
R . l2 When l is doubled, resistancewill become four times, or the newresistancewill be 4R.
ELECTRIC CURRENT www.physicsashok.in 6 VARIATION OF RESISTANCE WITH TEMPERATURE The resistance of a conductor varieswith temperature. The graph of variation of resistance of puremetalwith temperature is shown. Mathematically the dependence of resistance R on temperature is expressed as R0 0 T0 T R Slope = R R (T) = R O 0[1 + .(T � T0)] In this equationR(T) is the resistance at temperature T and R0 is the resistance at temperature T0, often taken to be 0ºC or 20ºC. The factor . is temperature coefficient of resistivity. Example 6. The resistance of thin silver wire is 1.0 ..at 20ºC. Thewire is placed in a liquid bath and its resistance rises to 1.2 ..What is the temperature of the bath ? ..for silver is 3.8 × 10�3 perº C. Sol. R (T) = R0[1 + .(T � T0)] Here, R (T) = 1.2 ., R0 = 1.0. ..= 3.8 × 10�3 perº C and T0 = 20ºC Substituting thevalues,we have 1.2 = 1.0[1 + 3.8 × 10�3(T � 20)] or 3.8 × 10�3 (T � 20) = 0.2 Solving this, we get T = 72.6º C Example 7. Aresistance Rof thermal coefficient of resistivity= . is connected in parallelwith a resistance = 3R, having thermal coefficient of resistivity= 2.. Find the value of .eff . Sol. The equivalent resistance at 0ºC is 10 20 0 10 20 R R R R R . . ...(i) The equivalent resistance at tºC is 1 2 1 2 R R R R R . . ...(ii) But R1 = R10 (1 + .t) ...(iii) R2 = R20(1 + 2.t) ...(iv) and R = R0(1 + .efft) ...(v) Putting the value of (i), (iii), (iv), (v) in eqn (ii), eff 54 . . . Example 8. (a) The current density across a cylindrical conductor of radius R varies according to the equation 0 J J 1 r R . . . . .. .. , where r is the distance fromthe axis. Thus the current density i
s a maximumJ0 at the axis r = 0 and decreases linearlyto zero at the surace r = R.Calculate the current interms of J0 and the conductor�s cross sectional area isA= .R2. (b) Suppose that instead the current densityis amaximumJ0 at the surace and decreases linearlyto zero at the axis so that 0 J J r . R . Calculate the current.
ELECTRIC CURRENT www.physicsashok.in 7 Sol. (a)We concider a hollowcylinder of radius r and thickness dr. The cross-sectional area of considered element is dA= 2.rdr The current in considerd element is 0 dI JdA J 1 r 2 rdr R . . . . . . .. .. or 0 dI 2 J 1 r rdr R . . . . . .. .. . R 0 0 I 2 J 1 r rdr R . . . . . . .. .. 2 0 0 R A I J J 3 3 . . . (b) R 2 0 0 r I 2 J dr R . . . 3 R 0 0 I 2 J r R 3 . . . . . . . . 3 0 0 2 J R 2A I J R 3 3 . . . Example 9.Anetwork of resistance is constructedwith R1&R2 as shown in the figure. The potential at the points 1, 2, 3,......, N are V1, V2, V3,........, VN respectivelyeach having a potential k time smaller than R2 R2 R2 R3 V0 R1 V VN N�1 R1 R1 V1= V0 k
I I1 V2= V0 k2 VN= V0 kN I´ I´1 I´2 I2 previous one. Find: (i) 12 RR and 23 RR in terms of k (ii) current that passes through the resistance R2 nearest to theV0 in termsV0, k&R3. Sol. R2 R2 R2 R3 V0 R1 V VN N�1 R1 R1 V1= V0 kI I1 V2= V0 k2 VN= V0 kN I´ I´1 I´2 (i) According to kcL, I = I1 + I2 or 0 0 0 0 0 2 1 2 1 V V V 0 V V k k k k R R R . . . . . or . . . . 0 0 0 2 1 2 1 k 1 V V k 1 V kR kR k R . . . .
ELECTRIC CURRENT www.physicsashok.in 8 . . .2 12 R k 1 R k. . Also, I´ = I´1 + I´2 or 0 0 0 0 N 2 N 1 N 1 N 1 1 2 1 3 V V V 0 V 0 k k k k R R R R . . . . . . . . . . After solving, 23 R k R k 1 . . (ii)Here, 0 1 1 2 2 V 0 V 0 k I R R. . . . . . 0 0 0 1 2 2 3 3 V V k 1 V I kR k k R k R k 1 . . . . . . .. . .. Example 10. Arod of length Land cross-sectionareaAlies along the x-axis between x= 0 and x= L.Thematerial obeysOhm�s lawand its resistivityvaries along the rod according to .(x) = .0 e�x/L. The end of the rod at x = 0 is at a potentialV0 and it is zero at x = L. (a) Find the total resistance of the rod and the current in thewire. (b) Find the electric potential in the rod as a function of x. Sol. (a) The resistance of considered element is 0 x / L dR dx e dx
A A . . . . . . 0 L x / L 0 x /L L0 0 R e dx L e A A . . . .. . . . . . . . x dx 0 1 0 L L 1 R e 1 1 A A e . . . . . . . .. . .. . . . . . . 0L e 1 R A e . . . . . . . . . . . . 0 0 0 0 I V 0 V AeV R R L e 1 . . . . . . (b) . E = .J or x /L 0 E . . e. J E = 0 x /L I Ie A A . . . , . . x / L 0 0 0 e AeV E A L e 1 . . . . . . . x /L 01 E e V L 1 e . . . . . 0 x E V V x. . . V0 � Vx = Ex
ELECTRIC CURRENT www.physicsashok.in 9 or . . . . x /L 1 0 x 0 1 V e e V V Ex 1 e . . .. . . . . ELECTRIC CELL An electric cell is a devicewhichmaintains a continuous flow of charge (or electric current) in a circuit by a chemicalreaction. In an electric cell, there are two rods of differentmetals called as electrodes. These electrodes are kept in a solution called electrolyte. On joining two electrodes by a wire the charge begins to flow in the wire, i.e., current flows inthewire. Inside the cell, a chemical reaction takes place in the electrolytewhichmaintains the charge onthe electrodes and the flowofcharge inthewire is continuouslymaintained. Thus, a cellconverts the chemicalenergyinto electricalenergy. Inthe light ofmodernviews inreference to a cell following terms need to be reviewed. Electromotive Force (EMF) The emf ofa cell is defined aswork done bythe cell inmoving unit positive charge in thewhole circuit including the cellonce. Therefore, ifWis thework done bya cell inmoving a charge q once around a circuit including the cell, emf E Wq . SI unit of emf is joule/coulomband is called volt. The emf of a cell in a circuit is taken to be positive, if circuit current inside a cell, is fromnegative to positive terminal, (i.e., cell is discharging) otherwise negative as shown inFig. E1 E2 i � + � + i E = E1 + E2 (a) E1 E2 i � + + � i E = E1 � E2 (b) E1 E2 i � + + � i E = E2 � E1 (c) NOTE : The term electromotive force is misleading introduced by Volta who thought it to be
force that causes the current to flow. Actually emf is not a force but work required to carry unit charge from lower potential to higher potential inside the cell. Internal Resistance (r) and Terminal Potential Difference The potential difference across a real source in a circuit is not equal to the emf of the cell. The reason is that chargemoving throughthe electrolyte ofthe cell encounters resistance.We cellthis the internalresistance ofthe source, denoted byr.As the currentmoves through r, it experiences an associated drops in potential equal to �ir�.Thus,whena current is drawnthrougha source, the potentialdifference between the terminalofthe source is, V = E � ir This can also be shown as below: E r i A B VA � E + ir =VB or VA � VB = E � ir Following three special cases are possible : (i) If the current flows in opposite direction (as in case of charging of a battery), thenV= E + ir (ii)V= E, if the current through the cellis zero
ELECTRIC CURRENT www.physicsashok.in 10 (iii) V= 0, if the cellis short circuited. This is because current in the circuit i Er . or E = ir Short Circuited E r . E � ir = 0 or V = 0 Thus,we can summarise it as follows : E r i V = E � ir or V < E E r i V = E + ir or V > E E r V = E if i = 0 E r i =ErV = 0 if short circuited COMBINATION OF RESISTANCES In Series Fig., represent a circuit consisting ofa source of emf and two resistors connected in series. AB R1 R2 V V1 V2 ABV R i Let equivalent resistor is R as shown. Then R = R1 + R2 This result can be readily extended to a network consisting of �n� resistors in series. . R = R1 + R2 + ....... + Rn In Parallel Fig., represents a circuit consisting of a source of emf and two resistors connected in paralle. AB V R1 ABV R ii1 i2 R2 i IfR be the equivalent resistance, then 1 2
1 1 1 R R R . . This result can be extended to a network consisting of n resistors in parallel. The result is
ELECTRIC CURRENT www.physicsashok.in 11 1 2 n 1 1 1 .......... 1 R R R R . . . . NOTE : In series combination, the same current flows through each resistance while in parallel combination, the voltage drop across each resistor is equal to the source voltage V. Example 11. Compute the equivalent resistance of the network shown in Fig., and find the current idrawn fromthe battery. 18V i i 4 6 3 Sol. The 6 . and 3 . resistances are in parallel. Their equivalent resistance is, 18V i 4 2 1 1 1 or R 2 R 6 3 . . . . Nowthis 2 . and 4 . resistances are in series and their equivalent resistance is 4 + 2 = 6 .. Therefore, equivalent resistance of the network = 6 .. i 18V 6 Current drawn fromthe batteryis, i = net emf = 18 = 3A net resistance 6 Star-Delta (.) Conversion For Fig. (a) . (Delta), to be equivalent to Fig. (b) (Star)Y, R2 A B C R1 R3 R1 R3 R2RA A B C RB RC A B C RA RB RC (a) (b) (c) 1 3 1 2 A B 1 2 3 1 2 3 R R R , R R R R R R R R R . . . . . .
ELECTRIC CURRENT www.physicsashok.in 12 and 2 3 C 1 2 3 R R R R R R . . . Example 12. Find the equivalent resistance betweenAand B in Fig. using Star-Delta Theorem. 10 A B CD 10 10 10 10 10 10 Sol. Using Star-Delta theorem, the equivalent circuit can be drawn as shown in fig. 10 10 10 RC = 5 RB = 5 RD = 2.5 15 22.5 5 9 5 A CD O B A O B A B CB CD C CB CD DB R R R 20 10 5 R R R 20 10 10 . . . . . . . . . . DC DB D DC DB CB R R R 10 10 2.5 R R R 10 10 20 . . . . . . . . . . BC BD B BC BD DC R R R 20 10 5 R R R 20 10 10 . . . . . . . . . . R´ 15 22.5 15 22.5 9 15 22.5 37.5
. .
. . . .
. Hence, RAB = R´ + 5 = 9 + 5 = 14 . Example 13.What willbe the change in the resistance of a circuit consisting of five identical conductors if two similar conductors are added as shown by the dashed line in figure. Sol.
ELECTRIC CURRENT www.physicsashok.in 13 r r r r r r r B A r r r r r r B A r 2r2r A r B A r r r B R2 = Req = 3r But before added the conductors, the equivalent resistance is 1 eq R . R´ . 5r . 21 R 3 R 5 . . KIRCHHOFF�S LAW Manyelectric circuits cannot be reduced to simple series-parallel combinations. For example, two circuits that cannot be so broken down are shown in fig. R1 R2 R3 E1 E2 A BD C F E (a) (b) E1 E2 E3 R1 R2 R3 R4 A B D E F G I H R5 C However, it is always possible to analyze such circuits byapplying two rules, devised byKirchhoff. First here are two terms that wewill use often. Junction : Ajunction ina circuit is a point where three ormore conductorsmeet. Junctions are also called nodes or branch points. For example, in figure (a) pointsDand Care junctions. Similarly, in figure (b) point B and F are junctions. Loop : Aloop is any closed conducting path. For example, in figure (a)ABCDA,DCEFDandABEFAare loops.
Similarly, in figure (b), CBFEC, BDGFBare loops. Kirchhoff�s rules consists of the following two statements. (i) Junction rule : The algebraic sumof the currents at any junction is zero. That is, junction . i . 0 i2 i2 i3 This lawcan also bewritten as, �the sumof all the curents directed towards a i4 point incircuit is equal to the sumof all the currents directed awayfromthat point.�
ELECTRIC CURRENT www.physicsashok.in 14 Thus, infigure i1 + i2 = i3 + i4 The junction rule is based on conservation of electric charge. (ii) Loop rule : The angebraic sumof the potentialdifferences in anyloop including those associated emf�s and those of resistive elements,must equal zero. That is, closed loop . .V . 0 The loop rule is based on the fact that the electrostatic field is conservative in nature. In applying the loop rule, we need sign convention as discussed below: (a) Whenwe travel through a source in the direction from� to +, the emf is considered to be positive. E A B Path (a) (b) Whenwe travel form+ to �, the emf is considered to be negative. E A B Path (b) (c) Whenwe travel through a resistor in the same direction as the assumed current, the iR termis negative because the current goes in the direction R A B Path (c) i of decreasing potential. (d) Whenwe travel through a resistor in the direction opposite to the assumed current, the iR termis positive because this represents a rise of potential. R A B Path (d) i Example 14. Find current in different branches ofthe electric circuit shown in figure. 2 A B CD 4 4 2 F E 2V 4V 6V Sol. ApplyingKirchhoff�s first law(junction law) at junctionB, i1 = i2 + i3 ...(i) ApplyingKirchhoff�s second lawin loop 1 (ABEFA), �4i1 + 4 � 2i1 + 2 = 0 ...(ii) ApplyingKirchhoff�s second lawin loop 2 (BCDEB), �2i3 � 6 � 4i3 � 4 = 0 ...(iii) Solving Eqs. (i), (ii) and (iii),we get i1 = 1A CD 2V 1 4V 2 6V
A BF E 2 4 4 2 i3 i2i1 i3 i1 2 i 8 A 3 . 3 i 5 A 3 . . Here, negative sign of i3 implies that current i3 is in opposite direction ofwhat we have assumed.
ELECTRIC CURRENT www.physicsashok.in 15 IMPORTANT FEATURES 1. Distribution of current in parallel connections : When more than one resistances are connected in parallel, the potential difference across themis equal and the current is i i2 3 i1 i R 2R distributed among themin inverse ratio oftheir resistance, as 3R i VR . or i 1R . for same value ofV e.g., inthe figure, 1 2 3 i : i : i 1 : 1 : 1 6 : 3 : 2 R 2R 3R . . 1 i 6 i 6 i 6 3 2 11 . . . . .. . . .. . 2 i 3 i 3 i 6 3 2 11 . . . . .. . . .. and 3 i 2 i 2 i 6 3 2 11 . . . . .. . . .. 2. Distribution of potential in series connections :Whenmore than one resistances are connected in series, the current through themis same and the potential distributed in the ratio of their resistance, as V = iR or V . R for same value of i. e.g., inthe figure, V1 V2 V3 R 2R 3R i V1 : V2 : V3 = R : 2R : 3R = 1 : 2 : 3 . 1 V 1 V V 1 2 3 6 . . . . .. . . .. 2 V 2 V V 1 2 3 3 . . . . .. . . .. and 3 V 3 V V 1 2 3 2 . . . . .. . . .. COMBINATION OF CELLS
Cells are usuallygrouped in following threeways : In Series Suppose �n� cells each of emfE and internal resistancer are connected in series as shown in figure. Then
ELECTRIC CURRENT www.physicsashok.in 16 r i R E E r E r Net emf = nE Total resistance = nr + R . Current in the circuit = Net emf Total resistance or i = nE nr +R NOTE : If palarity of m cells is reversed, then equivalent emf = (n � 2m)E while total resistance is still nr + R . .n - 2m. E i= nr+R In Parallel : Consider the following three cases : Ist case : Let �n�cells each of emfE and internal resistance �r�are connected in parallel. Net emf = E Total resistance r R n . . i R i E r E r E r . Current inthe circuit i = net emf total resistance or i = E R + r/n IInd case : Let �n�cells have different E and �r�. Net emf = Eeq . . . . E/r = 1/ r .. Total resistance = Req . . R 11/ r . . . i2 i3 i1 i R E1
E2 E3 r1 r2 r3 i A B F E Hence, C D eq eq E i R . or . . . . E / r i 1 R 1/ r . . . . IIIrd case : This ismost general case of parallel grouping inwhich E and �r�of different cells are different and the positive terminals of fewcells are connected to the negative terminals of the others as shown. Net emf = Eeq
ELECTRIC CURRENT www.physicsashok.in 17 . . . . . . 1 1 2 2 3 3 1 2 3 E / r E / r E / r 1 1 1 r r r . . . . . . . . . . . i i2 3 i1 i R E1 E2 E3 r1 r2 r3 i Total resistance = Req 1 2 3 R 1 1 1 1 r r r . . . . . . . . . . Hence, eq eq E i R . . . . . . . 1 1 2 2 3 3 1 2 3 E / r E / r E / r i 1 R 1 1 1 r r r . . . . . . . . . . . . In Mixed Grouping There are �n� identical cells in a rowand number of rows are �m�. Net emf = nE Total resistance = Req R nr m . . r i E R i
Hence, i nER nr m . . Example 15. Find the emf and internal resistance of a single batterywhich is equivalent to a combination of three batteries as shown in figure. 4V 10V 22 1 6V Sol. The givencombination consists of two batteries in parallelandresultant ofthese two inserieswiththe third one. For parallel combinationwe can apply, 1 2 1 2 eq 2 2 E E 10 4 E r r 2 2 3V 1 1 1 1 r r 2 2 . . . . . . . Further, eq 1 2 1 1 1 1 1 1 r r r 2 2 . . . . . . req = 1 . Nowthis is in serieswith the third one, i.e.,
ELECTRIC CURRENT www.physicsashok.in 18 1 1 6V 3V The equivalent emf of these two is (6 � 3)Vor 3Vand the internal resistancewill be, (1 + 1) or 2 .. r = 2 E=3V Example 16. Abatteryconsists of a variable number n of identical cells having internal resistance connected in series. The terminals of the batteryare short circuited and the current Imeasured. Which one of the graph belowshows the relationship between I and n ? (A) I/A O n (B) I/A O n (C) I/A O n (D) I/A O n (E) I/A O n Sol. nE nr I nE E nr r = = I n Example 17. In previous problem, if the cell had been connected in parallel (instead of in series) which of the above graphswould have shown the relationship between total current I and n ? (A) I/A O n (B) I/A O n (C) I/A O n (D) I/A O n (E) I/A O n Sol. En r I E nE r r n = = I nEr = I
n Example 18. Underwhat conditioncurrent passing through the resistanceR can be increased by short circuiting the battery of emfE2. The internal resistances of the two batteries are r1 and r2 respectively. E1 r1 E2 r2 R (A) E2r1 > E1(R + r2) (B) E1r2 > E2(R + r1) (C) E2r2 > E1(R + r2) (D) E1r1 > E2(R + r1) Sol. The current throughR before short circuit
ELECTRIC CURRENT www.physicsashok.in 19 1 2 1 2 I E E r r R = + + + After short circuit : 1 1I' E r R = + E2 E1 r1 r2 I� > I R 1 1 2 1 1 2 E E E r R r r R > + + + + E1r1 + E1r2 + E1R > E1r1 + E1R + E2r1 + E2R . 1 2 2 1 2 E r > E r + E R WHEATSTONE�S BRIDGE This is an arrangement of four resitances inwhich one resistance is unknown and rest known.TheWheatstone�s bridge is shown in fig. The bridge is said to be balancedwhen deflection in galvanometer is zero, i.e., ig = 0 and hence, the conditionof deflection is G R SP Q B A C E D i2 ig = 0 i1 i2 i1 i P R Q S . NOTE : In Wheatstone�s bridge, cell and galvanometer arms are interchangeable. G R S P Q B A C D G R S P Q B A C D
In both the cases, condition of balance bridge is, P = R Q S Example 19.Ahemisphere network of radius a ismade by using a conductingwire of resistance per unit length r. Find the equavalent resistance acrossOP. P B O D A C Sol. Point (AandC) and (DandB) are symmetricallylocatedwith respect to points Oand P. Hence, the circuit can be drawn as shown in figure. This is a balancedWheatstone bridge between P andO Hence, r3 can be removed.And, 1 2 PO R r r 4. . Here . . 1 PB PD a r r R R 2 . . . . r22 r22 r3 r21 r21 B C PO and r2 = ROB = (a)r
ELECTRIC CURRENT www.physicsashok.in 20 . . PO 2 ar R 8 . . . Ans. Example 20. In the circuit shown, what is the potential difference VPQ? (A) +3V (B) +2V 4V 2V Q P 1V 21 2 3 (C) �2V (D) none Sol. In ABCPA�2I + 4 � I + 2 = 0 I 6 2A 3 = = VP � VQ = �{algebriac sumof rise up and drop up of voltage} 2V 4V Q P 1V I A B C D II (I�I ) 1 I1 I1 I1 2 1 2 3 = � (2 � 2I) = � (2 � 4) . P Q V - V = 2 V Example 21. Two batteries one of the emf 3V, internal resistance 1 ohmand the other of emf 15 V, internal resistance 2 ohmare connected in series with a resistance R as shown. If the potential difference between a and b is zero a b R the resistance of R in ohmis (A) 5 (B) 7 (C) 3 (D) 1 Sol.According to loop rule, 3 � I � IR + 15 � 2I = 0 I 18 3 R = + . Va � Vb = � (�3 + I) a b
R 15V 3V I or 0 = 3 � I . I = 3 A . 3 18 3 R = + or 9 + 3R = 18 . R 9 3 3 . . . Example 22. Consider an infinite ladder network shown . in figureAvoltage V is applied between the points AandB. This applied velue of voltage is halved after each R1 R1 R1 R1 R1 R2 R2 R2 R2 R2 AB section. (A) R1/R2 = 1 (B) R1/R2 = 1/2 (C) R1/R2 = 2 (D) R1/R2 = 3 Sol. I = I1 + I2 1 2 1 v � v v � 0 v � v 2 2 2 4 R R R . . or 1 2 1 v v v 2R 2R 4R = +
ELECTRIC CURRENT www.physicsashok.in 21 or 1 2 1 1 1 v R R 2R = + I2 v v = 0 v/2 I I1 v/4 R2 R2 R2 R2 R2 v = 0 v = 0 v = 0 v = 0 v = 0 or 1 2 1 1 2R R = or 12 R 1 R 2 = Example 23. If the switches S1, S2 and S3 in the figure are arranged such that current through the batteryisminimum, find the voltage across pointsAand B. 6 3 6 S1 1 1 1 S3 S2 9 AB Sol. Forminimumcurrent through battery, equivalent resistance 24V across battery should bemaximum.Aswe know, in series resistance increases, but in parallel, resistance decreases. Fromthese points of view, all switches should be open. 9 6 6 3 1 A 7 A 24 V I I I0.5 B 4.5 1 1 24 V According to loop rule, 24 � 7I � 4.5I � 0.5I = 0 . I = 2A . VAB = VA � VB = 0.5 × I = 0.5 × 2 = 1 V Example 24. In the circuit shown in figure, calculate the following :
(i) Potentialdifference between points �a� and �b�when switch �S� is open. (ii) Current through�S� in the circuit when�S� is closed. 36 63 a S b 36V Sol. (i) Here 1 I 36 0 4 A 9. . . Also 2 I 36 0 4 A 9. . . Also, 36 � Va = I1 × 6 = 6 I1 or 36 � Va = 6 × 4 = 24 V . Va = 36 � 24 = 12 V 36 63 a b V = 0 V0 = 36V I1 I2 36 � Vb = 3 × 4 = 12 V and Vb = 36 � 12 = 24 V . Va � Vb = 12 � 24 = �12 V (ii) In loopABCIHA. �6(I � I1) + 3I1 = 0 or �6I + 6I1 + 3I1 = 0 or 9I1 = 6I
ELECTRIC CURRENT www.physicsashok.in 22 . 1 1 I 9 I 3 I 6 2 . . ...(i) In loop CDEJIC, 3 6 36 D CB A HI E J FI G (I � I1) I1 I2 (I � I1 + I2) �3(I � I (I1 � I2) 1 + I2) + 6(I1 � I2) = 0 or �3I + 3I1 � 3I2 + 6I1 � 6I2 = 0 or �3I + 9I1 � 9I2 = 0 ...(ii) In loop EFGHIJE, 36 � 3I1 � 6(I1 � I2) = 0 or 9I1 � 6I2 = 36 ...(iii) After solving eqn(i), (ii) and (iii) I2 = 3Afrom�b� to �a�. Example 25.An enquiring physics student connects a cell to a circuit andmeasures the current drawnfromthe cell to I1.When he joins a second identical cell is serieswith the first, the current becomes I2.When the cells are connected are in parallel, the current through the circuit is I3. Show that relation between the current is 3 I3 I2 = 2I1(I2 + I3) Sol. Let the equivalent resistance of circuit is R. The emf and internal resistance of cell is E and r respectively. . 1 I E r R . . and 2 I 2E 2r R . . and 3 I E r / 2 R . . L.H.S. = 3 2 . . 3I I 3E 2E r / 2 R 2r R . . . .. . .. . . .. . 2 3 2 6E 3I I
r / 2 R 2r R . . . R.H.S. = 2I1 (I2 + I3) 2E 2E E r R 2r R r / 2 R . . . . . . .. . .. .. . . .. . .. . 6E2 r / 2 R 2r R . . . Hence, L.H.S. = R.H.S. Example 26. Find the potential differenceVA �VB for the circuit shown in the figure. 1 1 1 1 1 1 1 1 1 1V 1V 1V 1V 1V 1V B 1V 1V A Sol. i1 + i2 = i 4i + i1 = 0
ELECTRIC CURRENT www.physicsashok.in 23 . x 49 . 4 B A 3 2 1 0 4i x ii2 i1 3i 2i x+4 x+3 x+2 x+1 i1 . A B V V 2 4 22 V 9 9 . . .. . . . . .. .. Example 27. Power generated across a uniformwire connected across a supply is H. If the wire is cut into n equal parts and all the parts are connected in parallel across the same supply, the totalpower generated in the wire is : (A) 2 Hn (B) n2H (C) nH (D) Hn Sol. Let the resistance of wire is R A = r . r = the resistance of each piece R nA n = r . = v2 H t R = ...(i) when all pieces are corrected in parallel, Then equivalent resistance is 0 2 r r R n n = = . 2 2 2 1 0 v n v H t t r R = = . n2H fromeqn. (i) Example 28. The ratio of powers dissipatted respectively in R and 3R, as shown is : (A) 9 (B) 27/4
3R 2R R (C) 4/9 (D) 4/27 Sol. P1 = Power dissiatted in resistance R 2 2I R 4 I2R 3 9 . . . . .. .. and P2 = Power dissipatted in 3R = I2(3R) = 3I2R 3R 2R R I/3 2I/3 I I . 2 1 2 2 P 4I R P 9 3I R . . . 12 P 4 P 27 = Example 29. In the figure shown the power generated in y is maximumwhen y = 5.. (A) 2 . (B) 6 . .......... 2R R y 10V, (C) 5 . (D) 3 . Sol. According to KOL : 10 � 2 I � I y � I R = 0 . I 10 2 y R = + + . P = I2y R y 10V I I I or 2 P 100 y (2 y R) = + + For maximumpower dissipatted,
ELECTRIC CURRENT www.physicsashok.in 24 dP 0 dy . . By solving, y = 2 + R or 5 = 2 + R . R . 3. Example 30. Find the current through 25Vcell&power supplied by20Vcell in the figure shown. 10 511 5 10V 5V 20V 30V 25V Sol. Fromfigure,1 I 55 0 5A 11. . . 2 I 5 0 1A 5. . . 10 511 5 10V 5V 20V 30V 25V 25V 25V 25V 25V 25V V=0 V=0 V=0 V=0 V=0 15V I4 I3 I2 I1 I=I +I +I +I4 1 2 3 4 30V 5V 55V 3 I 30 0 3A 10. . . 4 I 15 0 3A 5. . . . Electric current through 25Vcell = I1 + I2 + I3 + I4 = 5 + 1 + 3 + 3 = 12A The power supplied by 20Vcell is P = � 20I2= �20 × 1 = � 20 W Example 31. The current I through a rod of a certain metallic oxide is given by I = 0.2 V5/2, where V is the potential difference across it. The rod is connected in series with a resistance to a 6Vbattery of negligible internal resistance.What value should the series resistance have so that : (i) the current in the circuit is 0.44 (the value of (2.2)2/5 = 1.37)
(ii) the power dissipated in the rod is twice that dissipated in the resistance. Sol. (i) the potentialdifference across rod for current 0.44 is 20.44 5 V´ 0.2 . . . .. .. . .2V . 2.2 5 The potential difference across connected resistor isV´´= 0.44 R E = V´ + V´´ or . .26 . 2.2 5 . 0.44R or . .20.44R . 6 . 2.2 5 . . .26 0.2 5 R 0.44 0.44 . . R = 13.64 � 3.12 R = 10.52 . (ii)Total power supplied by battery is used byrod and resistor . E I = V I + I2 R
ELECTRIC CURRENT www.physicsashok.in 25 But VI = power dissipated in rod V I = 2I2R . EI = 2I2R + I2R = 3I2R or E = 3IR or 6 = 3IR . IR = 2 ...(i) Also, V´ + V´´ = E 2 I 5 IR 6 0.2 . . . . .. .. or 2I 5 2 6 0.2 . . . . .. .. or 2I 5 4 0.2 . . . .. .. or 2 5 5 I 4 2 32 0.2 . . . . . .. .. or I = 32 × 0.2 = 6.4A Fromeqn (i), IR = 2 . R 2 20 0.3125 6.4 64 . . . . Remarks : In the case of rod,V� I graph is not straight line. So, ohm�s lawis not applicable in the case of the givenrod. Example 32. Aperson decides to use his bath tubwater to generate electric power to run a 40watt bulb. The bath tub is located at a height of 10mfromthe ground&it holds 200 litres ofwater. Ifwe installawater driven wheelgenerator on the ground, at what rate should thewater drain fromthe bathtub to light bulb?Howlong canwe keep the bulb on, if the bath tubwas full initially. The efficiency of generator is 90%. (g = 10m/s�2) Sol. Power P gh dm dt . . or 40 gh dm 90 dt 100 . . or 40 0.9 gh dm dt . . dm 40 40 4 kg / sec.
dt 0.9 gh 9 10 9 . . . . . 0 m dm t dt . . . .. .. . 0 t m 200 450 sec. dm 4dt 9 . . . . . .. ..
ELECTRIC CURRENT www.physicsashok.in 26 Example 33. The circuit shown in figure ismade of a homogeneouswire of uniform cross-section. 1234 isasquare. Find the ratio Q12/Q34of the amounts of heat liberated per unit time in conductor 1�2 and 3�4. 3 4 1 2 Sol. Let us represent the central junction ofwires in the formof two junctions connected by thewire 5�6 as shown in figure. then in follows fromsymmetry that there is no current through thiswire. Therefore, the central junction can be removed fromthe initialcircuit. Further, R12 = R13 = R34 = R24 = r and 15 25 36 46 R R R R r2 . . . . Let Vbe the voltage between 1 and 2. 3 4 1 2 6 5 3 4 1 2 6 5 Then the amount of heat liberated in conductor 1 � 2. 2 12 V Q r . ...(i) Current through3 � 4, 3 4 . . i V r 2 3 . . . . . . 2 2 3 4 3 4 2 Q i r V r 2 3 . . . . . . . .2 1 2 3 4 Q 2 3 11 6 2 Q .. . . . . Ans. Metre Bridge Themetre bridge is the practicalapplication of theWheatstone�s network principle. G l (100 � l) D A C
P Q B R J S In such a bridge, the ratio of two resistances sayRand S, can be determined fromthe ratio of their balancing lengths. In Fig.,AC is a 1mlong uniformwire, Let AD = l cm, thenDC = (100 � l) cm Since, resistance . length . P AD Q DC 100 . . . l l If P is known, thenQcan be determined. Example 34. The potentiometer wire AB is 100 cm long. When AC = 40 cm, no deflection occurs in the galvanometer, findR.
ELECTRIC CURRENT www.physicsashok.in 27 G A B 10 R Sol. 10 AC R CB . . R 10 CB (10) 100 40 10 60 15 AC 40 40 . . . . . . . . . . . . . . . . . . . . . . . . Example 35. Abatteryof emf .0 = 10Vis connected across a 1 mlong uniform wire having resistance 10./m. Two cells of emf .1 = 2Vand .2 = 4Vhaving internal resistances 1 ..and 5 ..respectively are connected as shown in the figure. If a galvanometer shows no deflection at the point P, find the G 51 10 P A B 0 =10V 1=2V 2=4V distance ofpoint P fromthe pointA. Sol. The resistance inAP is R1 = x. = 10 x where xis inmetre. The resistance in PB is R2 = (1 � x)10 The equivalent circuit is According to loop rule 10 � IR1 � IR2 � 10I = 0 . 1 2 I 10 10 1 R R 10 10 10 2 . . . . . . G R1 P R2 I 10V I I A B E = r E1 r1 E2 r2 + = 56214+ 5 = 14 6 V r = 1 × 5 1 + 5 = 56But VA � VP = E = IR1 . 14 1 10x
6 2 . . . x 14 2 0.4667 m 46.67 cm 6 10 . . . . Example 36. In the figure shown for which values ofR1 and R2 the balance point for Jockey is at 40 cmfromA.WhenR2 is shunted by a resistance of 10 ., balance shifts to 50 cm. Find R1 and R2. (AB = 1m) : G R1 A B R2 Sol. Let resistance perunit lengthof potentiometer is ..Assume that P is contact point of potentiometer. . The resistance inAP is R3 = x. and the resistance in PB is R4 = (1 � x). ( . AB = 1 m) According to balance condition ofwheatstone bridge, R1R4 = R2R3 or R1(1 � x). = R2 x. or R1(1 � x) = R2x ...(i) whenR2 is shunted.
ELECTRIC CURRENT www.physicsashok.in 28 Then 2 2 2 2 2 R R 10 10R R 10 R 10 . . . . . . R1(1 � x). = R´2 x. . . 2 1 2 R 1 x 10R x R 10 . . . ...(ii) Fromeqn. (i) and (ii), 1 R 10 3 . . and R2 = 5.. IMPORTANT FEATURES 1. When batteryand galvanometer arms of aWheatstone�s bridge are interchanged, the balance position remains undisturbedwhile sensitivityof bridge changes. I � I G 1 R P I1 IG (I � I1 + IG) S I A B I (I1 � IG) Q 2. Two other common forms of balancedWheatstone�s bridge are shown. ABG P Q R S POTENTIOMETER Potentiometer is an ideal device to measure the potential difference betweentwo points. It consists of a long resistancewireAB of uniformcross-section inwhich a steadydirect current is set up bymeans of a battery. LE1 i2 = 0 A B i C G E2 , ri i Potentialgradient k Potential difference across AB Total length .
AB k VL . AB k iR i L . . .
ELECTRIC CURRENT www.physicsashok.in 29 where AB RL . . = resistance per unit lengthof potentiometerwire The emf of source balanced between pointsB and C CB 2 E . kl . i R . l l E2 = iRCB Applications (i) To find emf of an unknown battery : l E1 i2 = 0 A B i C G EK 1 E1 i2 = 0 A B i C G EU 2 1 l2 i We calibrate the device by replacing E2 by a source of known emfEK and then by unknown emfEU. Let the null points are obtained at lengths l1 and l2. Then, EK = i(. l1) and EU = i(. l2) Here, ..= resistance ofwireAB per unit length. . K 1 U 2 EE . ll or 1 U K 2 E E . . .. .. ll So, bymeasuring the lengths l1 and l2, we can find the emf of an unknown battery. (ii) To find the internal resistance of a call : Firstly the emfE of the cell is balanced against a lengthAD= l1. For this the switch S´ is left opened and S is closed.Aknown resistance R is then connected to the cell as shown.The terminalvoltageVis nowbalanced against a smaller lengthAD´ = l2.Here, no switch S Is opened and S´ is closed. Then,
12 EV . ll G (E, r) D´ D A S´ RE l2 S l1 Since, B E R r V R. . {. E = i(R + r) and V = iR} or 12 R r R. . ll or 12 r 1 R . . . . . . . . ll Example 37. Apotentiometer wire of length 100 cm has a resistance of 10 .. It is connected in series with a resistance and a cell of emf2Vand of negligible internal resistance.Asource of emf 10mVis balanced against a length of 40 cmof the potentiometer wire.What is the value of external resistance ?
ELECTRIC CURRENT www.physicsashok.in 30 Sol. Fromthe theoryof potentiometer, VCB = E, if no current is drawn fromthe battery or 1 CB AB E R E R R . . . . . . . . E1 E A B C G R i Here, E1 = 2V, RAB = 10 . CB R 40 10 4 100 . . . . . . .. .. and E = 10 × 10�3 V Substituting in above,we get R = 790 . MOVING COIL GALVANOMETER Moving coilgalvanometer is a device used to detect small current flowing in an electric circuit.With suitable modifications, it can be used to measure current and potentialdifference. Conversion of galvanometer into an Ammeter An ammeter is an instrument whichis used tomeasure current in a circuit in ampere (ormilli-ampere ormicroampere). Hence, it is always connected in series in the circuit. Since, the galvanometer coilhas some resistance of its own, therefore, to convert a galvanometer into an ammeter, its resistance is to be decreased so, to convert a galvanometer into ammeter a low resistance, called shunt (S) is connected is S a G b i Ammeter in parallelto the galvanometer as shown in figure. Here, g . . i S i S G . . and . .. . A . . S G R G S . . where RA = resistance of ammeter S = resistance of shunt G= resistance of galvanometer
Conversion of Galvanometer into Voltmeter AVoltmeter is an instrument which is used tomeasure the potential difference between two points of an electric circuit directly in volt (ormilli-volt ormicro-volt). Hence, it is connected in parallel across those two points of the circuit, betweenwhich the potentialdifference ig G R is to be measured.When it is connected. Voltmeter Since, the resistance of coilofgalvanometer of its ownis low, hence, to convert a galvanometer into a voltmeter, high resistance Ris series is connectedwith the galvanometer. Here, g i V R G . . where R +G= RV = resistance of voltmeter Example 38. Amoving coil galvanometer of resistance 10 . produces full scale deflection, when a current of 25
ELECTRIC CURRENT www.physicsashok.in 31 mAis passed through it. Describe showing full calculations, howwill you convert the galvanometer into a voltmeter reading upto 120V. Sol. Here, G = 10 .., ig = 25 mA= 25 × 10�3A To convert the galvanometer into voltmeter reading upto 120V: To convert a galvanometer into voltmeter of rangeV, a large resistance Rhas to be connected in series to it.The value ofR is given by g R V G i . . Here, V = 120 V . 3 R 120 10 4790 25 10. . . . . . Example 39. Amilliammeter of range 10 mAand resistance 9. is joined in a circuit as shown. Themetre gives full-scale deflection for current I whenAand B are used as its terminals, i.e., current enters at Aand leaves , 10 mA at B (C is left isolated). The value of I is A B C (A) 100 mA (B) 900 mA (C) 1 A (D) 1.1 A Sol. According to loop rule, �9 × 10 � 0.9 × 10 + 0.1 × (I � 10) = 0 or I 10 90 9 990 0.1 - = + = 10 mA I 10 mA I�10 I . I = 1000 mA . 1 A Example 40. In the circuit shown in figure reading of voltmeter is V1 when only S1 is closed, reading of voltmeter is V2 when only S2 is closed. The reading of voltmeter is V3 when both S1 and S2 are closed then V 6R 3R R E S2 S1 (A) V2 > V1 > V3 (B) V3 > V2 > V1 (C) V3 > V1 > V2 (D) V1 > V2 > V3 Sol. Step-I�When S1 is closed, I 4R . . . 1 v I 3R 3R 4R . . . . . R
3R 1 v 34 . . Step-II�When S2 is closed I 7R . . . 2 v . I . 6R R 6R 2 v 6R 6 7R 7 . . . . Step-III�When S1 and S2 both are closed. I 3R . . v3 = I × 2R R 2R
ELECTRIC CURRENT www.physicsashok.in 32 3 2 v 2R 3R 3 . . . . Example 41. In the circuit shown in figure the reading of ammeter is the samewith both switches openaswith both closed. then find the resistance R. (ammeter is ideal) + � 300 1.5V R 1000 50 A Sol. Step-I : Discuss the circuit when both switches open : According to loop rule : 1.5 � 300 I � 100 I � 50 I = 0 . I 1.5 15 1 A 450 4500 300 . . . Step-II :Discuss the circuit after closing the switch. In loopABCDEA �IR + 1.5 � 300I1 = 0 or 300I1 + IR = 1.5 ...(1) A 50 I I I I 100 300 1.5 V In loop BCGFB �100I + (I1 � I) R = 0 or (I1 � I)R = 100 I I1R = (100 + R)I . . . 1 100 R I I R. . ...(2) Fromeqn (1) and (2) .100 R. I 300 IR 1.5 R. . . 300 1.5V R 100 50 A no current I1 I1
I1 AFB I C DE (I1�I) G or .100 R. 1 R 300 1.5 R 300 300 . . . . R = 600 . Example 42. The battery in the diagramis to be charged by the generator G. The generator has a terminal voltage of 120 volts when the charging current is 10 amperes. The battery has an emf of 100 volts and an internal resistance of � + G R +� 1 ohm. In order to charge the battery at 10 amperes charging current, the resistance R should be set at (A) 0.1 . (B) 0.5 . (C) 1.0 . (D) 5.0 . Sol. VA � VB = � {algebraic sumof rise up and drop up of voltage} 120 = � {�IR � 1 × I � 100} 120 = IR + I + 100 100V I B A R I or 20 = 10R + 10 . R . 1 . Example 43. Agalvanometer having 50 divisions providedwitha variable shunt �s� is used tomeasure the current when connected inserieswith a resistance of 90 ..and a battery of internal resistance 10 .. It is observed that when the shunt resistance are 10., 50., respectivelythe deflection are respectively9&30 divisions.What is the resistance of the galvanometer ?
ELECTRIC CURRENT www.physicsashok.in 33 Sol. The electric current through galvanometer is proportionalto its deflection. I × . . gg I 9 3 I´ 30 10 . . But . . S g g S IR I R R . . Also, g S g S I ER R 100 R R . . . . . . S g g S g S g S ER I R R 100 R R R R . . . . . . . . . . or . . . . S g S g S g g Sg S g S g S g S R R R 100 R R I R R I´ R´ R R´ 100 R R´
R R´ . . . . . . . . . . . . . . . . . . . or . . . . . . g g g g g g 10 10R 100 R 10 3 R 10 10 50 R 50 50R 100 R 50 . . . . . . . . . . . . . . . . . . . . . . .. . .. . Rg = 233.3 .. IMPORTANT FEATURES 1. (a) The reading of an ammeter is always lesser than actual current in the circuit. E R i (a) E R i´ (b) A G i´ i´ S (c) For example, in Fig. (a), actual current throughRis, i ER . ...(i) while the current after connecting an ammeter of resistance A GS G S . . . .. . .. in serieswithRis, i´ E R A
.
. ...(ii) FromEqs. (i) and (ii), we see that i´< i and i´= i ,WhenA= 0, i.e., resistance of an ideal ammeter should be zero.
ELECTRIC CURRENT www.physicsashok.in 34 (b) Percentage error inmeasuring a current throughan ammeter is 1 1 i i´ 100 R R A 100 i 1R . . . . . . . . . . . . . . . . . . .. .. or % error A 100 R A . . . . .. . .. 2. (a) The reading of a voltmeter is always lesser than true value. VG R RV r i i i r For example, if a current �i�is passing through a resistance r, the actualvalue is, V = ir ...(i) Nowif a voltmeter of resistance RV(=G+R) is connected across the resistance �r�, the newvaluewill be . . VV i rR V´ r R . . . or V V´ ir 1 r R . . ...(ii) FromEqs. (i) and (ii), we can see that V´ < V and V´ = V if RV = . Thus, resistance of an idealvoltmeter should be infinite. (b) Percentage error inmeasuring the potentialdifference bya voltmeter is, V V V´ 100 1 100 V 1 r R . . . . . . . . . . . . . . . . . . . . . or V %error 1 100 1 r R . . . . . . . . . . . . . RC CIRCUITS
In precedingsectionswe dealt onlywith circuits inwhichthe current did not varywith time.Here,were discuss about time-varying currents. Charging of a Capacitor First we consider the charging of a capacitor without resistance. + �S V C + �V q = CV + � 0 Consider a capacitor connected to a battery of emfVthrough a switch S. On closing the switch, there is no time tag between connecting and charging.Acharge q0 = CVcomes in the capacitor as soon as switch is closed and q � t graph in this case is a straight line q0 q r parallel to t=axis. Ifwe employed a resistance in the circuit, charging takes some time Fig. The q � t equation in this case, q = q0(1 � e�t/.)
ELECTRIC CURRENT www.physicsashok.in 35 Here, q0 = CV SC C and . = CR= time constant V q � t graph is anexponentially increasing graph. The charge q increases expponentially from0 to q0. Fromthe graph and equationwe see that, At t = 0, q = 0 q0 q t 0.632q0 and at t = ., q = q0 At t = ., q = q0(1 � e�1) = 0.632 q0 Here, . can be defined as the time inwhich 63.2%charging is over in a C� R circuit. Current flows ina C� Rcircuit during charging of capacitor.Once charging as over or the steady state condition is reached the current becomes zero. The current at any time t can be calculated bay differentiating qwith respect to t. Hence, i0 i t Charging current is, i = i0e�t/. i.e., current decreases exponentiallywith time. Time i � t graph is as shown in fig. Discharging of a Capacitor Againwe consider the discharging of capacitor with resistance. Suppose a capacitor has a charge q0. The positive plate has a charge +q0 and negative plate �q0. S q = 0 + � q0 When the switch is closed, the extra electrons on negative plate immediately comes to the positive plate and net charge on both plates become zero. So, we cansay that discharging takes place immediately. S C R q0 + � In case ofC � Rcircuit discharging also takes time. The q � t equation in this case is, t / c 0 q . q e. . Thus, q decreases exponentially fromq0 to zero, as shown in Fig. At t = 0, q = q0 q0 q t 0.368q0 At t = ., q = 0 In case of discharging definition of . is charged. At time t = ., q = q0 e�1 = 0.368q0 Hence, in this case . can be defined as the timewhen charge reduces to 36.8%of itsmaximumvalue q0. During discharging current flows inthe circuit till�q� becomes zero.This current can be found bydifferentiating
�q�with respect to �t� butwithnegative signbecause charge is decreasingwith time.Hence, discharging current is, i = i0e�t/. This is an exponentially decreasing equation. Thus, i � t graph decreases exponentiallywith time fromi0 to 0. The i � t graph is as shown in Fig. i0 i t
ELECTRIC CURRENT www.physicsashok.in 36 Example 44. Acapacitor C= 100 µF is connected to three resistor each of resistance 1 k. and a battery of emf 9V.The switch S has been closed for long time so as to charge the capacitor.When switch S is opened, the capacitor discharges with 1k 1k 1k C9V S time constant (A) 33 ms (B) 5 ms (C) 3.3 ms (D) 50 ms Sol. 1k 1k at t = 0 C R = 0.5 k . Time constant . = RC = 0.5 × 103 × 100 × 10�6 = 50 × 10�3 S = 50 m/s Example 45. In the circuit shown in figure C1 = 2C2. Switch S is closed at time t = 0. Let i1 and i2 be the currents flowing throughC1 and C2 at any time �t�, then the ratio i1/i2 (A) is constant (B) increases with increase in time t C2 C1 V S RR (C) decreases with increase in time t (D) first increases then decreases Sol. Here 1 t RC 1 I V e R . . and 2 t RC 2 I V e R . . . 1 1 2 2 t RC t 1 1 1 R C C t
2 RC I e e I e . . . . . . . . . . . . . . 1 2 . 1 2 t C C 1 RC C 2 I e I . . . Hence option (B) is correct. Example 46. In the R�C circuit shown in the figure the total energy of 3.6 × 10�3 J is dissipated in the 10 . resistorwhen the switch S is closed. The initial charge on the 2µF 10 S capacitor is
ELECTRIC CURRENT www.physicsashok.in 37 (A) 60 µC (B) 120 µC (C) 60 2 µC (D) 602 µC Sol. According to conservationprinciple of energy: Total energy stored on capacitor appears as heat in resistor. . 20 3 q H 3.6 10 2C . . . . . or q02 = 2C × 3.6 × 10�3 q02 = 2 × 2 × 10�6 × 3.6 × 10�3 q02 = 14.4 × 10�9 . q0 = 12 × 10�5 q0 = 120 µC Example 47. The capacitors shown in figure has been charged to a potential difference ofVvolts, so that it carries a charge CVwith both the switches S1 and S2 remaining open. Switch S1 is closed at t = 0.At t = R1C switch S1 E S2 C + � R1 R2 S1 is opened and S2 is closed. Find the charge on the capacitor at t = 2 R1C + R2C. Sol. When t < R1C 1t q CV e R C . . At t0 = R1C, 1 0 q CV e CV e . . . when S2 is closed, q0 �q0 R1 R2 t = t E 0 At instant t (t > t0) . . 1 2 E q I R R 0 C . . . . or (R1 + R2)CI = EC � q or . . 1 2 R R C dq EC q dt . . . �q R1 R2
E + +q I I or . . 0 1 q t 1 2 CV t R C e R R C dq dt EC q . . . . . . or . . . . . . 1 q t 1 2 CV R C e R . R C...ln EC. q .. . t or . . . . 1 2 1 CV R R C ln EC ln EC q t R C e . . . . . . . . . . . . . . . . . . or . . 1 2 1 EC CV R R C ln e t R C EC q . . . . . . . . . . . . . . . But t = 2R1C + R2C Putting the value,
ELECTRIC CURRENT www.physicsashok.in 38 2 q CE 1 1 CV e e . . . . . .. .. Example 48. Inthe figure shown initiallyswitch is open for a long time. Nowthe switch is closed at t = 0. Find the charge on the rightmost capacitor as a function of time given that it was initiallyunchanged. V R R C C S Sol. Step-I :Discuss the circuit,when switch is open. Here I V 2R . Also, 0 q CIR CV 2 . . II I +q �q0 Step-II : Discuss the circuit after closing the switch : 1 V I R q 0 2 2 C . . . Also, . . 0 1 V R q q q I 0 2 2 C . . . . . or 0 1 V I R q q q 0 2 2 C C C . . . . . V R R V2R2 q q �q1 q1 q +q�q0 1 �(q +q�q0 1) or 0 V I R q q V IR 0 2 2 C C 2 2 . . . . . . or 0 V IR q q 0 C C . . . . or . . 0 q q IR V C.
. . or 0 dq q q R V dt C. . . . . . . . . or q t 0 0 0 Rdq dt V q q C . . . . . . . . . . . or q 0 0 q q RC ln V t C . . . . . .. . . . . . . .. . . . . . .. or 0 0 V q q RC ln C t V qC . . . . . . . . . . . . . . . . . . . .. .. or 0 t RC 0 V q q C e V qC . . . . .
ELECTRIC CURRENT www.physicsashok.in 39 or t V q0 q V q0 eRC C C . . . . . . . . . . . . . . . . . ...(i) But potentialdiffer accross each capacitors are sume. 0 1 1 q q q q C C . . . or q0 + q = 2q1 . 0 1 q q q 2. . Putting the value of q, t RC 1 q CV 1 1 e 2 2 . . . . . . . . . THINKING PROBLEMS 1. Is a current-carrying conductor electrically charged? 2. Is there an electric field near the surface of a conductor carrying direct current? 3. Is current a scallar or vector? 4. A potential difference V is applied to a copper wire of diameter d and length l.What is the effect on the electron drift velocity of (a) doubling l, (b) doubling V, and (c) doubling d? 5. A current i enters the top of a copper sphere of radius R and leaves through the diametrically opposite point.Are all parts equally effective in dissipating Joule heat? 6. Account for the increase in the resistance ofmetals with rise in temperature. 7. Answer briefly� how can three resistances of values 2., 3. and 6. be connected to produce an effective resistance of 4. ? 8. How can an electric heater designed for 220V be adopted for 110V without changing the length of the coil and also without a change in the consumed power ? 9. The brilliance of lamps in a roomnoticeably drops as soon as a highpower electric iron is switched on and after a short interval, the bulbs regain their original brilliance. Explain. 10. Acurrent is passed through a steelwire which gets heated to a dull red. Then half the wire is immersed in cold water. The portion out of the water becomes brighter.Why? 11. For manual control of the current of a circuit, two rheostats in parallel are preferable to a single rheostat. Why? 12. In a hollow nonconducting pipe, there are two streams of ions in opposite directions. The ions of one streamare negatively charged and constitute a current of strength l and those of
the other stream are positively charged and constitute a current of the same strength.What is the total current through the pipe? 13. The drift velocity of electrons is quite small. How then does a bulb light up as soon as the switch is turned on, although the bulbmay be quite far fromthe switch? 14. Consider a voltaic cell in the open circuit with the copper plate at a higher potential with respect to the zinc plate. The electrolyte between the plates is a conducting medium. So the charges on the plates should be neutralized at once as it happens when two charged conductors are joined by a wire.Why are the charges not neutralized immediately? 15. Does emf have electrostatic origin? 16. Why is it easier to start a car engine on a warmday than on a chilly day? 17. The resistance of the human body is about 10 k.. If the resistance of our body is so large, why does one
ELECTRIC CURRENT www.physicsashok.in 40 experience a strong shock from a live wire of 220 V supply? 18. Lay people have the notion that a person touching a high power line gets �stuck� to the line. Is this true? If not, what is the fact? 19. When a direct current flows through a conductor, the amount of energy liberated is VQ, where Q is the charge passing through the conductor and V is the potential difference, while an energy of VQ/2 is liberated when a capacitor is discharged.Why? 20. An ordinary cellwith a small emf can produce larger current than an electrostaticmachine which generates thousands of volts? Explain. THINKING PROBLEMS SOLUTIONS 1. No, the amount of positive and negative charge in any elementary volume remains the same through the electrons are inmotion. So a current-carrying conductor is electrically neutral. 2. There is an electric field inside the conductor. This is equal to the rate of fall of potentialalong the conductor. There is continuity of this field outside the surface of the conductor. 3. Current is a scalar but current density is a vector. 4. u = j/ne = .E/ne where s is the conductivity and E is the electric field (byOhm�s law j = .E), or u = (./ne) (V/l) (. E = V/l). Obviously from this expression (a) u is halved if l is doubled, (b) u is doubled if V is doubled, and (c) u remains uncharged on doubling d. 5. No, the resistance of elements at right angles to the diameter varies fromelement to element but current is the same through all sections and so heating effect varies fromsection to section. 6. metals have mobile electrons.With increase of temperature, the lattice vibrations increase in amplitude. Thus, the probability of electrons striking ionic cores increases. This amounts to increase in resistance with temperature. 7. Connect the 3. and 6. resistors in parallel and the 2. resistor in series with the combination of 3. and 6.. Then 3. and 6. in parallel will sumup to 3 6 2 3 6 . . . . and this, with 2. in series, willwork out to 4.. 8. Join the ends of the coil and apply the supply voltage (110V) between this common terminal and the midpoint of the coil as shown in the figure. Let P be the power of the entire coil when the coltage applied is V volts. TheP = V2/R where R is the resistance of the coil. When it is connected as shown in the figure, the two parts are in parallel and the resistance of each part becomes R/2. Therefore, P1 = (V/2)2/(R/2)=P/2 Total power = P/2 + P/2 = P 9. The cold resistance of the coil of the iron ismuch smaller thanwhen it is hot. So when the iron is switched
on, there is a large drop of voltage and consequently the bulbs do not receive the proper voltage.As the coil gets heated, its resistance increases and so the voltage drop is made up when it is fully heated. This is why after some time the bulbs region their brilliance. 10. The resistance of the immersed portion decreases and so the current through the entire wire increases. This is why the portion outside the water becomes brighter. 11. Suppose, to produce a certain current i in a circuit, the length of the single rheostat wire required is l. Then i = E/lr where E is the emf of the cell and r is the resistance per unit length of the wire. To produce the same current in the same circuit by using two rheostats in parallel, let l� be the length of each rheostat wire. Then i = E/(l�r/2) = 2E/l�r. Obviously, l�= 2l. Thusmore length of eachwire is required to produce the same current. This is definitely of advantage. 12. The current constituted by the negative ionsmoving opposite to the positive ions has the same sense as the current constituted by the positive ions. Hence the total current is 2I and not zero. 13. For the bulb to light up, it is not necessary for the same electron to travel from the switch to the bulb. When the switch is turned on, every electron in the circuit begins to move simultaneoulsy, including those in the filament of the bulb. 14. The charges are prevented frombeing neutralized because the electrostatic field due to the charges on the plates is opposed by the charges of ions that migrate into the solution.
ELECTRIC CURRENT www.physicsashok.in 41 REASONINGTYPE QUESTIONS A statement of assertion (A) is given and a Corresponding statement of reason (R) is given just below it of the statements, mark the correct answer as � (A) If both A and R are true and R is the correct explanation of A. (B) If both A and R are true but R is not the correct explanation of A. (C) If A is true but R is false. (D) If both A and R are false. (E) If A is false but R is true. 1. Assertion (A) : When a wire is stretched so that its thickness is halved, its resistance would become 16 times. Reason (R) : The data is insufficient to predict. 2. Assertion (A) : A current flows in a conductor only when there is an electric field within the conductor. Reason (R) : The drift velocity of electron in presence of electric field decreases. 3. Assertion (A) : A current carrying wire should be changed. Reason (R) : The current in a wire is due to flow of free electrons in a definite direction. 4. Assertion (A) : The wires supplying current to an electric heater are not heated appreciably. Reason (R) : Resistance of connected wires is very small and H . R . 5. Assertion (A) : In meter bridge experiment, a high resistance is connected in series with the galvanometer. Reason (R) : As resistance increases current through the circuit increases. 6. Assertion (A) : A 60 watt bulb has greater resistance than a 100 watt bulb. Reason (R) : V 2 P R . 7. Assertion (A) : The conductivity of an electrolyte is very low as compared to ametal at room temperature. Reason (R) : The number density of free ions in electrolyte is much smaller compared to number density of free electrons in metals. Further, ions drift much more slowly, being heavier. 8. Assertion (A) : Terminal voltage of a cell is greater than emf of cell, during charging of the cell. Reason (R) : The emf of a cell is always greater than its terminal voltage. 9. Assertion (A) : Material used in the construction of a standard resistance is constantan or manganin. Reason (R) : Temperature coefficient of these materials is very small. 10. Assertion (A) : If the current of a lamp decreases by 20%, the percentage decrease in the illumination of the lamp is 40%. Reason (R) : Illumination of the lamp is directly proportional tot he current through lamp. 11. Assertion (A) : Heater wire must have high resistance and high melting point. Reason (R) : If resistance is high, the electrical conductivity will be less. 12. Assertion (A) : The range of given voltmeter can be increased. Reason (R) : By adjusting the value of resistance in series with galvanometer the range of voltameter can be adjusted.
ELECTRIC CURRENT www.physicsashok.in 42 LEVEL # 1 1. The resistance of tungsten increases with increasing temperature. As a result, the relation between the current, . , flowing in the tungsten filament of an electric lamp and the potential difference, V, between its ends is of the form. (A) IO V (B) IO V (C) I O V (D) IO V 2. The current in a copper wire is increased by increasing the potential difference between its ends. Which one of the following statements regarding n, the number of charge carriers per unit volume in the wire, and Vd, the drift velocity of the charge carriers, is correct? (A) n is unaltered but vd is decreased. (B) n is unaltered but vd is increased. (C) n is increased but vd is decreased. (D) n is increased but vd is unaltered. 3. A potentiometer is to be calibrated with a standard cell using the circuit shown in the diagram. The balance point is found to be near L. To improve accuracy the balance point should be nearer M. This may be achieved by (A) replacing the galvanometer with one of lower resistance. (B) replacing the potentiometer wire one of higher resistance per unit length. (C) putting a shunt resistance in parallel with the galvanometer. (D) increasing the resistance R. 4. When the switch 1 is closed, the current through the 8. resistance is 0.75 A. When the switch 2 is closed (only), the current through the 2. resistance is 1 A. The value of . is (A) 5 V (B) 5 2 V (C) 10 V (D) 15 V 5. The time constant . for the shown RC circuit is (A) RC (B) 2 RC (C) 2 RC (D) not defined R R . 6. Two cells of equal emf of 10 V but different internal resistances 3. and 2. are connected in series to an external resistance R. The value of R that makes the potential difference zero across the terminals of one of any cells is (A) 5. (B) 6. (C) 1. (D) 1.5. 7. The resistor in which the maximum heat is produced is given by (A) 2. (B) 3. (C) 4. (D) 12.
8. n resistances each of resistance R are joined with capacitors of capacity C (each) and a battery of emf E as shown in the figure. In steady state condition ratio of charge stored in the first and last capacitor is E R C R C R (A) n : 1 (B) (n�1) : R (C) (n2 + 1) : (n2 � 1) (D) 1 : 1
ELECTRIC CURRENT www.physicsashok.in 43 9. What is the equivalent capacitance between A and B in the circuit shown. (A) 6.F (B) 1.5 .F (C) Zero (D) 2.F 10. A resistance R = 12. is connected across a source of emf as shown in the figure. Its emf changes with time as shown in the graph. What is the heat developed in the resistance in the first four second. R = 12 . . Source . (Volt) 24 4 t(s) (A) 72 J (B) 64 J (C) 108 J (D) 100 J 11. A source of constant potential difference is connected across a A B conductor having irregular cross-section as shown in the Figure. Then P Q (A) electric field intensity at P is greater than that at Q (B) rate of electric crossing per unit area of cross section at P is less than that at Q (C) the rate of generation of heat per unit length at P is greater than that at Q (D) mean kinetic energy if free electorn at P is greater than that at Q 12. Suppose a voltmeter reads the voltage of a very old cell to be 1.40 V while a potentiometer reads its voltage to be 1.55 V. What is the internal resistance of the cell (A) 20. (B) 30. (C) 10. (D) 40. 13. In the above question, What is the current it would supply to a 5. resistor. Assume the voltmeter resistance be be 280.. (A) 44 A (B) 0.044 A (C) 4.4 A (D) None of the above. 14. If 1 . , 2 . and 3 . are conductances of three conductors then their equivalent conductance when they are joined in series will be (A) 1 . + 2 . + 3 . (B) 1 2 3 1 1 1 . . . . . (C) 1 2 3 1 2 3 . . . . .. .. (D) None of these 15. A conductor is made of an isotropic material (resistivity .) has rectangular cross-section. Horizontally dimension of the rectangle decreases linearly from 2x at one end to x at the other end and vertial dimension increases from y to 2y as shown in Figure. Length of the conductor along the axis is equal to . . A battery is connected across this conductor then (A) resistance of the conductor is equal to 4.. 9xy . (B) rate of generation of heat per unit length is maximum at middle cross-section. (C) drift velocity of conduction electrons is minimum at middle section. (D) at the ends of the conductor, electric field intensity is same.
16. Find the current through 4. resistor just after making the circuit (A) 0 A (B) 6 A (C) 12 A (D) 2 A 17. 1 m long metallic wire is broken into two unequal parts P and Q. P part of the wire is uniformly extended into another wire R. Length of R is twice the length of P and the resistance R is equal to that of Q. Find the ratio of the resistance of P and R (A) 1 : 4 (B) 1 : 3 (C) 1 : 2 (D) 1 : 1
ELECTRIC CURRENT www.physicsashok.in 44 18. Which of the following statements is/are correct for potentiometer circuit (A) Sensitivity varies inversely with length of the potentiometer wire (B) Sensitivity is directly proportional to potential difference applied across the potentiometer wire. (C) Accuracy of a potentiometer can be increased only by increasing length of the wire (D) Range depends upon the potential difference applied across the potentiometer wire. 19. In the given circuit the ammeter reading is zero. What is the value of resistance R ? (A) R = 100. (B) R = 10. (C) R = 0.1. (D) None of these 20. What is the equivalent resistance between A and B in the given circuit diagram. (A) 2. (B) 12. (C) 20. (D) 10. 21. A cell of emf. 1.5 V and internal resistance 0.5. is connected to a (non-linear) conductor whose . -V graph is shown in Figure. Find the current drawn from the cell and its terminal voltage (A) 1.5 A and 2 V (B) 1 A and 1 V (C) 1 A and 2 V (D) 2 A and 1.5 V 22. You are given several identical resistances each of value R = 10. and each capable of carrying a maximum current of 1A. It is required to make a suitable combination of these resistances to produce a resistance of 5. which can carry a current of 4 A. The minimum number of resistances required is (A) 4 (B) 10 (C) 8 (D) 20 23. A conductor or area of cross section A having charge carriers, each having a charge q is subjected to a potential V. the number density of charge carriers in the conductor is n and the charge carriers (along with their random motion) are moving with a velocity v. If . is the conductivity of the conductor and . is the average relaxation time, then (A) . . . nq2 m (B) nq2 m. . . (C) 2 nq2m. . . (D) 2 2nq m. . . 24. A vacuum diode consists of plane parallel electrodes separated by a distance d and each having an area A. On applying a potential V to the anode with respect to the cathode a current I flows through the diode. Assume that the electrons are emitted with zero velocity and they do not change the field between the electrodes. The electron velocity is v and the charge density is . at any point between the electrodes at a distance x from the cathode. If I is the equivalent current, m is the mass of each charge carrier, then (A) md v . 2eVx (B) 2eVxA 2 . . md (C) 2md
v . eVx (D) eVxA 2 . . 2md 25. A long round conductor of cross-sectional area A is made of a material whose resistivity depends on the radial distance r from the axis of the conductor as 2 r. . . , . is a constant. The total resistance per unit length of the conductor is R and the electric field strength in the conductor due to which a current I flow in it is . . (A) 2 A R 2 .. . (B) 2 A R 4 .. . (C) 2 A 2 .. I . . (D) 2 A 4..I . .
ELECTRIC CURRENT www.physicsashok.in 45 26. A 100 W bulb B1 and two 60 W bulbs B2 and B3 are connected to a 250 V source as shown in figure. NowW1, W2 and W3 are the output powers of the bulbs B1, B2 and B3 respectively. Then (A) W1 > W2 = W3 (B) W1 > W2 > W3 (C) W1 < W2 = W3 (D) W1 < W2 < W3 [JEE� 2002 (Scr)] 27. Variation of current passing through a conductor as the voltage applied across its ends is varied as shown in Fig. 9.55. If the resistance is determined at points A, B, C and D. We will find that : (A) Resistance at C and D are equal (B) Resistance at B is higher than at A (C) Resistance at C is higher than at B (D) Resistance at A is lower than at B 28. The P. D. between the points A and B in the circuit shown here is 16 V. Which is/are the correct statement(s) out of the following ? (A) the current through the 2 . resistor is 3.5 A (B) the current through the 4 . resistor is 2.5 A 4. 1. 1. 3. 2. A B 9V 3V (C) the current through the 3 . resistor is 1.5 A (D) the P.D. between the terminals of the 9 V battery is 7 V. 29. A and B are two points on a uniform ring of resistance R. The .ACB = ., where C is the centre of the ring. The equivalent resistance between A and B is (A) . . . . . (2 ) 4R2 (B) . .. . . .. . . . . 2 R 1 (C) R . . 2 (D) R . . . . 4 2 30. The charge flowing through a resistance R varies with time t as Q = at � bt2. The total heat produced in R is (A) 6ba3R (B) 3ba3R (C) 2ba3R (D) ba3 R 31. A resistance R carries a current I . The heat loss to the surroundings is . (T�T0) where . is a constant, T
is the temperature of the resistance, and T0 is the temperature of the atmosphere. If the coefficient of linear expansion is ., the strain in the resistance is (A) proportional to the length of the resistance wire (B) equal to .. I2 R (C) equal to 21 .. I2 R (D) equal to . . (IR) 32. The potential of the point O is (A) 3.5 V (B) 6.5 V (C) 3 V (D) 6 V 33. In the circuit shown in Figure the heat produced in the 5 ohm resistor due to the current flowing through it is 10 calories per second. The heat generated in the 4 ohms resistor is 4. 6. (A) 1 calorie/sec (B) 2 calories/sec 5. (C) 3 calories/sec (D) calories/sec [JEE� 1981]
ELECTRIC CURRENT www.physicsashok.in 46 34. A battery of internal resistance 4. is connected to the network of resistance as shown. In order that the maximum power can be delivered to the network, the value of R in . should be [JEE�1995] (A) 49 (B) 2 (C) 83 (D) 18 35. A steady current flows in a metallic conductor of non-uniform cross-section. The quantity/quantities constant along the length of the conductor is/are [JEE�1997] (A) current, electric field and drift speed. (B) drift speed only (C) current and drift speed (D) current only. 36. In the circuit P . R, the reading of the galvanometer is same with switch S open or closed. Then [JEE�1999] (A) R G . . . (B) P G . . . (C) Q G . . . (D) Q R . . . 37. A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by .T in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. The temperature of the wire is raised by the same amount .T in the same time t, the value of N is [JEE�2001] (A) 4 (B) 6 (C) 8 (D) 9 38. The effective resistance between points P and Q of the electrical circuit shown in the figure is [JEE�2002] (A) 2Rr R . r (B) 8 . . 3 R R r R r . . (C) 2 r + 4 R (D) 5 2 2R . r 39. Express which of the following set ups can be used to verify Ohm�s law? (A) A V (B) A V (C) A V (D) A V 40. The three resistance of equal value are arranged in the different combination shown below. Arrange them in increasing order of power dissipation. [JEE�2003] (I) a (II) b (III) c (IV) c (A) i (B) i (C) i (D) i R R R
R R 4R E 6R 4. P Q R S G v P Q r r 2R2R 2R 2R2R 2R
ELECTRIC CURRENT www.physicsashok.in 47 41. In the givencircuit, no current is passing through the galvanometer. If the crosssectional diameter ofAB is doubled then for nullpoint ofgalvanometer the value ofACwould [JEE�2003 (Scr)] G A x (A) x (B) x/2 (C) 2x (D) None C B 42. Six equal resitances are connected between points P, Qand R as shown in the figure. Thenthe net resistancewillbemaximumbetween [JEE�2004 (Scr)] P Q R (A) P and Q (B) Q and R (C) P and R (D) any two points 43. For the post office box arrangement to determine the value of unknown resistance, the unknown resistance should be connected between [JEE�2004 (Scr)] B1 C1 B C D (A) B and C (B) C and D (C) Aand D (D) B1 and C1 44. A capacitor is charged using an external battery with a resistance x in series. The dashed line shows the variation of .n . with respect to time. If the resistance is changed to 2x, the new graph will be [JEE�2004] (A) P (B) Q (C) R (D) S 45. A moving coil galvanometer of resistance 100. is used as an ammeter using a resistance 0.1.. Themaximum deflection current in the galvanometer is 100.A. Find the minimum current in the circuit so that the ammeter shows maximum deflection. [JEE�2005] (A) 100.1 mA (B) 1000.1 mA (C) 10.01 mA (D) 1.01 mA 46. In the figure shownthe current through 2. resistor is (A) 2 A (B) 0 A 10V 20V 10 2 5 (C) 4 A (D) 6 A [JEE�2005 (Scr)] 47. Agalvanometer has resistance 100 ..and it requires current 100 µAfor full scale deflection.Aresistor 0.1 . is connectedtomake it anammeter. the smallest current required inthe circuit to produce the fullscale deflection is [JEE�2005 (Scr)] (A) 1000.1 mA (B) 1.1 mA (C) 10.1 mA (D) 100.1 mA 48. Consider a cylindricalelement as shown in the figure. Current flowing through the element is I and resistivityofmaterialof the cylinder is .. Choose the correct option out the following. 4r I l/2 l/2 2r A B C (A) Power loss in second half is four times the power loss in first half. (B) Voltage drop in first half is twice of voltae drop in second half. (C) Current density in both halves are equal.
(D) Electric field in both halves is equal. [JEE�2006]PQRS t ln I
ELECTRIC CURRENT www.physicsashok.in 48 49. Aresistance of 2 . is connected across one gap of a meter-bridge (the length of thewire is 100 cm) and an unknown resistance, greater than 2., is connected across the other gap. When these resistances are interchanged, the balance point shifts by20 cm. Neglecting any corrections, the unknown resistance is [JEE�2007] (A) 3 . (B) 4 . (C) 5 . (D) 6 . MORE THAN ONE CHOICE MAY BE CORRECT 50. Capacitor C1 of capacitance 1micro-farad and capacitor C2 of capacitance 2microfarad are separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at time t = 0. [JEE�1989] (A) The current in each of the two discharging circuits is zero at t = 0. (B) The currents in the two discharging circuits at t = 0 are equal but not zero. (C) The currents in the two discharging circuits at t = 0 are unequal. (D) Capacitor C1, losses 50% of its initial charge sooner than C2 loses 50% of its initial charge. 51. In the circuit shown in Figure the current through (A) the 3. resistor is 0.50 A. (B) the 3. resistor is 0.25 A (C) the 4. resistor is 0.50 A (D) the 4. resistor is 0.25 A 52. Which of the following is/are wrong? (A) A current carrying conductor is electrically charged. (B) There is an electric field inside a current carrying conductor. (C) For manual control of the current of a circuit, two rheostats in series are preferable to a single rheostat. (D) None of these 53. A constant voltage is applied between the two ensd of a uniform metallic wire. Some heat is produced in it. The heat developed is doubled if: (A) both the length and radius of the wire are halved. (B) both the length and radius of the wire are doubled. (C) the radius of the wire is doubled. (D) the length of the wire is doubled and the radius of the wire is halved. 54. All the resistance in the given Wheatstone bridge have different values and the current through the galvanometer is zero. The current through the galvanometer will still be zero if, (A) the emf of the battery is doubled. (B) all resistance are doubled. (C) resistance R1 and R2 are interchanged. (D) the battery and the galvanometer are interchanged. 55. Two cells of equal emf and having different internal resistance R1 and R2 (R2 > R1) are connected in series. If resistance of connecting wires is equal to R, which of the following statements is/are correct? (A) At a particular value of r, potential difference across second cell can be equal to zero. (B) If R = 0, negative terminal of second cell will be at higher potential than its positive terminal (C) Negative terminal of first cell can never be at higher potential than its positive terminal. (D) None of these 56. When electric bulbs of same power but with different marked voltages are connected in series across a power line, their brightness will be:
(A) proportional to their marked voltages. (B) inversely proportional to their marked voltages. (C) proportional to the squares of their marked voltages. (D) inversely proportional to the square of their voltages. 57. Two conductors made of the same material have lengths L and 2 L, but have equal resistances. The two are connected in series in a circuit in which current is flowing. Which of the following is/are correct? (A) The potential difference across the two conductors is the same. (B) The electric drift velocity is larger in the conductor of length 2L. (C) The electric field n the first conductor is twice that in the second. (D) The electric field in the second conductor is twice that in the first. 2. 2. 2. 2. 2. 4. 3. 9V 8. 8. R1 R2 R3 R4
ELECTRIC CURRENT www.physicsashok.in 49 58. Mark correct statements: (A) Heat is always generated in a battery whether it charges or discharges. (B) When a battery supplies current in an external circuit, heat generated in it is always less than electrical power developed in it. (C) Potential difference across terminals of a battery is always less than its emf. (D)None of these. 59. A mocrometer has a resistance of 100. and a full scale range of 50. A. It can be used as a voltmeter or as a higher range ammeter provided a resistance is added to it. Pick the correct range and resistance combinations. (A) 50 V range with 10 k . resistance in series. (B) 10 V range with 200 k . resistance in series. (C) 5 mA range with 1. resistance in parallel. (D) 10 mA range with 1. resistance in parallel. 60. An electric current is passed through a circuit containing three wires arranged in parallel. If the length and radius of the wires are in ratio 2 : 3 : 4 and 3 : 4 : 5, then the ratio of current passing through wires would be: (A) 54 : 64 : 75 (B) 9 : 16 : 25 (C) 4 : 9 : 25 (D) 3 : 6 : 10. 61. When a galvanometer is shunted with a 4. resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with a 2. wire, the further reduction in the deflection will be (the main current remains the same). (A) 8/15 of the deflection when shunted with 4. only. (B) 5/13 of the deflection when shunted with 4. only. (C) 3/4 of the deflection when shunted with 4. only. (D) None of these 62. During the charging of a storage battery, the current is 22 A and the voltage is 12 V. The rate of heat generated in the battery is 12 W. The rate of change of internal energy is: (A) 240 J/s (B) 252 J/s (C) 264 J/s (D) 126 J/s. 63. A cell of emf 5 V, internal resistance 1. will give maximum power output to: (A) a single resistor of 1. (B) two 1. resistors connected in series. (C) two 1. resistors connected in parallel. (D) two 2. resistors connected in parallel. 64. In the circuit shown in figure, the current through: (A) the 3. resistor is 0.50 A. (B) the 3. resistor is 0.25 A. 3. 2. 2. 2. 2. 2.(C) the 4. resistor is 0.50 A. (D) the 4. resistor is 0.25 A. 9V 8. 8. 4. 65. For the circuit shown in the figure 6k 1.5k 2k R1 R2 24V RL I [JEE�2009] (A) the current I through the battery is 7.5 mA (B) the potential difference across RL is 18 V (C) ratio of powers dissipated in R1 and R2 is 3 (D) if R1 and R2 are interchanged, magnitude of the power dissipated in RL will decrease by a factor of 9
ELECTRIC CURRENT www.physicsashok.in 50 FILL IN THE BLANKS 1. An electric bulb rated for 500 watts at 100 volts is used in a circuit having a 200 volts supply. The resistance R that must be put in series with the bulb, so that the bulb delivers 500 Watts is ........... ohms. (1997) 2. The equivalent resistance between points A and B of the circuit (Figure) given below is ........... .. 3. In the circuit (Figure) shown above, each battery is 5 V and has an internal resistance of 0.2 ohm. A 2R 2R R B V Fig - 1 Fig - 2 The reading in the ideal voltmeter V is ......... V. [JEE� 1997] TRUE / FALSE 4. Electrons in a conductor have no motion in the absence of a potential difference across it. [JEE� 1982] 5. The current-voltage graphs for a given metallic wire at two different temperatures T1 and T2 are shown in the figure. [JEE� 1985] The temperature T2 is greater than T1. TABLE MATCH 6. Column I Column II (A) Current (P) Mircoscopic quantity (B) Current Density (Q) Macroscopicquantity (C) Electric field (R) Parallel to the conductor boundaries (D) Resistance (S) Flux associated with current density 7. Two bulbs A and B consume same power P when operated at voltage VA and VB respectively. Bulbs are connected with a supply of d.c source then: Column I Column II (A) In series connection, the ratio of (P) RA/RB potential difference across A and B (B) In series connection, the ratio of (Q) V2A/V2B power consumed by A and B (C) In parallel connection, the ratio of (R) RB/RA current in A and B (D) In parallel connection, the ratio of (S) V2B/V2A power consumed in A and B 8. In a R-C circuit. Column I Column II (A) Charging current at tiem t = 0 (P) 1 2 2 CV (B) Discharging current at t = 0 (Q) Maximum (C) While charging energy stored (R) Capacitor becomes short circuit (D) While charging energy dissipated as heat (S) Exponential law VT1 T2 I
ELECTRIC CURRENT www.physicsashok.in 51 9. A battery has an emf E and internal resistance r. A variable resistor R is connected across the terminals of the battery. Column I Column II (A) Current in the circuit is maximum (P) R . . (B) Potential difference across the (Q) R = 0 terminals is maximum (C) Power delivered to the resistor is maximum (R) i Er . (D) Power delivered to the load is zero (S) r = R 10. Consider two identical cells each of emf E and internal resistance r connected to a load resistance R. Column I Column II (A) For maximum power transfer to load (P) 2 4Er if cells are connected in series (B) For maximum power transfer to load if (Q) 2 2Er cells are connected in parallel (C) For series combination of cells (R) eq E . E , eq 2r . r (D) For parallel connection of cells (S) 2 eq E . E , 2 eq r . r PASSAGE TYPE QUESTIONS PASSAGE # I A physics instructor devises a simple electrical circuit setup in which one can easily insert various resistors and capacitors in series and parallel combinations. One can have only resistor combinations, only capacitor combinations, or capacitor-resistor combinations. The circuit is usually used for DC (direct current studies but can also be used for AC (alternating current) studies. The DC battery voltage is 6 volts. The AC rms voltage is 120 volts (at 60 Hz.) The student inserts the resistors and /or capacitors as instructed and has available suitable ammeters and voltmeters for both DC and AC experiments. (There are also three resistors, each of 2 ohms resistance. There are also three capacitors, each of 1 microfarad capacitance.) 1. All three resistors are connected in series and the combination is connected to the 6-volt DC battery. What voltage drop occurs across each individual resistor as measured by the voltmeter? (A) 0.33 V (B) 1.0 V (C) 2.0 V (D) 6.0 V 2. One capacitor and one resistor are connected in parallel. The ends of this combination are then connected t the 6-V DC battery, what are the final current and voltage, respectively, across the 1 microfarad capacitor? (A) 0 A, 6 V (B) 0.33 A, 3 V (C) 0.33 A, 6 V (D) 6 A, 6 V 3. Two of the 2-ohm resistors are connected in parallel and the 120-V AC voltage
is applied to the ends of this parallel combination. What current will the AC ammeter measure if connected so it measures only the current through one of the resistors? (A) 1 A (B) 2 A (C) 60 A (D) 120 A 4. Two resistors are connected in parallel and the set of parallel resistors is then connected in series with the third resistor. If this series parallel combination is connected across the 6-volt battery. What total DC current is drawn from the battery? (A) 0.5 A (B) 2 A (C) 3 A (D) 6 A 5. One capacitor and one resistor are connected is series. The 120 V, 60 Hz, A.C. voltage is then applied to this series �RC� circuit. What is the AC current through this series circuit? (A) 0.045 A (B) 0.50 A (C) 0.72 A (D) 40.0 A
(A) 0.9 W (B) 2.0 W (C) 4.4 W (D) 18 W A 6-votl battery is connected across a 2-ohm resistor. What is the heat energy dissipated in the resistor in 5 minutes? (A) 430 joules (B) 560 joules (C) 4300 joules (D) 5400 joules minutes is used to heat 2 kg of water (which is thermally insulated so that no heat escapes). The initial (A) 0.9 W (B) 2.0 W (C) 4.4 W (D) 18 W A 6-votl battery is connected across a 2-ohm resistor. What is the heat energy dissipated in the resistor in 5 minutes? (A) 430 joules (B) 560 joules (C) 4300 joules (D) 5400 joules minutes is used to heat 2 kg of water (which is thermally insulated so that no heat escapes). The initial ELECTRIC CURRENT PASSAGE # II
A set of experiments in the physics lab is designed to develop understanding of simple electrical circuit principles for direct current circuits. The student is given a variety of batteries, resistors, and DC meters; and is directed to wire series and parallel combinations of resistors and batteries making measurements of the currents and voltage drops using the ammeters and voltmeters. The student calculates expected current and voltage values using Ohm�s law and Kirchoff�s circuit rules and then checks the results with the meters.
6. A student connects a 6-volt battery and a 12-volt battery in series and then connects this combination across a 10-ohm resistor. What is the current in the resistor? (A) 0.8 A (B) 0.9 A (C) 1.8 A (D) 3.6 A 7. Resistorsof 4ohmsand8ohmsareconnectedinseries.Abatteryof 6voltsisconnectedacrosstheseries combination. How much power, in watts, is consumed in the 8-ohm resistor? (D) 24 W (A) 0.67 W (B) 2W (C) 12 W 8. Two 4-ohm resistors are connected in series and this pair is connected in parallel with an 8-ohm resistor. A 12 volt battery is connected across the ends of this parallel set. What power, in watts, is consumed in the 8-ohm resistor in this case? 9. 10. A 12-volt battery is connected across a 4-ohm resistor and the heat energy dissipated in the resistor in 10 temperature of the water is 20°C and the specific heat of the water is 4184 joule/kg-°C. What is the final temperature of the water at the end of the 10 minutes of heating? (A) 22.6 °C (B) 28.4°C (C) 34.2°C (D) 56.4°C PASSAGE # III In the laboratory, the voltage across a particular circuit element can be measured by a voltmeter. A voltmeter has a very high resistance and should be connected in parallel to the circuit element whose voltage is being measured. Connected improperly, the voltmeter will affect the circuit, interrupt
ing it and preventing current from flowing through the circuit element that it is meant to measure. An experiment is conducted in which a voltmeter is used to investigate voltages in a circuit containing a capacitor and a light bulb. The bulb and the capacitor are connected in series with a battery and the voltmeter is placed in different position: in the first case across the capacitor in the second case across the light bulb, and in the third case across the battery (see figure 1). The voltmeter reading is recorded every 10 seconds. The voltage for Case 1 as a function of time is shown in figure 2. Vvoltmeter capacitor bulb battery Case 1 V voltmeter capacitor bulb battery Case 2 Vvoltmeter capacitor bulb battery Case 3 Voltage Time Fig. 2
A capacitor consists of two conducting plates separated by a nonconducting material. When a battery is connected to a circuit containing a capacitor and a light bulb in series, a current will flow, causing positive charge to accumulate on one capacitor plate and an equal amount of negative charge to accumulate on the other. After the current has flowed for a finite time, the capacitor will be fully charged. The ratio of the absolute amount of charge on one plate to the voltage across the plates is defined as the capacitance; this is constant for a given capacitor. A light bulb is a resistor that, when enough current flows through it, becomes hot enough to emit energy in the form of light. (Note : Assume that the battery has no internal resistance and that the resistance of the light bulb is constant).
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ELECTRIC CURRENT ELECTRIC CURRENT 11. In the circuit shows in figure 1, which of the following conditions would indicate that the capacitor was fully charged � I. A voltmeter connected across the capacitor reads a constant voltage. II. The light bulb in the circuit stops shining. III. The voltage across the bulb equals the voltage across the battery (A) I Only (B) III Only (C) I and II only (D) I, II, and III 12. Which one of the following graphs could correctly represent the voltage across the battery as a function of time during the experiment described in the passage � Time (D) Time Voltage Voltage Voltage Voltage After the experiment described in the passage is completed, the battery is taken out of the circuit and the wires are reconnected. Which of the following graphs, represents the voltage across the capacitor as a (A) (B) (C) Time
Time
13. function of time �
(A) Time Voltage (B) Time Voltage (C) Time Voltage (D) Time Voltage 14. How will the voltage across the light bulb vary with time as the capacitor is charging � (A) It will decrease, because as the capacitor plates fill will charge, they will impede further charge, which will decrease the current and the voltage across the bulb. (B) It will remain the same, because as the capacitor plates fill with charge and impede the current, the voltage output of the battery will increase to keep the current constant. (C) It will increase, because as the capacitor plates fill with charge, they will induce further charge, which will create a greater voltage across the bulb. (D) It will increase, because as the capacitor plates fill with charge, the volt
age across the capacitor will decrease, and therefore the voltage across the light bulb will increase. 15. The light bulb shown in figure 1 is replaced first with two identical resistor in series, and then with the same two resistors in parallel. The total time taken for the capacitor to charge is measured in both cases, and found to be longer for the first case. It can be deduced that � (A) When the resistance of the circuit si increased, the capacitance of the capacitor increases. (B) the presence of resistors affects the final voltage across the capacitor plates. (C) the more charge is absorbed by the resistors as the resistance of the circuit increases. (D) the presence of resistors hinders the flow of charge, thus reducing the current in the circuit. 16. withoutremovingthevoltmeter� Inthediagram below, avoltmeterisconnectedinseriestoacircuit that includesabatteryandtwobulbsin series. The bulbs, which had been shining in the absence of the voltmeter immediately stop shining. How might the circuit be modified in order to make the bulbs shine steadily again with their former brilliance voltmeter
bulb
V bulb
battery
(A) V (B) V (C) V (D) V www.physicsashok.in 53
ELECTRIC CURRENT www.physicsashok.in 54 LEVEL # 2 1. In the figure shown, 2. 2. 3. 3. 1 .F 3 .F 6V, 0.5 . find the charge on 2 .F each capacitor in steady state. [in .C ]. 2. In the given circuit diaram, 1. 4. 2. 3. AB CD EF 50 V find the current passing through wire CD [in Ampere] 3. It is required to send a current of 8 A through a circuit whose resistance is 5 . . What is the least number of cells which must be used for their purpose and how should they be connected? Emf of each cell is 2V and internal resistance is 0.5 . . 4. In the circuit shown, the capacitor is charged by a battery of emf 100 V and 1. internal resistance by closing the switch. Calculate the heat generated across 99 . resistance during the charging of 0.1 F 100 V, 1. 99 . s capacitor. [in Joule]. 5. In a Wheatstone�s bridge a battery of 2 volt and internal resistance 2 ohm is used. Find the value of the current through the galvanometer in that unbalanced condition of the bridge when P = 1 ohm, Q = 2 ohm, S = 30 ohm and resistance of galvanometer is 4 ohm. 6. A 20 volt battery with an internal resistance of 6 . is connected to a resistor of x ohms. If an additional 6 . resistance is connected across the battery find the value of x so that external power supplied by battery remains the same. 7. Find how the voltage across the capacitor C varies with time t after capacitors C varies with time t after the shorting of the switch S at C S R R E the moment t = 0. 8. A homogeneous poorly conducting medium of resistivity . fills up the space between two thin coaxial ideally conducting cylinders. The radii of the cylinders are equal to a and b, (a < b) the length of each
cylinder is . . Neglecting the edger effects. Find the resistance of the medium between cylinders.
ELECTRIC CURRENT www.physicsashok.in 55 9. Find the current flowing through the resistance R in the circuit shown in figure. The internal resistance of the batteries are negligible. 10. A circuit shown in figure has resistance R1 = 20. and R2 = 30.. At what value of the resistance Rx will the thermal power generated in it be practically independent of small variation of that resistance ? The AB R1 Rx R2 voltage between the points A and B is supposed to be constant in this case. 11. An ammeter and voltmeter are connected in series to a battery with an emf E = 6.0v. When a certain resistance is connected in parallel with the voltmeter, the reading of the latter decreases . = 2.0 times whereas the reading of ammeter increases the some number of time. Find the voltameter reading after the connecting of the resistance. 12. In the circuit shown in the figure the current through 3 . resistance is 2A. If E1 = 12V, E2 = 14V, what is the value E1 E2 E3 3. 1. 1.5. 10. 1. of E ? Internal resistance of each battery is 1 . . 13. The circuit shows a capacitor C two batteries, two resistors and a switch S. Initially S has been open for a longtime. It is then closed for a long time by how much does the change on the capacitor change over this time period ? Assume C = 10 ., E1 = 1.0 V, E3 = 3.0v, R1 = 0.20 . , R2 = 0.40 . . 14. In the given network switch S is closed at t = 0. Find the current through the 10 ohm resistor at t = 75 . sec. 10V 20V S 2.F 10. .. .. 15. Two coils connected in series have resistance of 600 . and 300 . and temperature co-efficient of 0.001 and 0.004 (°C)�1 respectively at 20°C. Find resistance of the combination at a temperature of 50°C. What is the effective temperature co-efficient of combination. 16. The resistance of the galvanometer G in the circuit is 25.. The meter deflects full scale for a current of 10 mA. The meter behaves as an ammeter of three different ranges. The range is 0�10 A, if the terminals R1 R2 R3 G O and P are taken; range is 0�1 A between O and R. Calculate the resistance R1, R2 and R3.
ELECTRIC CURRENT www.physicsashok.in 56 17. Calculate the steady current in the 2-ohm resistor shown in the circuit in the figure. The internal resistance of the battery is negligible and the capacitance of the conductor C is 0.2 microfarad. OR Two resistors, 400 ohms, and 800 ohms are connected in series with a 6-volt battery. It is desired to measure the current in the circuit. An ammeter of a 10 ohms resistance is used for this purpose. what will be the reading in the ammeter? similarly, If a voltmeter of 10,000 ohms resistance is used to measure the potential difference across the 400-ohms resistor, What will-be the reading in the voltmeter. LEVEL # 3 1. A part of circuit in a steady state along with the currents flowing in the branches, the values of resistances etc., is shown in the figure. Calculate the energy stored in the capacitor C (4.F) [JEE� 1986] 2. An infinite ladder network of resistances is constructed with a 1 ohm and 2 ohm resistances, as shown in figure. [JEE� 1987] The 6 volt battery between A and B has negligible internal resistance: (i) Show that the effective resistance between A and B is 2 ohms. (ii) What is the current that passes through the 2 ohm resistance nearest to the battery? 3. An electrical circuit is shown in Figure. Calculate the potential difference across the resistor of 400 ohm, as will be measured by the voltmeter V os resistance 400 ohm, either by applying Kirchhoff�s rules or otherwise. [JEE� 1996]
ELECTRIC CURRENT www.physicsashok.in 57 4. Find the emf (V) and internal resistance (r) of a single battery which is equivalent to a parallel combination of two batteries of emfs V1 and V2 and internal resistance r1 and r2 respectively, with polarities as shown in figure. [JEE� 1997] 5. A leaky parallel plane capacitor is filled completely with a material having dielectric constant K = 5 and electrical conductivity . = 7.4 x 10�12 .�1 m�1. If the charge on the plane at instant t = 0 is q = 8.85 mC, then calculate the leakage current at the instant t = 12 s. [JEE� 1997] 6. In the circuit shown in Figure, the battery is an ideal one, with emf V. The capacitor is initially uncharged. The switch S is closed at time t = 0. [JEE� 1998] (A) Find the charge Q on the capacitor at time t. (B) Find the current in AB at time t. What is its limiting value as t ... : 7. A thin uniform wire AB of length 1m, an unknown resistance X and a resistance of 12. are connected by thick conducting strips, as shown in the figure. A battery and a galvanometer (with a sliding A B C D jockey connected to it) are also available. Connections are to be X 12. made to measure the unknown resistance X using the principle of Wheatstone bridge. Answer the following questions. [JEE� 2002] (A) Are there positive and negative terminals on the galvanometer? (B) Copy the figure in your answer book and show the battery and the galvanometer (with jockey) connected at approxiate points. (C) After appropriate connections are made, it is found that no deflection takes place in the galvanometer when the sliding jockey touches the wire at a distance of 60 cm from A. Obtain the value of the resistance of X. 8. How a battery is to be connected so that the shown rheostat will behave like a potential divider? Also indicate the points out A B C R which output can be taken. [JEE� 2003] 9. Draw the circuit diagram to verify Ohm�s Law with the help of amain resistance of 100. and two galvanometers of resistances 106 . and 10�3 . and a source of varying emf. Show the correct positions of voltmeter and ammeter. [JEE� 2004] 10. An unknown resistance is to be determined using resistances R1, R2 or R3. Their corresponding null points are A, B and C. Find which of the above will give the most accurate reading and why? [JEE� 2005] A B C G X R R = R1 or R2 or R3
ELECTRIC CURRENT www.physicsashok.in 58 ANSWER KEY Reasoning Type Que. 1 2 3 4 5 6 7 8 9 10 11 12 Ans. C C D A C A A C A D B A Level # 1 Q. 1 2 3 4 5 6 7 8 9 10 Ans. C B D D A C C D D B Q. 11 12 13 14 15 16 17 18 19 20 Ans. B B B D CD A A ABCD A C Q. 21 22 23 24 25 26 27 28 29 30 Ans. B C B AB C D A ACD A A Q. 31 32 33 34 35 36 37 38 39 40 Ans. B A B B D A B A A A Q. 41 42 43 44 45 46 47 48 49 50 Ans. A A C B A B D A A BD Q. 51 52 53 54 55 56 57 58 59 60 Ans. D AC B ABD ABC C AC AB BC A Q. 61 62 63 64 65 Ans. B B AD D AD Fill in the Blanks / True�False / Match Table 1. 20 2. R 2 3. 0 4. F 5. T 6. A .Q, R and S, B .P and R, C .P and R, D .Q 7. A .P and Q, B .P and Q, C .R and S, D .R and S 8. A .Q, R and S, B .Q and S, C .P and S, D .P and S 9. A .Q and R, B .P, C .S, D .Q, P and R 10. A .Q, B .Q, C .S, D .R Que. 1 2 3 4 5 6 7 8 9 10 Ans. C A C B A C B D D A Que. 11 12 13 14 15 16 Ans. C A B A D A Passage Type
ELECTRIC CURRENT www.physicsashok.in 59 Level # 2 1. 72µC 11 2. 2 A 3. 160 4. 495 J 5. 19 Amp 6. x = 7.5. 7. V = . 2t RC. E 1 e 2 . . 8. ... ... .. an b 2 . . 9. I = . . . . . . 2 3 2 32 3 0 3 R R R R RE R R E R . .. . 10. Rx = 1 2 1 2 R R R R . = 12. 11. 1 E. . = 2.0 V 12. E = 7V 13. decreases by 13.3 .c 14. I = 20 mA 15. 954 . , 0.002 (°C)�1 16. R1 = 0.025 . , R2 = 0.2275 . , R3 = 2.5275 . 17. 0.9A or 4.96 x 10�3 A, 1.95 V Level # 3 1. 2.88 x 10�4 J 2. (ii) 1.5 A 3. 10/3 V 4. 1 2 2 1 1 2 V r V r r r .. , 1 2 1 2 r r r . r 5. 0.199.A 6. (a) . . 1 2 3 2 CV . e. t RC ; (b) 2 3 2 6 V V e t RC R R . . ; 2
VR 7. (a) No (b) A B C D X 12. G J (c) 8. 8. Battery should be connected across A and B. Out put can be taken across the terminals A and C or B and C. 9. G1 G2 103 Ammeter Voltmeter 106 100E 10. This is true for r1 = r2; So R2 given most accurate value. �X�X�X�X�
MAGNETIC FIELD
DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125
MAGNETIC FIELD www.physicsashok.in 1 THEORY OF MAGNETICEFFECTOFELECTRICCURRENT CONCEPT OF MAGNETIC FIELD The space around a current carrying conductor, inwhich itsmagnetic effect can be experienced, is called magnetic field. Inmagnetics, there are basicallytwomethods ofcalculatingmagnetic field at some point.One isBiot Savert�s lawwhich gives themagnetic field due to an infinitesimallysmall current carryingwire at some point and the other isAmpere�s law, which is useful in calculating themagnetic field of a highly symmetric configuration carrying a steadycurrent. BIOT-SAVART�S LAW According to this law, the magnetic field dB.. at the point P due to the small current element of length d.lis given by 0 2 2 dB µ id sin Wb /m or tesla 4 r . . . l where µ0 is a constant and is called, permeabilityof free space. i r P µ0 = 4. × 10�7 Wb/A�m Rules to Find the Direction of Magnetic Field (i) Right hand palm rule no. 1 : Ifwe spread our right hand in such a way that thumbis towards the directionof current and fingures are towards that point wherewe have to find the direction of field then the direction of field i Current carrying conductor P B will be perpendicular to the palm (ii) Maxwell�s right handed screw rule : If a right handed cork screw is rotated so that its tip moves in the direction offlowofcurrent throughthe conductor, then the rotationof the head ofthe screwgives the direction ofmagnetic lines of force. × P2 P1 Magnetic line of force i Current carrying conductor NOTE : By convention the direction of magnetic field ..Bperpendicular to paper going inwards is shown by . and the direction perpendicular to the paper coming out is shown by . . Applications of Biot-Savart�s Law
Let us consider fewapplications ofBiot-Savart�s Law: (i) Magnetic field due to a straight thin conductor is 0 . . 1 2 B µ i sin sin 4 d . . . . . d 12 p i (a) For aninfinitelylong straight wire, .1 = .2 = 90º
MAGNETIC FIELD www.physicsashok.in 2 . B µ0i 2 d . . (b)Whenwire is semi-infinite, 1 2 0 and 2. . . . . i . d 0 µ i B sin 0 sin 4 d 2. .. . . . . . . . 0 µ i B 4 d . . (c) B 1d . , i.e., B�d graph for aninfinitely long straight wire is a rectangular hyperbola as shown in figure. B (ii) Magnetic field on the axis ofa circular coil havingNturns is d . . 2 0 2 2 3/ 2 B µ NiR 2 R x . . Here, R = radius of the coil O x R P x = the distance of point P fromcentre and i = current in the coil (a)At the centre of the loop, x = 0 . 0 B µ Ni 2R . (b) For x > > R, x2 + R2 . x2 . 2 2 0 0 0 3 3 3 µ NiR µ 2Ni R µ 2M B 2x 4 x 4 x . . . . . . . . . . . . . . . . . . . . . .. . . . .. . . Here, M=magneticmoment of the loop M = NiA = Ni .R2 NOTE : This result was expected since, the magnetic field on the axis of dipole is 0 3 µ 2M
4p x Example 1. Two circular coilsAand Bof radius 52 cmand 5 cmrespectively carry current 5Amp and 52 Amp respectively.The plane ofB is perpendicular to plane ofAand their centres coincide. Find themagnetic field at the centre. Sol. Themagnetic field due to first coil is 7 0 1 1 1 2 µ I 4 10 5 B 5 2r 2 10 2 . . . . . . . . .7 1 2 20 3.14 10 B 5 1.441 10. . . . . . . B1 = 8.88 × 10�5 web/m2
MAGNETIC FIELD www.physicsashok.in 3 and 7 0 2 2 2 2 µ I 4 10 5 B 2r 2 5 10 2 .. . . . . . . . . 5 5 2 2 3.14 10 B 4.44 10 1.414 . . . . . . . . 2 2 B . B1 . B2 B . (8.88)2 . (4.44)2 .10.5weber /m2 B . 78.85 . 19.71 . 9.93 wb / m2 Example 2. Three rings, each having equal radius R, are placedmutuallyperpendicular to each other and each having its centre at the origin of co-ordinate system. If current �I� is flowing through each ring then themagnitude ofthemagnetic field at the common centre is (A) 0 3 µ I 2R (B) zero (C) . . 0 2 1 µ I 2R . (D) . . 0 3 2 µ I 2R . Sol. Themagnetic field due to ring inx�yplane is 0 1 B µ I k� 2R . .. themagnetic field due to ring iny�z plane is 0 2 B µ I �i 2R . .. and themagnetic field due to ring in x�z plane is 0 3 B µ I �j
2R . .. . B . B1 .B2 .B3 .. .. .. .. 0 . . B µ i k� i� �j 2R . . . .. . 0 B 3µ I 2R . Hence, option (A) is correct (c)Magnetic field due to an arc of a circle at the centre is 0 0 µ i µ i or B 2R 4 R . . . . . . . . . . . . . . . . .. . .. . . B= 2 O R Inwards i (iii) Field along the axis of a solenoid is 1 2 R x L O
MAGNETIC FIELD www.physicsashok.in 4 0 . . 2 1 B µ Ni cos cos 2 . . . . (a) For a long solenoid (L >>R), i.e., .1 = 180º and .2 = 0º B = µ0Ni (b)At the ends of solenoid, .2 = 0º, .1 = 90º we get, 0 B 1 µ Ni 2 . (for L>>R) Example 3. In a high tensionwire electric current runs fromeast to west. Find the direction ofmagnetic field at point above and belowthewire. Sol. When the current flows fromeast to west,magnetic field lines are circular round it as shown infigure (a).And so, themagnetic field above thewire is towards north and belowthewire towards south. W NE S B i (a) (b) Example 4.A0.5 mlong solenoid has 500 turns and has a flux density of 2.52 × 10�3 T at its centre. Find the current in the solenoid. Given, µ0 = 4. × 10�7Hm�1. Sol. Here, B = 2.52 × 10�3 T; µ0 = 4. × 10�7 Hm�1 Length of the solenoid, l = 0.5m; Total number ofturns in the solenoid,N= 500 Therefore, number ofturns per unit lengthof the solenoid, n N 500 1000 m 1 0.5 . . . . l If �i� is the current throughthe solenoid, then B=µ0ni or 3 7 0B 2.52 10 i 2.0 A µ n 4 10 1000 . .. . . . . . . Example 5. Arectangular polygon of �n�sides is formed by bending a wire of total length 2.R which carries a current �i�. Find themagnetic field at the centre of the polygon. Sol. One side of the polygon is,
a 2 R n. . 2n2 n . .. .. .. . . . . . . d a i . . d cot a / 2 . . . d a cot R cot 2 n n . . . . . . .. . . . . . . . . . . . . . . .
MAGNETIC FIELD www.physicsashok.in 5 All sides of the polygon produce the magnetic field at the centre in same direction (here . ). Hence, net magnetic field, B = (n) (magnetic field due to one side) 0 . . µ i B n sin sin 4 d . . . . . . . . . . . or 0 µ in B n tan 2sin 4 R n n . . . . . .. . .. . . . . . . . . . . .. . . . . . . . or . . 2 0 sin µ i n n B 2R cos / n . . .. . . . .. .. . . . . . . .. .. . Example 6. Infinite number of straight wires eachcarrying current �I� are equally placed as shown in the figure.Adjacent wires have current in opposite direction. Netmagntic field at point P is 30º 1 2 3 4 5 y x z a a P 30º (A) 0 µ I n 2 �k 4. 3 a l (B) 0 µ I n 4 �k 4. 3 a l (C) 0 µ I n 4 ( k� ) 4 3 a . . l (D) Zero Sol. . . 1 2 3 4 B . B . B . B . B ........... k� .. Here 0 . . 1 B µ I sin 30º sin 30º 4 a . . . 0 1 µ I 1 1 B 4 a cos30º 2 2 . . . . . .. ..
0 1 2µ I B 4 3 a . . Similarly, 0 2 2µ I B 8 3 a . . and so on . 0 2µ I 1 1 1 B 1 ........ 4 3 a 2 3 4 . . . . . . .. . .. .. .. 0 2µ I � B ln 2k 4 3 a . . .. 0 µ I � B ln 4k 4 3 a . . .. Hence, option (B) is correct. Example 7. Along straight wire, carrying current I, is bent at itsmidpoint to formanangle of 45º.Magnetic field at point P, distanceRfrompoint of bending is equal to : 45º I I P R (A) . . 0 2 1 µ I 4 R .. (B) . . 0 2 1 µ I 4 R .. (C) . . 0 2 1 µ I 4 2 R . . (D) . . 0 2 1 µ I 2 2 R . .
MAGNETIC FIELD www.physicsashok.in 6 Sol. 0 . . B µ I sin 90º sin135º 4 . R2 . . . 0 . . B µ I 2 1 4 R . . . Hence option (A) is correct Example 8. Find themagnetic field at P due to the arrangement shown (A) 0 µ i 1 1 2 d 2 . . . . . . . . . (B) 0 2µ i 2 d . . 90º 45º d P (C) 0 µ i 2 d . . (D) 0 µ i 1 1 2 d 2 . . . . . . . . . Sol. B . B1 . B2 .. .. 0 B µ I sin sin 2 4 d 4 2 2 . . .. . .. . . . . . . . 0 B µ I 1 1 2 d 2 2 . . . . . .. .. . 90º 45ºP45º /4 d2d20 2 µ I 2 1 B 2 d 2 . . . . . . . . . . . 0 . . B µ I 2 1 2 d . . . . 0 B µ i 1 1 2 d 2 . . . . . . .. .. Hence option (A) is correct Example 9.What is themagnitude ofmagnetic field at the centre �O� of loop of radius 2 mmade of uniformwirewhen a current of 1 amp enters in the loop and taken out of it by two longwires as shown in the figure. 45º
1 amp 90º 1 amp O Sol. 1 B .. =magnetic field due to left wire is 0 1 µ I � B sin sin k 4 d 2 4 . . .. . . . . . . . O/4 d 2 B .. =magnetic field due to right wire 0 µ I � sin sin ( k) 4 d 2 4 . . .. . . . . . . . . In circularwire, I 1R . . 12 II 2 . . . . . I1 I2 � . 1 2 I I 2 . . . . . . . . . ..
MAGNETIC FIELD www.physicsashok.in 7 . 3 B .. =magnetic field due to circularwire 0 1 0 2 µ I µ I (2 ) 4 r 4 r . . . . . . . . 0 2 0 2 µ I µ (2 ) I 0 4 r 4 r(2 ) . . . . . . . . . . . . . . B . B1 . B2 . B3 . 0 .. .. .. .. AMPERE�S CIRCUITAL LAW It states that the line integral of B.. around any closed path or circuit is equal to µ0 times the total current crossing the area bounded bythe closed path provided the electric field inside the loop remains constant. Thus, . . 0 net C . B. d . µ i .. .l Its simplified formis Bl = µ0 inet This equation canbe used onlyunder following conditions : (a) at every point of the closed path B|| dI .. . (b)magnetic field has the samemagnitudeB at all places on the closed path. Applications of Ampere�s Circulatal Law (i) Magnetic field due to a longmetal rod of radiusR carrying a current �i� : (a) If r < R, 0 2 µ i B r 2 R . . . .. . .. , i.e., B .r (b) If r = R (i.e., at the surface) 0 B µ i 2 R . . (ii) Magnetic field of a solenoidwounded in the formof a helix is B = µ0Ni NOTE : Ampere�s law is valid only for steady currents. Further more, it is useful only for calculating the magnetic fields of current configurations with high degrees of symmetry, just as gauss�s law is useful only for calculating the electric fields of highly symmetric charge distributions. Example 10. Two long conductors are arranged as shown above to form overlapping cylinders, each of raidus �r�, whose centers are separated by a distance �d�. Current of densityJ flows into the plane of the page along the shaded part of one conductor and an equal current flows out d Conductor
y x Vacuum A of the plane of the page along the shaded portion of the other, as shown. What are themagnitude and direction of themagnetic field at pointA? (A) (µ0/2.).dJ, in the +y-direction (B) (µ0/2.)d2/r, in the +y-direction (C) (µ0/2.)4d2J/r, in the �y-direction (D) (µ0/2.)Jr2/d, inthe �y-direction Sol. B . B1 . B2 .. .. .. 0 0 µ d � µ d � B j i j i 2 2 2 2 . . . . . . . . . . . . . .. . . . . . . . . .. . .
MAGNETIC FIELD www.physicsashok.in 8 B µ0 jk� d �i µ0 jk� d i� 2 2 2 2 . . . . . . . . . .. .. .. .. .. 0 0 B µ jd �j µ jd �j 4 4 . . .. 0 0 B µ jd �j µ d j 2 2. . . . .. along y-axis Hence, option (A) is correct. MAGNETIC FIELD OF A MOVING POINT CHARGE The magnetic field vector B.. at point P at position vector r . , fromthe charge qmovingwith a velocityv . is found to be 0 . . 3 B µ q v r 4 r . . . . .. . . Note downthe following points regarding this equation. (a)Magnitude ofBis, 0 2 B µ qvsin 4 r . . . r vq p It is zero at . = 0º and 180º andmaximumat . = 90º (b) Direction ofB.. is along v r . . . if q is positive and opposite to v . r . . if q is negative. (c) Suppose a charge q1 is moving with velocity and 1 v . another charge q2 is moving with velocity 2 v . at position vector r . relative to q1, then force on q2will be, . . 2 2 F . q v . B . . .. . . 0 1 . . 2 3 1
µ q F v . v r 4 r . . . . . . . . . . . . . . . . . . . . 0 1 2 . . . . 3 2 1 F µ . q q v v r 4 r . . . . . . . . . . . . This corresponds to Coulomb�s electricalforce between the charges q1 and q2movingwith velocities 1 v . and 2 v . respectivelyrelative to an observer at rest. Example 11. For a charge �q�movingwith velocityv . , find the relation between electric andmagnetic fields. Sol. 3 0 E q 4 r . .. .. ...(i) 0 3 B µ qv r 4 r. . ... . . ...(ii) and 0 0 c 1 µ . . or 2 0 0 c 1 µ . .
MAGNETIC FIELD www.physicsashok.in 9 FromEqs. (i) and (ii), we get B . µ0.0 v . E .. . .. . 2 B v E c. . .. . .. FORCE ON A CURRENT CARRYING CONDUCTOR IN MAGNETIC FIELD When a current carrying conductor is placed in amagnetic field, the conductor experiences a force in a directionperpendicular to both the direction ofmagnetic field and the direction of current flowing in the conductor. This force is also called B F i Lorentz force. i The direction of this force can be found out either by Fleming�s left hand rule or by right handpalmrule. Themagnetic force is F = ilB sin . In vector form, F . i. . B. . . .. l Where B= intesityofmagnetic field i= current in the conductor l = lengthof the conductor and . = angle between the length of conductor and directionofmagnetic field. Case : (i) If ..= 90º or sin ..= 1 then F = ilB(maximum) Therefore, forcewillbemaximumwhen the conductor carrying current is perpendicular tomagnetic field. (ii) If ..= 0º or sin ..= 0, Then F = ilB × 0 = 0 Thus, the forcewill be zero,when the current carrying conductor is parallel to the field. Example 12. Astraight current carrying conductor is placed in such away that the current in the conductor flows in the direction out of the plane of the paper.The conductor is placed between two poles of two magnets, as shown. The S N Q R S P conductor willexperience a force inthe direction towards. (A) P (B) Q (C) R (D) S Sol. The direction ofmagnetic field on the conductor is along SR. But F . i . B . . .. l F . i k� . (�B�i) . l F . .i B�j . i B(.�j) .
l l Hence, the direction ofmagnetic force on thewire is towards Q. Hence option (B) is correct Example 13. In the figure shown a semicircularwire loop is placed in uniformmagnetic fieldB = 1.0T.The plane of the loop is perpendicular to themagnetic field. Current i= 2Aflows inthe loop in the direction shown. Find themagnitude of themagnetic force in both the cases (a) and (b). The radius of the loop is 1.0m. i = 2A i = 2A × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × (a) (b) 1m
MAGNETIC FIELD www.physicsashok.in 10 Sol. Refer figure (a) : It forms a closed loop and the current completes the loop. Therefore, net force on the loop in uniformfield should be zero. Refer figure (b) : In this case although it forms a closed loop, but current does not complete the loop.Hence, net force is not zero.FACD . FAD . . . Floop . FACD . FAD . 2 FAD . . . . × × C × × × × × × × × × × × × × × B A D . | Floop | . 2| FAD | . . | Floop | . 2i B sin . . l (l = 2r = 2.0 m) = (2) (2) (2) (1) sin 90º = 8 N Example 14. An arc of a circular loop of radius �R�is kept in the horizontal plane and a constantmagnetic field �B� is applied in the verticaldirection as shown in the figure. If the arc carries current �I� then find the force in the arc. × B 90º I × × × × × × × × × × × × × × × × × × × × × Sol. Aswe know,magnetic force on a closed loop placed in uniformmagnetic field is zero. × × × × × × . B1 . B2 . 0 .. .. . 1 2 B . .B . .I B(.�j) .. .. l 1 B . I B�j .. l /2 R R (1) (2) 1 B . I B( 2R) �j .. . B1 . 2 IBR .. Example 15. Aconducting wire bent in the formof a parabola y2 = 2x carries a current i = 2Aas shown in figure. thiswire is placed in a uniformmagnetic field B . .4 k� .. Tesla. Themagnetic force on thewire is (in newton) y(m) AB
2 x(m) (A) .16�i (B) 32�i (C) .32�i (D) 16 �i Sol. The netmagnetic force on closed loop is zero. . Force on parabola + force on straight wireAB = 0 . Force on the parabola = � force on straight wireAB . ..I�j. B. .. AB . .2..4�j. ..4k� .. F . .32 �i Hence option (C) is correct. Example 16. Aconductor of length �l � and mass �m�is placed along the east-west line on a table. Suddenly a certain amount of charge is passed through it and it is found to jump to a height �h�. The earth�smagnetic induction isB. The charge passed throughthe conductor is : (B is horizontal) (A) 1 Bmgh (B) 2gh Blh (C) gh Blh (D) m 2gh Bl Sol. Themagnetic force is F = I lB or F dq B dt . l
MAGNETIC FIELD www.physicsashok.in 11 or dq Fdt B . l or 0 q Fdt mv B B . . . l l But 02 = v02 � 2gh . 0 v . 2gh . m 2gh q B . . l Hence option (D) is correct Example 17. AU-shapedwire ofmassmand length l is immersedwith its two ends inmercury(see figure).Thewire is ina homogeneous field ofmagnetic inductionB. If a charge, that is, a current pulse q . .idt , is sent through thewire, thewirewill jump up. × × × × × × × × × × × × × × × × × × × × × l m B i Calculate, fromthe height �h� that thewire reaches, the size of the Hg charge or current pulse, assuming that the time of the current pulse is verysmall incomparisionwiththe time offlight.Make use ofthe fact that impulse offorce equals . Fdt ,which equalsmv. Evaluate �q�for B = 0.1Wb/m2, m= 10 gm, l = 20 cm&h = 3 meters. [g = 10 m/s2] Sol. I l B = F . Fdt = mv or I l Bdt = mv But 1 mv2 mgh 2 . . v . 2gh . Idt mv m 2gh B B . . l l . 3 2 q m 2gh 10 10 2 10 3 15 coulmb. B 20 10 0.1 . . . . . . . . l . .
Example 18.Ametal ring of radius r = 0.5mwith its plane normal to a uniformmagnetic fieldB ofinduction 0.2T carries a current I = 100A. The tension in newtons developed in the ring is : (A) 100 (B) 50 (C) 25 (D) 10 Sol. We consider a small portion .l of the loop. For equilibrium, F 2T sin 2T T 2 2 . . . . . . or I.lB = T. ......... Tcos /2 /2 /2 Tcos /2 T T F + B or Ir.B=T. . T = IrB = 100 × 0.5 × 0.2 = 10 N Hence option (D) is correct
MAGNETIC FIELD www.physicsashok.in 12 FORCE ON A MOVING CHARGE IN MAGNETIC FIELD : LORENTZ FORCE If a charged particlemoves in amagnetic field, then a force acting on it is given by F = qvB sin . In vector form, F . q.v . B. . . .. × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × FCases : v (i) If v= 0, then F = 0 i.e., no force is exerted on a stationary charge, in a magnetic field. (ii) If ..= 0, then F = 0 i.e., when the charge ismoving parallel to the field then no forcewillbe exerted bythe field. (iii) If ..= 90º, then sin ..= sin 90º = 1 F = qvB × 1 = qvB i.e., when the charged particle is moving parpendicular to the field, the force exerted by the field will be maximum. Rules of Find the Direction of Force (i) Right hand palm rule : Ifwe stretch the right hand palmsuch that the fingers and the thumb are mutually perpendicular to each other and the fingers point the direction ofmagnetic field and the thumb points the direction ofmotionof positive charge, the direction of forcewillbe along the outward normalon the palm. Field B Force F Current or motion of positive charge (ii) Fleming�s left hand rule : Ifwe spread the forefinger, central finger and thumb of our left hand insuch away that these three are perpendicular to each other then, if first forefinger is in the direction ofmagnetic field, second centralfinger is in the direction of current, then thumbwill represent the direction of force. Current or motion of positive charge Force F Field B v NOTE : To learn this rule, remember the sequence of Father, Mother, child. Thumb . Father . Force Forefinger . Mother . Magnetic field Central finger . Child . Current or direction of positive charge Example 19.Whena protonhas a velocity v . .2�i . 3�j..106 m/ s . it experiences a force F . ..1.28 .10.13 k� .N. . When its velocityis along the z-axis, it experiences a force along the x-axis.What is themagnetic field ? Sol. Substituting proper values in,
F . q.v . B. . . .. We have, . 13 . . 19 . . . . . 6 0 . 1.28.10. k� . 1.6 .10. . 2�i . 3�j . .B �j . .10 . .
MAGNETIC FIELD www.physicsashok.in 13 . 1.28 = 1.6 × 2 × B0 or 0 B 1.28 0.4 3.2 . . Therefore, themagnetic field is, B . ..0.4�j.T .. MOTION OF CHARGED PARTICLE IN MAGNETIC FIELD The pathofchargedparticle inuniformmagnetic field dependsonangle betweenv . andB.. .Therefore, following cases are possible : Case-I .When . is 0º or 180º : The magnetic force is F = Bqv sin 0º or sin 180º = 0. Hence, path of the charged particle is a straight line (undeviated) when it enters parallel or antiparallel to magnetic field. Case-II.When . = 90º : Themagnetic force is F = Bqv sin 90º = Bqv.Thismagnetic force is perpendicular to the velocity at every instant. Hence, path is circle. The necessarycentripetal force is provided by the magnetic force hence, if �r�be the radius of the circle, then + + q v v q or B B Fm= 0 mv2 qBv r . . r mv qB . This expression of �r�can bewritten in following different ways : mv p 2Km 2qVm r qB qB qB qB . . . . Here, P =momentumof particle K= KE of particle p2 or p 2Km 2m . . Further, time period of the circular pathwill be 2 mv 2 r qB 2 m T v v qB . . .. . . . . . . . . or T 2 m qB . . or the angular speed (.) of the particle is 2 qB
T m . . . . . qB m . . Frequencyof rotation is,1f or f qB T 2 m . . .
MAGNETIC FIELD www.physicsashok.in 14 IMPORTANT FEATURES If angle . is other than 0º, 180º or 90º, then velocity of charged particle can be resolved in two components one along B.. and another perpendicular to B.. . Let the two components be | | v and v. . Then v| | . v cos. and v vsin . . . q, m + B v v sin v cos The component perpendicular to field ( v. ) gives a circular path and the component parallelto field ( | | v ) gives a straight line path. The resultant path is a helix as shown in figure. The radius ofthis helicalpath is, r mv mvsin qB qB . . . . Time period and frequency do not depend on velocity and so theyare given by T 2 m and f qB qB 2 m . . . . There is onemore termassociatedwitha helicalpath, that is pitch (p) of the helicalpath. Pitch is defined as the distance travelled alongmagnetic field in one complete cycle, i.e., p = v| |T or p .vcos . 2 m qB . . . . . p 2 mvcos qB . . . Example 20. What is the smallest value of B that can be set up at the equator to permit a proton of speed 107m/s to circulate around the earth ? [R = 6.4 × 106 m, mp = 1.67 × 10�27 kg]. Sol. Fromthe relation R mv qB . We have B mv qR . Substituting thevalues,we have . .. . . .. . 27 7 8 19 6 1.67 10 10
B 1.6 10 T 1.6 10 6.4 10 . . . . . . . . . Example 21. Ablock ofmassm&charge �q�is released on a long smooth inclined planemagnetic field Bis constant, uniform, horizontal and parallelto surface as shown. Find the time fromstart when block loses contact with the surface. B m q (A) mcos qB . (B) m cosec . qB (C) m cot . qB (D) none
MAGNETIC FIELD www.physicsashok.in 15 Sol. For losing the contects N = 0 . qvB=mgcos. . v mg cos qB . . But v = u + at or v = 0 + gt sin . . gt sin mg cos qB . . . . t mcot qB . . Hence option (C) is correct. Example 22.An electronhaving kinetic energyTismoving in a circular orbit ofradiusRperpendicular to a uniform magnetic inductionB.. . If kinetic energyis doubled andmagnetic induction tripled, the radiuswillbecome (A) 3R2 (B) 3R 2 (C) 2R 9 (D) 4R 3 Sol. R mv P qB qB . . But kinetic energy P2 T 2m . . P . 2mT . 2mT R qB . . 2m(2T) R´ q(3B) . R´ 2 R 2 R 3 9 . . Hence, option (C) is correct. Example 23. Acharged particle (charge q, massm) has velocity v0 at origin in +x direction. In space there is a
uniformmagnetic fieldB in �z direction. Find the �y� coordinate of particlewhenis crosses y�axis. Sol. The parth ofcharged particle is circularwhose radius is 0 mv r qB . O C + B . 0 2mv y 2r qB . . Example 24. Amass spectrometer is a devicewhich select particle of equal mass.An ironwith electric charge q > 0 starts at rest froma source S and is accelerated through a potentialdifferenceV. It passes through a hole into a region of constant magnetic field B.. perpendicular to the B V S plane of the paper as shown in the figure. The particle is deflected by themagnetic field and emerges throughthe bottomhole at a distance �d� fromthe top hole. Themass of the particle is (A) qBd V (B) qB2d2 4V (C) qB2d2 8V (D) qBd 2V
MAGNETIC FIELD www.physicsashok.in 16 Sol. The speed of charged particle just before entering themagnetic field isV0. 20qV 1 mv 2 . 0 v 2qV m . The radius of circular path inmagnetic field is 0 mv r qB . or 0 d mv 2 qB . or 2 2 2 2 0 2 2 2 2 d m v m 2qV 4 q B q B m . . or 2 2 d 2mV 4 qB . . qB2d2 m 8V . Hence, option (C) is correct Example 25. Acyclotron is operatingwith a flux density of 3Wb/m2. The ionwhich enters the field is a proton havingmass 1.67 × 10�27 kg. If themaximumradius of the orbit of the particle is 0.5m, find : (a) themaximumvelocityof the proton, (b) the kinetic energy of the particle, and (c) the period for a half cycle. Sol. (a)As in case ofmotion of a charged paricle in amagnetic field, r mv i.e., v qBr qB m . . So, max max v qBr m . So, 19 8 max 27 1.6 10 3 0.5 v 1.43 10 m/ s
1.67 10 . . . . . . . . . (b) . .K 1 mv2 1 1.67 10 27 1.43 108 2 2 2 . . . . . . . i.e. 11 11 19 1.71 10 K 1.71 10 J 107 MeV 1.6 10 . . . . . . . . . (c) In case of circularmotion, as 8 8 T 2 r 2 0.5 2.19 10 s v 1.43 10 . . . . . . . . . So, time for helf cycle, T 1 (T) 1.09 10 8 s 2 . . . .
MAGNETIC FIELD www.physicsashok.in 17 Example 26. The region between x = 0 and x = Lis filledwith uniformsteadymagnetic field 0 B k� .Aparticle of mass �m�, positive charge �q� and velocity 0 v �i travels along x-axis and enters the region of themagnetic field. Neglect the gravitythroughout the question (a) Find the value of�L�iftheparticle emerges fromthe regionofmagnetic fieldwithits finalvelocityat anangle 30º to its initialvelocity. (b) Find the finalvelocity of the particle and the time spent by it in themagnetic field, if themagnetic field now expands upto 2.1 L. Sol. (A) . = 30º, sin LR . . Here 0 0 R mv qB . . 00 sin 30º L mv qB . ×+ + + + + + + + + + + + + + + + + + + + + + + P v0 X Y qACx=0 x=L v0 LR B = B0 k or 00 1 qB L 2 mv . . 00 L mv 2qB . (b) In part (i) sin 30º L
R . or 1 L 2 R . or L = R/2 × × × × × × × × × × × × B R v0 v0 L´ =2.1R > R Nowwhen L´ = 2.1 L or 2 2.1R 2 . L´ > R Therefore, deviation of the particle is . = 180º is as shown. . f 0 B v . .v i� . v . . and AB 0 t T m 2 qB . . . Example 27. Awire loop carrying a current �I�is placed in the x�y plane as shown in fig. (a) If a particlewith charge +Qandmass �m� is placed at the centre �P� and given a velocityv . alongNP (see figure), find its instantaneous acceleration. V P+Q 120º M I N a y O x (b) If an external unfiormmagnetic induction field B . B�i .. is applied find the force and the torque acting on the loop due to this field. Sol. Themagnetic field at the centre P due to current inwireNMis 0 . . 1 B µ I sin 60º sin 60º 4 r . . .
MAGNETIC FIELD www.physicsashok.in 18 0 1 µ B I 3 3 4 a / 2 2 2 . . . . . . . . . v P Q 120º MN a y 60º 30º 30º Q x V sin60º V0 V cos60º 1 B µ 2I 3 4 a . . directed awayfromthe reader perpendicular to the plane of paper. sin 30º ra . . r a2 . cos30º MS a . 30º S r P a MN a . MS 3 a 2 . . MN . 3 a Themagnetic field at the centre P due to current in arcMNis 0 0 0 2 µ 2 I µ 2 I 2 / 3 µ 2 I B 2 a 2 4 a 2 4 3a . . . . . . . . . . . . . . . . . . .. . . . . . directed towards the reader perpendicular to the plane of paper
The netmagnetic field 0 0 1 2 B B B µ 2 3I µ 2 I 4 a 4 3a . . . . . . . 0 0 µ 2I µ 2I B 3 (.68) 4 a 3 4 a . . . . . . . . . . . . . . (directed away fromthe reader perpendicular to the plane of paper) The force acting on the charged particleQwhen it has a velocity v and is instantaneously at the centre is F = QvB sin . = QvB sin 90º = QvB The acceleration produced 0 F QvB Qv µ 2I A (0.68) M m m 4 a . . . . . . . .. . .. 0 A 0.11µ IQv ma . y x P B v M R QQ N120º The direction of acceleration is given by the vecotr product v . B . .. or byapplying Fleming�s left hand rule
MAGNETIC FIELD www.physicsashok.in 19 ..RPN = 90º and .MPN = 120º . .MPR = 120º � 90º = 30º Since, .MPQ = 60º . .RPQ = 30º i.e.,wthe acceleration vectormakes an angle of 30º with the negative x-axis. The torque acting on the loop in themagnetic field is giving. . . M . B . ... .. where M= IA where A= (area of PMQNP)�(area of trangle PMN) A 1 . a2 . 1 MN PS 3 2 . . . . . 2 A a 1 3a a a2 3 3 2 2 3 4 . .. . . . . . . . . . . . A a2 3 k� 3 4 .. . . . . . . . .. . Ia2 3 k� �iB 3 4 .. . . . . . . . . . . BIa2 3 �j 0.614 BIa2J� 3 4 . . . . . . . . . . . . The force acting on the loop is zero. Example 28. An electron gunGemits electrons of energy 2 keVtravelling in the positiveX-direction. The electrons are requeired to hit the spot Swhere GS = 0.1m, and the lineGS make an angle of 60º with the x-axis as shown in figure.Auniformmagnetic field B.. parallel toGS exists in the region outside 60º B SG X the electron gun. Find theminimumvalue ofBneeded tomake the electrons hit S. Sol. Kinetic energyof electron, K 1 mv2 2keV 2 . . . speed of electron, v 2K m . 60º B S
G v 16 31 v 2 2 1.6 10 m/ s 9.1 10 . . . . . . . v = 2.65 × 107 m/s Since the velocity (v) . of the electronmakes an angle of . = 60º with themagnetic field B.. , the pathwillbe a helix. So, the particlewillhit S if GS = nP Here n = 1, 2, 3, ............. p pitch of helix 2 m v cos qB . . . .
MAGNETIC FIELD www.physicsashok.in 20 But for B to beminimum, n = 1 Hence, GS p 2 m vcos qB . . . . . min B B 2 mv cos q(GS) . . . . Substituting thevalues,we have 31 7 min 19 (2 )(9.1 10 )(2.65 10 ) 1B 2 tesla (1.6 10 )(0.1) . . . . . . . .. .. . . or Bmin = 4.73 × 10�3 tesla Example 29. An electronmovingwith a velocity 1 V . 2 i� .. m/s at a point in a magnetic field experiences a force 1 F . .2�jN . . If the electron ismoving with a velocity 2 V . 2�j .. m/s at the same point, it experiences a force 2 F . .2�i N . . The force the electronwould experience if it weremovingwith a velocity 3 V . 2 k� .. m/s at the same point is (A) zero (B) 2k.N (C) .2k.N (D) informationis insufficient Sol. F1 . qv1 . B . . ..1 .2 �j . .e(v ) . B . .. .2�j . .e(2 i�) . B.. �j . 2e( �i . B) .. �j . eB( �i) . (.k� ) . eB = 1 2 F . .e(2 �j) . (.Bk� ) . 2�i . 2eBi� eB = 1 F3 . .e(v3 . B) . . .. 3 F . .e(2 k� ) . (.Bk� ) . 0 . .
Hence option (A) is correct Charged particle in uniform & ... .. E B When a charged particlemoveswith velocityv . in an electric field E.. andmagnetic field B.. , then.Net force experienced byit is given byfollowing equation. F . qE . q(v . B) . .. . .. Combined forceF . is known as lorentz force (i) E || B||v .. .. . E B v In above situation particle passes underviated but its velocitywillchange due to electric field andmagnetic force on it is zero.
MAGNETIC FIELD www.physicsashok.in 21 (ii) E || B .. .. and uniform, particle is releasedwith velocity v0 at an angle .. v0 +q E, B v0 x E, B v0 cos +q v0 siny z 0 mv sin 2 m r ; T qB qB . . . . x . . 2 . . . .. . 0 r v cos t 1 qE t i� Rsin t �j R R cos �t k� 2 m . . . . . . . . . . . . . . . . R c y v0 sin z Cycloid motion Suppose that B.. points inthe x-direction, and E.. inthe z-direction. E x B z a b c y 0 F . q.E . v . B. . q.Ez� . Bzy� . Byz�. . ma . m.yy� . zz�. . .. . .. . . . .. .. qB m . . .. y z, z E y B . . . . .. . . .. . .. .. ... Their generalsolution is 1 2 3 2 1 4 y(t) C cos t C sin t (E / B)t C z(t) C cos t C sin t C . . . . . . .. . . . . . . y(t) E ( t sin t), B . . . . . z(t) E (1 cos t) B
. . .
. R EB . . (y � R.t)2 + (z � R)2 = R2 v R EB . . . The particlemoves as through it were a spot on the rimof awheel, rolling down the y axis at speed, v.The curve generated in thisway is called a cycloid. Notice that the overallmotion is not in the direction ofE.. , but perpendicular to it.
MAGNETIC FIELD www.physicsashok.in 22 Example 30. Aparticle of specific charge (charge/mass) . startsmoving fromthe origin under the action of an electric field 0 E . E �i .. andmagnetic field 0 B . B k� .. . Its velocity at (x0, y0, 0) is (4�i . 3�j) . The value of x0 is : (A) 0 0 13 E 2 B . (B) 0 0 16 B E. (C) 0 25 2.E (D) 0 5 2B. Sol. Theworkdone bymagnetic force isWB = 0, because,magnetic force is always perpendicular to instantaneous displacement. E 0 0 0 0 0 W . qE .S . qE �i . (x �i . y �j) . qE x .. . The speed of particle at (x0, y0) is v . 42 . (.3)2 . 5m/ s According to work � energytheorm,W T 1 mv2 1 mu2 2 2 . . . . or 2 E B W W 1 m5 0 2 . . . or 0 0 qE x 25m 2 . . 0 0 0 x 25m 25 2qE 2E . . . qm . . . . . .
. .
. Hence, option (C) is correct. Example 31. Aparticle of specific charge (q/m) is projected fromthe origin of coordinates with initial velocity [u �i . v �j] .Uniformelectricmagnetic fields exist inthe regionalong the+ydirection, ofmagnitude EandB. The particlewilldefinitelyreturn to theorigin once if (A) [vB/2.E] is an integer (B) (u2 + v2.1/2 [B/.E] is an integer (C) [vB/.E] in an integer (D) [uB/.E] is an integer Sol. y a qE m . mu2 quB r . r mu qB . also, T 2 m qB . . y vt qE t2 2m . . . For origin, x = 0, and y = 0 0 vt qE t2 2m . . .
MAGNETIC FIELD www.physicsashok.in 23 . t 2mV qE . For returning at the origin,nT 2mV qE . 2n m 2mV qB qE . . n 2mVqB qE(2 m) . . . n VBE . . Hence, option (C) is correct FORCE BETWEEN PARALLEL CURRENT CARRYING WIRES Consider two longwires 1 and 2 kept parallel to each other at a distance �r�and carrying currents i1 and i2 respectively in the same direction. The force per unit length of thewire 2 due towire 1 is : 0 1 2 F µ i i 2 r . l . F i1 l i2 1 2 The same force acts onwire 1 due to wire 2. r NOTE : The wires attract each other if currents in the wires are flowing in the same direction and they repel each other if the currents are in opposite directions. Example 32. Aconductor of length 2 mcarrying current of 2Ais held parallel to an infinitely long conductor carrying current of 10Aat a distance of 100mm. Find the force on small conductor. Sol. We knowthat force per unit length of short conductor due to long conductor is given by 0 1 2 f µ 2i i 4 r . . . Total force on length l of the short conductor is 0 1 2 F f µ 2i i 4 r . . . l l 7 10 2 2 10 2 5 F 8 10 N
0.1 . . . . . . . . . Force is attractive if the direction of current is same in both the parallel conductors and is repulsive if the direction of current is opposite intwo parallel conductors. Example 33. Astraight segment OC (of lengthL) of a circuit carrying a current �i�is placed along theX-axis as shown infigure.Two infinitely long straight wiresAand B, each extending fromz = � . to + ., are fixed at y = �a and y = + a respectively. as shown in the figure. If the ×× AB Y X O i C Z wiresAandB each carry a current �i�into the plane of the paper, obtain the expression for the force acting on the segment OC.What will be force ofOC if the current in the wire Bis reversed ?
MAGNETIC FIELD www.physicsashok.in 24 Sol. (a) Let us assume a segment of length dx at a point P, a distance x fromthe centre shown in figure. ×× AB y x O i y´ C I aa I dx x LBnet BB BA is the positive Z-direction and × is the negative Z-direction and Magnetic field at P due to current in wiresAand B will be in the directions perpendicular to AP and BP respectivelyas shown. 0 A B | B | | B | | B| µ i 2 AP . . . . .. .. .. Therefore, net magnetic field at Pwillbe along negative y= axis as shown and 0 net µ i x B 2 | B| cos 2 2 AP AP . . . . . . . .. ... .. .. .. 0 0 net 2 2 2 µ i.x µ i x B . (AP) (a x ) . . . . .. . .. . . Therefore, force on the element will be (F = ilB) 0 2 2 µ i x dF i dx a x . . . . . . . . . (in negative z-direction) . Total force onthewirewill be x L 2 L 0 2 2 x 0 0 F dF µ i xdx x a
.
.
. .
. . . . 2 2 2 0 2 µ i L a F ln 2 a . . . . . . . . . (innegative z-axis) Hence 2 2 2 0 2 µ i L a � F ln k 2 a . . . . . . . . . . . A× BY X i Bnet C BA BB P (b) If current inwire Bis reversed, thenmagnetic fields due to AandBwillbe in the directions shown in figure. i.e., net magnetic field net B .. willbe along positive x-axis and since current is also along positive x-axis, force onwireOCwill be zero. Note : A B .. is not necessarily parallel to BP. CURRENT LOOP IN UNIFORM MAGNETIC FIELD . . M . B . ... .. | . |. MBsin . . ; whereM= NIA M B I Work done in rotating loop in uniformfield from.1 to .2 W=MB (cos .1 � cos .2)
MAGNETIC FIELD www.physicsashok.in 25 Example 34. Themagneticmoment of a circular orbit of radius �r�carrying a charge �q�and rotatingwith velocity �v� is given by (A) qvr 2. (B) qvr 2 (C) qv.r (D) qv.r2 Sol. The convectioncurrent is I q qv 2 2 r . . . . . . Themagneticmoment is M I r2 qv r2 2 r . . . . . . M qvr 2 . Hence option (B) is correct Example 35. Qcharge is uniformly distributed over the same surface of a right circular cone of semi-vertical angle . and height �h�. The cone is uniformlyrotated about its axis at angular velocity.. Calculated associatedmagnetic dipolemoment. Sol. I = Themoment of gnertia3 MR2 10 . . Angularmomentum L I 3 mR2 10 . . . . . Butmagneticmoment P QL 2m . . P Q 3 mR2 2m10 . . . P 3Q R2 20 . . But tan Rh . . . R = h tan. . 3QR2 3Qh2 tan2 P 20 20 . . . . . Example 36. Figure shown a square current carrying loopABCDof side 10 cmand
current i= 10A. Themagneticmoment M... of the loop is (A) (0.05).�i . 3 k� .A.m2 (B) (0.05).�j. k� .A.m2 y x z 30º DC BA i = 10 (C) (0.05). 3 �i . k� .A.m2 (D) . �i . k� .A.m2 Sol. Themagnitude ofmagneticmoment is M = Il2 = 10 × (10 × 10�2)2Am2 = 10 × 10�2 = 0.1 Am2
MAGNETIC FIELD www.physicsashok.in 26 The normal on the loop is in x � z plane. It makes 60º angle with x - axis. . M . Mcos60º i� .Msin 60º �j ... . M M �i 3 M�j 2 2 . . ... . . . M 0.1 �i 3 �j 2 . . ... M . (0.05).i� . 3 �j.Am2 ... Example 37. Arectangular coil PQhas 2n turns, an area 2a and carries a current 2I, (refer figure). The plane of the coil is at 60º to a horizontaluniformmagnetic field of flux densityB. The torque onthe coil due tomagnetic force is 60ºB coil 2n, 2a, 2I (A) BnaI sin 60º (B) 8 BnaI cos 60º (C) 4 BnaI sin 60º (D) none Sol. . . M. B . ... .. Here M= 2n(2I) (2a) M= 8 nIa . . =MB sin(90º � 60º) . = MB cos 60º ..= 8 nIa cos 60º Hence option (B) is correct Example 38. (a)Arigid circular loop of radius �r�&mass �m�lies in the �xy�plane on a flat table and has a current �I�flowing in it.At this particular place, the earth�smagnetic field is x y B . B �i . B �j. .. .. .. Howlargemust �I�be before one edge of the loopwill lift fromtable ? (b) Repeat if, x y B . B �i . B k�. .. .. .. Sol. (a)Torque due to magnetic force should be greater than torque due to weight. . I.r2B . mgr or 2 2 2 x y I.r B . B . mgr .(b) Since, Bz is parallel to dipolemoment . 2 x I.r B . mgr . x I mg rB . . Example 39. Aconducting ring ofmass 2 kg and radius 0.5mis placed on a smooth horizontal plane.The ring carries a current i= 4A.Ahorizontalmagnetic field
B = 10T is switched on at time t = 0 as shown in figure. The initial angular B acceleration of the ringwillbe (A) 40. rad/s2 (B) 20. rad/s2 (C) 5. rad/s2 (D) 15. rad/s2 Sol. Aswe know, if a coilor a closed ofany shape is placed in uniform electric field,magnetic force on the coil or the loop is zero.
MAGNETIC FIELD www.physicsashok.in 27 So, the centre ofmass of the coil remains in rest. the torque on the coil is . . = IAB = 4.r2B or 2 mr 2 4 r B 2 . . . . 8B 8 10 40 rad / s2 m 2 . . . . . . . . Hence option (A) is correct Example 40. In the figure shown a coilof single turn iswound on a sphere of radius R andmass �m�. The plane of the coil is parallel to the plane and lies in the equatorial plane of the sphere. Current in the coil is �I�. The value of B if the sphere is in B equilibriumis (A) mg cos IR . . (B) mg .IR (C) mg tan IR . . (D) mg sin IR . . Sol. For equilibrium, mg sin . = f Also, net torque should be zero. Themagneticmoment of the loop is perpendicualr to the plane. . The torque due tomagnetic force is .B = PB sin(180 � .) 180º � P B . .B = PB sin. = I.R2B sin. This torque balances the torque due to friction about centre ofmass. . FR = I.R2Bsin. or mg sin . = .IRB sin . . B mg IR . . Hence option (B) is correct Example 41. Asquare current carrying loopmade of thinwire and having a massm= 10 g can rotatewithout frictionwith respect to the vertical axisOO1, passing throughthe centre of the loop at right angles to two opposite sides of the loop. The loop is placed in a homogeneousmagnetic
fieldwith an induction B = 10�1 T directed at right angles to the plane O+ B I O1 of the drawing.Acurrent I =2Ais flowing in the loop. Find the period of smalloscillations that the loop performs about its position of stable equilibrium. Sol. . = � PB sin. or . = � IAB. or I0. = � IAB. or 0 IAB I . . . . . . . . . . 2 0 IAB I . .. . . . or 0 IAB I . .
MAGNETIC FIELD www.physicsashok.in 28 or 0 2 IAB T I . . . T 2 I0 IAB . . Here 2 2 2 2 0 0 0 0 0 I m m m m 2 2 12 12 . . . . . . . . .. .. .. .. l l l l 2 2 0 0 0 m m I 2 6 . . l l But 3 0 m m 2.5 10 kg 4 . . . . l2 =A= area of loop, I = 2 amp After putting the value, T = 0.57 sec. Example 42.Auniformconstantmagnetic fieldB.. is directed at an angle of 45º to theX-axis inX�Yplane. PQRS is a rigid squarewire frame carrying a steady current I0, with its centre at the originO.At time t =0, the frame is at rest in the position shown in the figurewith its sides parallel toXandYaxes. Each side of the frame is of massMand length L. (a) What is the torque . . about Oacting on the frame due to the magnetic field ? (b) Find the angle bywhich the frame rotates under the action of this torque in a short intervalof time .t, and the axis about which this rotation occurs (.t is so short that any variation in Y 0 X P Q S R I the torque during this intervalmay be neglected).Given : the moment of inertia of the frame about an axis throughits centre perpendicular to its plane is 4ML2 3 . Sol. Magneticmoment of the loop, 2
0 M . (iA)k� . (I L )k� ... Magnetic Field, B (Bcos 45º )�i (Bsin 45º )�j B (�i �j) 2 . . . . .. (a)Torque acting on the loop, 2 0 M B (I L k� ) B (�i �j) 2 . . . . . . . . .. .. . ... .. . 2 0 I L B(�j �i) 2 . . . . or 2 0 | . |. I L B . (b)Axis of rotationcoincideswith the torque and since torque is in �j. �i directionor parallel toQS.Therefore, the loopwill rotate about an axis passing throughQand S as shown along-side : Angular acceleration, | | I. . . .
MAGNETIC FIELD www.physicsashok.in 29 Where I =moment of inertia of loop about QS. IQS + IPR = IZZ (Fromtheoremof perpendicular axis) Y X P Q S R But IQS = IPR . 2 QS ZZ 2I I 4ML 3 . . . 2 QS I 2ML 3 . . 2 0 0 2 | | I L B 3 I b I 2 ML 2 M 3 . . . . . . . Angle bywhich the frame rotates in time .t is 1 . ( t)2 2 . . . . or 0 2 3 I B . ( t) 4 M . . . MOVING COIL OR SUSPENDED COIL OR D´ ARSONVAL TYPE GALVANOMETER Principle : When a current-carrying coil is placed in magnetic field, it experiences a torque. Construction : It consists ofa narrowrectangular coilPQRS consisting of a large number of turns of fine insulated copperwirewound over a framemade of light, non-magneticmetal.Asoft ironcylinder known as the core is placed symmetricallywithin the coil and detached fromit. The coil is suspended between the two cylindrical pole-pieces (Nand S of a strong permanent horse-shoemagnet) by a thin flat phosphor bronze strip, the upper end ofwhichis connected to amovable torsion head T.The lower end of the coil is connected to a hair-spring �s� of phosphor bronze having onlya fewturns. N Core S P S Q R s T T1 T2 m Moving coil galvanometer In order to eliminate air-disturbance, thewhole arrangement is enclosed in a brass case having a glasswindow
on the front. Levelling screws are provided at the base.The torsion head T is connected to a binding terminal T1. So, the phosphor-bronze strip acts as one �current lead� to the coil. The lower end of the spring �s� is connected to a binding terminalT2.Aplanemirror or a concavemirror of larger radius of curvature is rigidly attached to the phosphor bronze strip. This helps to measure the deflection of the coil by lamp and scale arrangement. Radialmagnetic field : The magnetic field in the small air gap between the cylinderical pole-pieces is radial. Themagnetic lines of forcewithin the air gap are along the radii. On account of this, the plane of the coil remains N S always parallel to the direction ofthemagnetic field i.e., the angle between Radial magnetic field the plane of the coil and themagnetic field is zero in all the orientations of the coil.
MAGNETIC FIELD www.physicsashok.in 30 Theory : Let I = current flowing through the coil; B=magnetic field induction l = lengthof the coil; b = breadth of the coil N= number of turns in the coil; A(= l × b) = area of the coil Since the field is radial, therefore, the plane ofthe coilremains parallel to themagnetic field inall the orientations of the coil. So, the sides SP and QR remain parallel to the direction of the magnetic field. So, they do not experience any force.The sides PQandRS remain perpendicualr to the direction of themagnetic field. These sides experience forces perpendicular to the plane of the coil. b F F P S Q R l Current-carrying loop in magnetic field Force on PQ, F = NBIl Applying Fleming�s left hand rule,we find that this force is normal to the plane of the coiland directed outwards, i.e., towards the reader. Force on RS, F = NBIl Applying Fleming�s left hand rule,we find that this force is normal to the plane of the coiland directed inwards, i.e., awayfromthe reader. The forces onthe sides PQandRS are : (i) equal inmagnitude; (ii) opposite in direction; and (iii) act at different points. So, the two forces constitute a couple. This couple tends to deflect the coil and is known as deflecting couple. Moment of deflecting couple =NBIl × b =NBIA [The field is radial. The forces onthe sides PQandRS always remain perpendicular to the plane ofthe coil. So, the perpendicular distance between the forces is always equal to �b� as in fig.] b FP SF When the coildeflects, the suspension fibre gets twisted. On account ofelasticity, a restoring couple is set up in the fibre.This couple is proportional to the twist. If . be the angular twist, then Moment of restoring couple = k. where �k� is the restoring couple per unit angular twist. It is also known as torsional constant. For equilibriumof the coil, NBIA= k. or I k NBA . . . . .. .. or I . K. where K k NBA . . . .. .. is the galvanometer constant. Now, I .. or . . I
So, the deflection of the coilis proportional to the current flowing through the coil. This explains as towhywe can use a linear scale in a galvanometer.The scale is calibrated to give direct values of current.
MAGNETIC FIELD www.physicsashok.in 31 CURRENT SENSITIVITY OF A GALVANOMETER Agalvanometer is said to be sensitive if it gives a large deflection for a small current. The current sensitivity of a meter is the deflection of the meter per unit current, i.e., I . . It is given by NBA I k . . [. NBIA= k.] The sensitiveness can be increased by increasing N,Aand B and decreasing the value of �k�. But N andA cannot be increasedmuch because thiswill increase the length and consequentlythe resistance of the coil. In that case, the galvanometer will not respond toweak electric currents. B can be increased by using a strongmagnet. �k� can be decreased by using phosphor-bronze for suspension. �k� can be further reduced by using quartz suspension fibre. VOLTAGE SENSITIVITY OF A GALVANOMETER It is defined as the deflection of the meter per unit voltage, i.e., V. . Now, V RI . . . or NBA V kR . . Advantages : (i) The galvanometer can bemade extremely sensitive. (ii) Since themagnetic field B is very high, therefore, the externalmagnetic fields cannot appreciably after the deflection of the coil. So, the galvanometer can be used in any position. (iii) Since the deflection of the coil is proportional to current, therefore, linear scale canbe used. (iv) Since the coil is wound over metallic frame, therefore, damping is produced by eddy currents. So, the galvanometer coilcomes to rest quickly.This typeofgalvamometer is called aperiodic or dead beat galvanometer. The galvanometer can bemade ballistic bywinding the coil on a non-conducting frame of ivory or ebonite. (v) The lamp and scale arrangement used to measure the deflection of the coilmakes the galvanometer very sensitive. POINTER TYPE OR WESTON OR PIVOTED MOVING COIL GALVANOMETER The suspended typemoving coil galvanometers are very sensitive. Theycanmeasure currents of the order of 10�9 ampere. But these require very careful handling. So, for general use in the laboratory and for those experimentswhose sensitivityis not required, pointer type galvanometers are used. T1 T2 N S 30 20 10 0 10 20 30
MAGNETIC FIELD www.physicsashok.in 32 Inthis typeof galvanometer, the coilis pivoted between two ball-bearings.Alight aluminiumpointer is attached to themoving coil. The controlling couple is providedwith the help of a spring. NOTE : Numerical Examples based on Moving Coil Galvanometer Formulae used : 1. . = NBIA 2. k. = NBIA 3. Current sensitivity, a = NBA I k 4. Voltage sensitivity, a = NBA V kR Units used : B in tesla, A in m2, R in ohm, k in N m rad�1. EARTH�S MAGNETISM Afreelysuspendedmagnet always points inthe north-south direction evenin the absence ofanyothermagnet. This suggests that the earth itself behaves as amagnet which causes a freely suspendedmagnet (ormagnetic needle) to point always in a particular direction : northand south.The shape ofearth�smagnetic field resembles that of a barmagnet of length one-fifth of earth�s diameter buried at its centre. SN Magnetic axis Magnetic N-pole Geographic S-pole Magnetic equatorEquator Magnetic S-pole Geographic N-pole Geographic axis The south pole of earth�smagnet is towards earth�s north pole (eographical north), while the north pole of earth�smagnet is towards earth�s south pole (geographical south). Thus, there is a magnetic S-pole near the geographicalnorth, and amagneticN-pole near the geographical south.The positions of the earth�smagnetic poles are not well defined on the globe, they are spread over an area. Magnetic equator : The great circlewhose place is perpendicular to the earth�smagnetic axis is called earth�s magnetic equator. Geographical equator : The great circlewhose plane is perpendicular to geographical axis is called geographical equator. Magnetic meridian : The line joining the earth�smagnetic poles is called themagnetic axis and a vertical plane passing throughit is called themagneticmeridian. Geographicalmeridian : The line joining the geographical north and south poles is called the geographic axis and a vertical plane passing through it is called the geographicalmeridian. Magnetic Elements To have a complet knowledge of earth�smagnetismat a place, the following three elementsmust be known :
(i)Angle of declination (ii)Angle ofdip or inclination (iii)Horizontal component of earth�s field. (i) Angle of declination : The angle between the magnetic meridian and geographicalmeridian at a place is called the angle of declination (or simplythe declination) at that place.
MAGNETIC FIELD www.physicsashok.in 33 Magnetic meridian B H C D C´B´ Geographical meridian Geographic north Magnetic north In fig.ABCDis themagneticmeridian andAB´C´Dis the geographicalmeridian. The angleB´AB = . is the angle of declination. (ii) Angle of dip or inclination : The anglewhich the axis of needlemakeswith the horizontal, is called angle of dip (.). In otherwords, the angle ofdip at a place is the anglewhich the resultantmagnetic field ofearth at that placemakeswith the horizontal. N S C i V DA B H In fig.ACshows the direction of resultantmagnetic field of earth and the angleBAC(=..) between it and the horizontalABis the angle of dip. (iii) Horizontal component of earth�s field : The direction of earth�s field at themagnetic poles is normal to the earth�s surface (i.e., in vertical direction) and at magnetic equator it is parallel to the earth�s surface, (i.e., in horizontaldirection).Thus, the resultant earth�s field can be resolved in two components as shown in fig. (a) the horizontal component HalongABand (b) the vertical componentV, alongAD. Fromfig. Horizontalcomponent H = Be cos . ...(i) and vertical component V = Be sin . ...(ii) . ee V B sin tan H B cos. . . . . or V =Htan . AgainEqs. (i) and (ii) give 2 2 2 . 2 2 . e H . V . B cos . . sin . or 2 2 e B . H . V
MAGNETIC FIELD www.physicsashok.in 34 THINKING PROBLEMS 1. Of the three vectors in the equation F . qv.B, . . .. which pairs are always at right angles?Whichmayhave any angle between them? 2. If an electron is not deflected in passing through a certainspace, canwe be sure that there is no magnetic field in that region? 3. Abreamof protons is deflected side ways. Could this deflection be caused (a) by an electric field? (b) by a magnetic field ? (c) If either is possible, howcan you tellwhich one is present? 4. Arectangular current loop is in an arbitary orientation in an externalmagnetic field. Is anywork required to rotate the loop about an axis perpendicular to its plane? 5. Will a tangent galvanometer work in the polar region ? 6. At that deflections is a tangent galvanometermost sensitive? 7. Whyis themagnetic needle short in a tangent galvanometer ? 8. Why is the field in amoving coilgalvanometer radialin nature? 9. What is the greatest disadvantagewith a suspended-typemoving coilgalvanometer? 10. An ammeter is aWeston galvanometer whose resistance ismade negligible byshunting the coil. Is this true or false? 11. Avoltmeter is aWeston galvanometer ofveryhigh resistance. Is this true of False? 12. Which gives amore accurate value of a potential difference, a potentiometer or a voltmeter? 13. Why is an ammeter connected in series? 14. What is a faraday? 15. There is no charge in the energyof a charged particlemoving in amagnetic field althrough amagnetic force is acting on it. Is this true or false?Give reasons in support of your answer. 16. Acurrent-carrying circular conductor is placed in a uniformmagnetic fieldwith its plane perpendicular to the field.Does it experience anyforce? Ifit does,what is this force if its radius is a and the current passing through it equals? 17. Howdo you knowthat the current inside a conductor is constituted by electrons and not by protons? 18. Acopper pipe is filledwithan electrolyte.When a voltage is applied, the current in the electrolyte is constituted bythemovement of positive and negative ions in opposite directions.Willsuch a pipe experience a forcewhen placed in amagnetic field perpendicular to the current? 19. Two parallel wires currying current in the same direction attract each other while two beams of electrons travelling in the same direction repel each other.Explainwhy? 20. Can a charged particle entering a uniformmagnetic field normally fromoutside complete a circle? 21. Acylindrical electrolytic bath containing a solution of copper sulphate between two electrodes ismounted above the northpole of a strong electromagnet.One electrode is at the axis (�) and the other electrode is at the edge of the bath (+).What happens to the electrolyte in these circumstances? 22. Averystrong current ismade to flowfor a short time through a solenoid.Willthere be anychange in its length
and diameter?Explain. 23. Cosmic rays are charged particles that strike the atmosphere fromsome external source. It is found thatmore low-energycosmic rays eachthe earth at the northand southmagnetic poles thanat themagnetic equator.Why is this so ?
MAGNETIC FIELD www.physicsashok.in 35 SOLUTION OF THINKING PROBLEMS 1. The pairsF . , B.. and F . , v . are always at right angle. B.. and v . may have any angle between them. 2. No, we cannot be sure that there is no magnetic field because the force will be zero when the direction of motion is along the direction ofthe field. 3. (a)Yes, it could be due to an electric field directed perpendicular to the motion. (b)Ys, it could be due to a magnetic field. (c) Stop the proton and keep it stationary. Ifit sill experiences a force in the same direction, it is due to an electric field. If the force vanishes of stopping themotion, there is amagnetic field. 4. No, no work is done in rotating the coil becausework doneW=mB (1 �cos.). Here there is no charge in . and so nowork is done. 5. No, because the earth�smagnetic field is vertical there. 6. Atengent galvanometer ismost sensitivewhen the deflections are near about 0º. 7. The field due to the circular coil is uniformover a very small region about the centre of the coil. So the needle must be short so that it maybe assumed tomove in uniformmegnetic fields. 8. The field ismade radial in order to have a linear relation between the current and the deflection. 9. The greatest disadvntage is that this type of galvanometer is not portable. 10. It is true. 11. It is true. 12. Apotentiometer gives amore accurate value of a potential difference than a voltmeter as it draws no current fromthe cell. 13. Since the current following through the circuit has to pass through the ammeter, it must be connected in series because it is a characteristic property ofseries connection that the same current passes through allparts of the circuit. 14. Afaradayis the amount of charge required to liberate 1 gramequivalent of any sustance. 15. True.Acharged particle experiences a force at right angles to the velocity and so it moves in a circle with a constant speed. So there is no charge in the energyfo the particle though amagnetic force acts on it. 16. It does not experiencemayforce because the forces on the elements are directed radially away and sumup to zero. 17. BytheHall effect.When a current-carrying flet conductor is placed inamagnetic field perpendicular to the flat face, a transversevoltage is developed. Fromthe directionofthis voltage calledHallvoltage, it canbe determined whether the current is due to negative or positive charge carriers. 18. The current constituted bynegtive ions in the positive directionis the same as the current formed bythe positive ions and so both form current in the same direction; they do not cancel out each other. So the pipe will experience a force. 19. Acurrent inonwire produces onlyamagnetic field and on electric field (because a current-carrying conductor is electricallyneutral) over the other current-carrying conductor and so onlyama
gnetic forcewhich is attractive is nature arises between them. But an electron beamis a source of both, an electric and amagnetic field. So there arise bothmagnetic force (attractive) and electric force (repulsive) between them. The repulsive force being in excess of the attractive force, theyrepel each other. 20. No, the boundaryline of themagneitc field has to be the diameter of the circle,whatever be the velocity of the charged particle.Hence, it will complete a semi-circle. 21. It moves counterclockwise (viewed fromabove) 22. The current in the adjacent turns flow in the same direction and so they attract each other. Thus there is contraction along the length ofthe solenoid.As the same currents in the diametricallyopposite elements flowin opposite directions, they repel each other. On account of this repulsive force the diameter of the solenoid increases. 23. Because at the poles, themagnetic field is parallel to the direction ofmotion of the cosmic rayparticles, both being vertical. Hence the cosmic ray particles do not experience anyforce. The force experienced is given by F = qvb sin(v . ,B.. ).At the equator these particles experiencemaximumdeflecting force and hence low-energy particles cannot reach the earth.
MAGNETIC FIELD www.physicsashok.in 36 ASSERTION&REASON Astatement of Statement-1 is given and a Corresponding statement of Statement-2 is given just belowit of the statements,mark the correct answer as � (A) If both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation ofStatement-1. (B) If both Statement-1 and Statement-2 are true and Statement-2 isNOT correct explanation ofStatement-1. (C) If Statement-1 is true but Statement-2 is false. (D) If both Statement-1 and Statement-2 are false. (E) IfStatement-1 is false but Statement-2 is true. 1. Statement-1 : Magnetic field interacts with a moving charge and not with a stationary charge. Statement-2 : A moving charge produces a magnetic field. 2. Statement-1 : If an electron is not deflected while passing through a certain region of space, then only possibility is that there is no magnetic field in this region. Statement-2 : Force is directly proportional to magnetic field applied. 3. Statement-1 : Free electrons always keep on moving in a conductor even then no magnetic force act on them in magnetic field unless a current is passed through it. Statement-2 : The average velocity of free electron is zero. 4. Statement-1 : Electron cannot be accelerated by the cyclotron. Statement-2 : Cyclotron is suitable only for accelerating heavy particles. 5. Statement-1 : The coil is wound over the metallic frame in moving coil galvanometer. Statement-2 : The metallic frame help in making steady deflection without any oscillation. 6. Statement-1 : In electric circuits, wires carrying currents in opposite directions are often twisted together. Statement-2 : If the wire are not twisted together, the combination of the wires forms a current loop. The magnetic field generated by the loop might affect adjacent circuits or components. 7. Statement-1 : If an electron and proton enter in an electric field with equal energy, then path of electron is more curved than that of proton. Statement-2 : Electron has a tendency to form large curve due to small mass. 8. Statement-1 : If a proton and an . -particle enter a uniform magnetic field perpendicularly, with the same speed, the time period of revolution of . -particle is double that of proton. Statement-2 : In a magnetic field, the time period of revolution of a charged particle is directly proportional to the mass of the particle and is inversely proportional to charge of particle. 9. Statement-1 : If an electron while coming vertically from outer space enter the earth�s magnetic field, it is deflected towards west. Statement-2 : Electron has negative charge. 10. Statement-1 : An electron and proton enters a magnetic field with equal velocities, then, the force experienced by proton will be more than electron. Statement-2 : The mass of proton is 1837 times more than the mass of electron. 11. Statement-1 : The magnetic field produced by a current carrying solenoid is independent of its length and cross sectional area. Statement-2 : The magnetic field inside the solenoid is uniform.
12. Statement-1 : The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil. Statement-2 : Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized. [JEE 2008]
MAGNETIC FIELD www.physicsashok.in 37 Level # 1 1. An electron is accelerated to a high speed down the axis of a cathode ray tube by the application of a potential difference of V volts between the cathode and the anode. The particle then passes through a uniform transverse magnetic field in which it experiences a force F. If the potential difference between the anode and the cathode is increased to 2 V, the electron will now experience a force (A) F 2 (B) F 2 (C) 2 F (D) 2 F 2. In a hydrogen atom, an electron of mass m and charge e is in an orbit of radius r making n revolutions per second. If the mass of the hydrogen nucleus is M, the magnetic moment associated with the orbital motion of the electron is (A) M .ner2 m (B) m .ner2M (C) (M m) ner2 m . . (D) .ner2 3. An electron of charge e moves in a circular orbit of radius r around a nucleus. The magnetic field due to orbital motion of the electron at the site of the nucleus if B. The angular velocity . of the electron is (A) 4 r2 0 eB . . . . (B) r0 eB . . . . (C) e4 rB 0 .. . . (D) e2 rB 0 .. . . 4. Three long, straight and parallel wires C, D and G carrying currents are arranged as shown in Figure. The force experienced by a 25 cm length D C G 30 A 10 A 20 A 3 cm 10 cm of wire C is (A) 0.4 N (B) 0.04 N (C) 4 x 10�3 N (D) 4 x 10�4 N 5. A charged particle of specific charge s passes through a region of space shown.
(A) Velocity of the particle in REGION 1 is B. . . . (B) Work done to move the charged particle in REGION 1 is ZERO. (C) The radius of the trajectory of the charged particle in REGION 2 is s..0 . . (D) The particle emerges from REGION 2 with a velocity ' . . v .1 . 2 B S E B0 xxxxxxxx xxxxxxxx xxxxxxxx xxxxxxxx x x x x x x x x REGION 1 REGION 2 where ' . . = ... . 6. A wire of resistance R in the form of a semicircle lies on the top of a smooth table. A uniform magnetic field B is confined to the region as shown. The ends of the semicircle are attached to springs C and D whose other ends are fixed. E C DB B = 0 below this line . X If r is the radius of the semicircle and k is the force constant for each spring, then the extension x in each spring is (A) kR x . 2EBr (B) kr x . 2EBR (C) kR x . EBr (D) 2kR x . EBr 7. An infinite collection of current carrying conductors each carrying a current . outwards perpendicular to paper are placed at x = x0, x = 3x0, x = 5x0, ........ ad infinitum on the x axis. Another infinite collection of current carrying conductors each carrying a current . inwards perpendicular to paper are placed at x = 2x0, x = 4x0, x = 6x0, ....... ad infinitum Here x0 is a positive constant. The magnetic field at the origin due to the above collection of current carrying conductors is (A) ZERO (B) 4 x log 2 0 e 0 . . . (C) . (D) 0
0 e 2 x log 2 . . .
MAGNETIC FIELD www.physicsashok.in 38 8. A conductor ABCDEF, with each side of length L, is bent as shown. It is carrying a current . in a uniform magnetic induction (field) B, parallel to the positive y-direction. The force experienced by the wire is B A C D E L L Z B Y X O F (A) BIL in the positive y-direction. (B) BIL in the negative z-direction. (C) 3 BIL (D) zero 9. The square loop ABCD, carrying a current . , is placed in a uniform magnetic field B, as shown. The loop can rotate about the axis XX�. The plate of the loop makes an angle . ( . < 90°) with the direction of B. Through what A B B X X� Y Z CD . angle will the loop rotate by itself before the torque on it becomes zero ? . (A) . (B) 90° (C) 90° + . (D) 180° � . 10. An electron is fired from the point A with a velocity V0 = 2 x 108 m/s. The magnitude and direction of magnetic field that will cause the A B 15 cm x V0 y electron to follow a semicircular path from A to B is (A) 1.5 x 10�4 T out of the page. (B) 1.5 x 10�3 T into the page (C) 1.5 x 10�2 T into the page (D) 1.5 x 10�2 T out of the page 11. Two long straight parallel wires 2 m apart, are perpendicular to the plane of the paper. Wire A carries a current of 10 A directed into the plane of the paper. Wire B carries a current such that the magnetic field at P at a distance of 0.8 m from this wire is zero.The magnitude and direction of the current in wire is zero. The AB S 1.2 m
1.6 m 2 mP magnitude and direction of the current in wire B is (A) 2.8 A into the page (B) 2.8 A out of the page (C) 2.8 x 10�6 A into the page. (D) 2.08 A out of the page. 12. Two particles each of mass m and charge q, are attached to the two ends of a light rigid rod of length 2 . . The rod is rotated at a constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is (A) 2mq (B) mq (C) m2q (D) mq . 13. If a charged particle is describing a circle of radius r in a magnetic field with a time period T then, (A) T 2 . r 3 (B) T 2 . r (C) T . r 2 (D) T . r 0 14. A non-conducting rod AB of length . has a linear charge density . . The rod is rotated about an axis passing through point A with constant angular velocity . as shown in the figure. The magnetic moment of A +++ B . the rod is (A) 2 2 ... (B) 2 3 ... (C) 3 3 2 .. . (D) 3 6 ...
MAGNETIC FIELD www.physicsashok.in 39 15. A charged particle of mass m and charge q is released from rest from (x0, 0) along an electric field 0 E j� . The angular momentum of the particle about origin (A) is zero (B) is constant (C) increases with time (D) decreases with time 16. Two particles Y and Z emitted by a radioactive source at P made tracks in a could chamber as illustrated in the figure. A magnetic field acted downward into the paper. Careful measurements showed that both tracks P Y were circular, the radius of Y track being half that of the Z track. Z Which one of the following statements is certainly true? (A) Both Y and Z particles carried a positive charge. (B) The mass of Z particle was one half that of the Y particle. (C) The mass of the Z particle was twice that of the Y particle. (D) The charge of the Z particle was twice that of the Y particle. 17. The resistances of three parts of a circular loop are as shown in the figure. The magnetic field at the centre O is (A) 0 6 I a . (B) 0 3 I a . (C) 0 23 I a . (D) Zero A BR C R a 2R O I 120° 120° 18. A particle with a specific charge s is fired with a speed v towards a wall at a distance d, perpendicular to the wall. What minimum magnetic field must exist in this region for the particle not to hit the wall? (A) v/sd (B) 2v/sd (C) v/2sd (D) v/4sd 19. Current . flows through a long conducting wire bent at right angle as shown in figure. The magnetic field at a point P on the right bisector of the angle XOY at distance r from O is (A) 0r .. .
(B) 0 2 r . . . X YO I P x 45° r (C) 0 . . 2 2 1 4 r .. . . (D) 0 . 2 1. 2 r .. . . 20. Two circular coils X and Y having equal number of turns and carry equal currents in the same sense and subtend same solid angle at point O. If the smaller coil X is midway d X Y d O be between O and Y, then if we represent the magnetic induction due to bigger coil Y at O as By and that due to smaller coil X at O as Bx, then (A) 1 yx BB . (B) y 2 x BB . (C) 12 yx BB . (D) 14 yx BB . 21. Two mutually perpendicular conductors carrying 1 . and 2 . lie in the x-y plane. Find locus of points at which magnetic induction is zero? (A) 22 2
y x . . . (B) 12 x y . . . (C) 12 y x . . . (D) 2 2 12 x y . . . .
MAGNETIC FIELD www.physicsashok.in 40 22. A coil in the shape of an equilateral triangle of side 0.02 m is suspended from a vertex such that it is hanging in a vertical plane between the pole pieces of a permanent magnet producing a horizontal magnetic field of 5 x 10�2 T. Find the couple acting on the coil when a current of 0.1 N S ampere is passed through it and the magnetic field is parallel to its plane. (A) 5 2N .m (B) 5 2N .10.4N .m (C) 5 3 .10.7N . m (D) 10.7N . m 23. A particle of charge q and mass m starts moving from the origin under the action of an electric field 0 E . E i� . and magnetic field 0 B . B k� . . It s velocity at (x, 3, 0) is .4i� . 3 �j. . The value of x is: (A) 0 0 36E B qm (B) 0 25 2 m q E (C) 0 10m q E (D) 0 0 25E B m 24. Two charged particles A and B enter a uniform magnetic field with velocities normal to the field. Their paths are shown in the Figure. The possible reasons are: (A) The momentum of A is greater than that of B (B) the charge of A is greater than that of B. (C) The specific charge of A is greater than that of B (D) the speed of A is less than that of B. 25. Two charged particle M and N enter a space of uniform magnetic field, with velocities, perpendicular to the magnetic field. the paths are as shown in the figure. The possible reason are: (A) the charge of M is greater than that of N. (B) the momentum of M is greater than that of N. (C) specific charge of M is greater than that of N. (D) the speed of M is less than that of N. 26. A current-carrying ring is placed in a magnetic field. The direction of the field is perpendicular to the plane of the ring, then: (A) there is no net force on the ring. (B) the ring may tend to expand. (C) the ring may tend to contact (D) none of these. 27. A charge q is moving with a velocity 1 v. .1i� m s at a point in a magnetic field and experiences a force . . 1 F . q .1�j .1 k� . N. If the charge is moving with a velocity 2 v. .1 �j ma/s at the same point, it experiences
MAGNETIC FIELD www.physicsashok.in 41 a force 2 F . q .1i� .1k�. N . . . . The magnetic induction B . at that point is: (A) .i� . �j . k�.Wb m2 (B) .i� . �j . k�.Wb m2 (C) ..i� . �j . k�.Wb m2 (D) .i� . �j . k�.W28. A parallel beam of electrons is shot into a uniform electric field, initially parallel to and against the field with a small initial speed. Then: (A) the beam will pass through the field accelerating down the field without changing its width. (B) the beam tends to spread out at the beginning and to narrow down later. (C) the beam tends to narrow down at the beginning and to spread out later. (D) the total energy of the beam is conserved. 29. The ratio of the energy required to set up in a cube of side 10 cm a uniform magnetic field of 4 wb/m2 and a uniform electric field of 106 V/m is: (A) 1.4 x 107 (B) 1.4 x 105 (C) 1.4 x 106 (D) 1.4 x 103 30. A charged particle is fired at an angle . in a uniform magnetic field directed along the x-axis. During its motion along a helical path, the particle will: (A) never parallel to the x-axis. (B) move parallel to the x-axis once during every rotation for all values of . . (C) move parallel to the x-axis at least once during every rotation if . � 45°. (D) never move perpendicular to the x-direction. 31. A charged particle q enters a region of uniform B . (out of the page) and is deflected a distance d after travelling a horizontal distance a. The magnitude of the momentum of the particle is: (A) 2 2 qB a d d . . . . . . . (B) 2 qBa (C) Zero (D) not possible to be determined as it keeps changing. 32. Two insulated rings, one of slightly smaller diameter then the other, are suspended along their common diameter as shown in the figure, initially the planes of the rings are mutually perpendicular. When a steady current is set up in each of them: (A) The two rings rotate towards a common plane. (B) The inner ring will oscillate about its initial position. (C) The inner ring stays stationary while the outer one moves into the plane of the inner ring. (D) The outer ring stays stationary while the inner one moves into the plane of the outer ring. 33. A particle with charge +Q and mass m enters a magnetic field of magnitude B, existing only on the right of the boundary YZ. The direction of the motion of the particle is perpendicular to the direction of B. Let T 2 m
QB . . . The time spent by the particle in the field will be: (A) T. (B) 2T. (C) 2 2 T . . . . . . . . . . (D) 2 2 T . . . . . . . . . . 34. The current I flows through a square loop of a wire of side a. The magnetic induction at the centre of the loop is: (A) 0 2a . . . (B) 0 2 2a. . . (C) 0 2a. . (D) 0 2 a . . .
MAGNETIC FIELD www.physicsashok.in 42 35. Suppose a uniform electric field and a uniform magnetic field exist along mutually perpendicular directions in a gravity free space. If a charged particle is released from rest, at a point in the space: (A) particle can not remain in static equilibrium. (B) first particle will move along a curved path but after some time its velocity will become constant. (C) particle will come to rest at regular interval of time. (D) acceleration of the particle will never become equal to zero. 36. A circular loop of radius R carries a charge q uniformly distributed on it. It is rotated at a frequency f about one of the diameters. A uniform magnetic field B exists along its diameter . The maximum and minimum torques acting on the loop due to the magnetic field are, respectively: (A) . qfR2B, 0 (B) 0, 0 (C) 2. qfR2B , . qfR2B (D) None of these 37. A semi-circular wire of radius R is connected to a wire bent in the form of a sine curve to form a closed loop as shown n the figure. If the loop carries a current . and is placed in a uniform magnetic field B, then the total force acting on the sine curve is: B (A) 2B.R (downward) (B) 2B.R (upward) (C) B.R (upward) (D) Zero 38. A charged particle of unit mass and unit charge moves with velocity of v . .8i� . 6 �j.m s . in a magnetic field of B . 2 k�T . . Choose the correct alternative(s): (A) The path of the particle be x2 + y2 � 4x � 21 = 0. (B) The path of the particle may be x2 + y2 = 25. (C) The path of the particle may be y2 + z2 = 25. (D) The time period of the particle will be 3.14 s. 39. A proton, a deuteron and an .-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If rp, rd, and r. denote respectively the radii of the trajectories of these particles, then. (JEE 1997) (A) p d r r r . . . (B) d p r r r . . . (C) d p r r r . . . (D) p d r r r. . . 40. A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a (JEE 1999) (A) straight line (B) circle (C) helix (D) cycloid 41. Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper, as shown. The variation of the magnetic field B along the line XX� is given by (JEE 2000) (A) d d X X� B (B) d d X X� B (C) d d X X�
B (D) d d X X� B 42. An infinitely long conductor PQR is bent to form a right angle as shown in figure. A current . flows through PQR. The magnetic field due tot his current at the point M is H1. Now, another infinitely long straight conductor QS is connected at Q so that current is . 2 in QR as well as in QS, the current in PQ remaining unchanged. The magnetic field at M is now H2. The ratio 1 2 H H is given by (JEE 2000) (A) ½ (B) 1 (C) 2 3 (D) 2
MAGNETIC FIELD www.physicsashok.in 43 43. An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the +x direction and a magnetic field along the +z direction, then (JEE 2000) (A) positive ions deflect towards +y direction and negative ions towards �y direction (B) all ions deflect towards +y direction. (C) all ions deflect towards �y direction (D) positive ions deflect towards �y direction and negative ions towards +y direction. 44. A non-planar loop of conducting wire carrying a current . is placed as shown in the figure. Each of the straight sections of the loop is of length 2a. The magnetic field due to this loop at the point x y z I 2a P(a, 0, a) points in the direction (JEE 2001) (A) 1 . � �. 2 . j . k (B) 1 . � � �. 3 . j . k . i (C) . . 1 � � � 3 i . j . k (D) 1 .� �. 2 i . k 45. Two particles A and B of masses mA and mB respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to Athis plane. The speeds of the particles are V B A and VB respectively and the trajectories are as shown in the figure. Then (JEE 2001) (A) mA vA < mB vB (B) mA vA > mB vB (C) mA < mB and vA < vB (D) mA = mB and vA = vB 46. A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current . passes through the coil, the magnetic field at the center is (JEE 2001) (A) 0 N b . . (B) 0 2 N b . . (C) . . 0 ln 2 N b b a a . . . (D) . . 0 ln 2 N b b a a . .
. 47. A particle of mass m and charge q moves with a constant velocity v along the positive x direction. It enters a region containing a uniform magnetic field B directed along the negative z direction, extending from x = a to x = b. The minimum value of v required so that the particle can just enter the region x > b is (JEE 2002) (A) qbB m (B) q .b a. B m. (C) qaB m (D) . . 2 q b a B m . 48. A long straight wire along the z-axis carries a current . in the negative z direction. The magnetic vector field B . at a point having coordinates (x, y) in the z = 0 plane is (JEE 2002) (A) . . . . 0 2 2 � � 2 y i x j x y ... .. (B) . . . . 0 2 2 � � 2 x i y j x y ... .. (C) . . . . 0 2 2 � � 2 x j y i x y ..
. .
. (D)
. .
. . 0 2 2 � � 2 x i y j x y ... .. 49. For a positive charged particle moving in a x-y plane initially along the x-axis, there is a sudden change in its path due tot he presence of electric and/or magnetic fields beyond P. The curved path is shown x y in the x-y plane and is found to be non-circular. Which one of the P following combinations is possible? (JEE 2003) (A) E . 0; B . b i� . c k� . . (B) E . a i�; B . c k� . a i� . . (C) E . 0; B . c �j . b k� . . (D) E . a i�; B . c k� . b �j . .
MAGNETIC FIELD www.physicsashok.in 44 50. A conducting loop carrying a current . is placed in a uniform magnetic field pointing into the plane of the paper as shown. x Y I B The loop will have a tendency to (JEE 2003) (A) contract (B) expand (C) move towards +ve x-axis (D) move towards �ve x-axis. 51. A current carrying loop is placed in a uniform magnetic field in four different orientations, I, II, III & IV, arrange them in the decreasing order of Potential Energy (JEE 2003) B � n I B � n II B �n III B � n IV (A) I > III > II > IV (B) I > II > III > IV (C) I > IV > II > III (D) III > IV > I > II 52. An electron traveling with a speed u along the positive x-axis enters into a region of magnetic field where . . 0 B . .B k� x . 0 . It comes out B x ye u � of the region with speed v then (JEE 2004) (A) v = u at y > 0 (B) v = u at y < 0 (C) v > u at y > 0 (D) v > u at y < 0 53. A magnetic needle is kept in a nonuniform magnetic field. It experiences (JEE 1982) (A) a force and a torque (B) a force but not a torque (C) a torque but not a force (D) neither a force nor a torque 54. A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E and B represent the electric and magnetic fields respectively. This region of space may have:(JEE 1985) (A) E = 0, B = 0 (B) E = 0, B . 0 (C) E . 0, B = 0 (D) E . 0, B . 0 55. A rectangular loop carrying a current i is situated near a long straight wire such that the wire is parallel to one of the sides of the loop and is in the plane of the loop. If steady current . is established in the wire as shown i I in the figure, the loop will : (JEE 1985) (A) rotate about an axis parallel to the wire (B) move away from the wire (C) move towards the wire (D) remain stationary 56. Two particle X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii R1 and R2 respectively. The ratio of the mass of X to that of Y is (JEE 1988)
(A) . .1 2 1 2 R R (B) 2 1 R R (C) . .2 1 2 R R (D) 1 2 R R 57. A particle of charge +q and mass m moving under the influence of a uniform electric field E i� and uniform magnetic field B k� follows a trajectory from P to Q as shown in Figure. The velocities at P and Q are v i� and .2 �j 2v P a Q x 2a B V E which of the following statement(s) is/are correct? (JEE 1991) (A) 3 2 4 E mv qa . . . . . . . (B) Rate of work done by the electric field at P is 3 3 4 mv a . . . . . . (C) Rate of work done by the electric field at P is zero (D) Rate of work done by the electric field at Q is zero
MAGNETIC FIELD www.physicsashok.in 45 58. A current 1 flows along the length of an infinitely long, straight, thin-walled pipe. Then: (JEE 1993) (A) The magnetic field at all points inside the pipe is the same, but not zero (B) The magnetic field at any point inside the pipe is zero. (C) The magnetic field is zero only on the axis of the pipe (D) The magnetic field is different at different points inside the pipe. 59. H+ , He+ and O++ all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. The masses of H+, He+ and O2+ are 1 amu, 4 and 16 amu respectively. The: (JEE 1994) (A) H+ will be deflected most. (B) O2+ will be deflected most (C) He+ and O2+ will be deflected equally (D) All will be deflected equally. 60. Two very long, straight, parallel wires carry steady currents . and � . respectively. The distance between the wires is d. At a certain instant of time, a point charge q is at a point equidistant from the two wires, in the plane of the wires, Its instantaneous velocity v is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is (JEE 1998) (A) 0 2 qv d . .. (B) 0 qv d ... (C) 0 2 qv d .. . (D) 0 61. An infinite current carrying wire passes through point O and in perpendicular to the plane containing a current carrying loop ABCD as shown in the figure. Choose the correct option (s) (JEE 2006) (A) Net force on the loop is zero C B O O´ A D (B) Net torque on the loop is zero. (C) As seen from O, the loop rotates clockwise. (D) As seen from O, the loop rotates anticlockwise. 62. A magnetic field 0 B . B �j .. exists in the region a < x < 2a and 0 B . .B �j ..
, in the region 2a < x < 3a, where B0 is a positive constant. A positive point charge moving with a velocity 0 v . v �j . , where v0 is a positive constant, enters the 0 a 2a 3a x �B0 B0 magnetic field at x = 0. The trajectory of the charge in this region can be like, (JEE 2007) (A) z a 2a 3a x (B) z x a 2a 3a (C) z a 2a 3a x (D) z x a 2a 3a 63. A particle of mass m and charge q, moving with velocity v enters Region II normal to the boundary as shown in the figure. Region II has a uniform magnetic field B perpendicular to the plane of the paper. The length of the Region II is l. Choose the correct choice(s) × × × × × × × × × × × × × × × × × × × × × × × × Region I Region II Region III 0 v (A) The particle enters Region III only if its velocity l v q B m . l (JEE 2008) (B) The particle enters Region III only if its velocity v q B m . l (C) Path length of the particle in Region II is maximum when velocity v q B m . l (D) Time spent in Region Ii is same for any velocity V as long as the particle returns to Region I
MAGNETIC FIELD www.physicsashok.in 46 FILL IN THE BLANKS 1. A neutron a proton, and an electron and alpha particle enter a region of constant magnetic field with equal velocities. The magnetic field is along the inward normal to the plane of the paper. The tracks of the X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X B C particles are labelled in Figure. The electron follows track ________ D and the alpha particle follows track ________. (JEE 1984) 2. A wire of length L meters carrying a current i amperes is bent in the form of a circle. The magnitude of its magnetic moment is ________ in MKS units. (JEE 1987) 3. In a hydrogen atom, the electron moves in an orbit of radius 0.5Å making 1016 revolutions per second. The magnetic moment associated with the orbital motion of the electron is ________ . (JEE 1988) 4. The wire loop PQRSP formed by joining two semicircular wires of radii R1 and R2 carries a current 1 as shown. The magnitude of II R2 S R C Q P R1 the magnetic induction at the centre C is ________ . (JEE 1988) 5. A wire ABCDEF (with each side of length L) bent as shown in figure and carrying a current . is placed in a uniform magnetic induction B B A C DE F Z X Y I parallel to the positively-direction. The force experienced by the wire is ________ in the ________ direction. (JEE 1990) 6. A metallic block carrying current . is subjected to a uniform magnetic induction as B . as shown in Figure. The moving charges experience a force F. given by _____ which results in the lowering of the potential of the face _____ Assume the sped of the carriers to be v. (JEE 1996) 7. A uniform magnetic field with a slit system as shown in figure is to be used as momentum filter for high-energy charged particles. With a field B Tesla, it is found that the filter transmits . -particles Source DetectorB each of energy 5.3 MeV. The magnetic field is increased to 2.3 B Tesla and deuterons are passed into the filter. the energy of each deuteron transmitted by the filter is _____ MeV. (JEE 1997)
MAGNETIC FIELD www.physicsashok.in 47 TRUE / FALSE 8. No net force acts on a rectangular coil carrying a steady current when suspended freely in a uniform magnetic field. (JEE 1981) 9. There is no change in the energy of a charged particle moving in magnetic field although a magnetic force is acting on it. (JEE 1983) 10. A charged particle enters a region of uniform magnetic field at an angle of 85° to the magnetic line of force. The path of the particle is a circle. (JEE 1983) 11. An electron and proton are moving with the same kinetic energy along the same direction. When they pass through a uniform magnetic field perpendicular to the direction of their motion, they describe circular paths of the same radius. (JEE 1985) TABLEMATCHING 12. Column I Column II (A) Electric field (P) Stationary charge (B) Magnetic field (Q) Moving charge (C) Electric force (R) Changes the kinetic energy (D) Magnetic force (S) Does not change kinetic energy 13. Regarding the trajectory of a charged particle, match the following: Column I Column II (A) In electric field (P) Straight line path (B) In magnetic field (Q) Circular path (C) In crossed field (R) Helical path (D) In mutually perpendicular electric and (S) Parabolic path magnetic field, charge being at rest (T) Parabolic path 14. A current flows along length of a long thin cylindrical shell: Column I Column II (A) Magnetic field at all points lying inside the shell (P) Inversely proportional with distance from axis of shell (B) Magnetic field at any point outside the shell (Q) Zero (C) Magnetic field is maximum (R) Just outside the shell (D) Magnetic field on the axis of the shell 15. Column I Column II (A) Unit of magnetic field (P) Am2 (B) Unit of magnetic permeability (µ0) (Q) N Am (C) Unit of magnetic flux (.) (R) N A2 (D) Unit of magnetic dipole moment (S) Nm A 16. Two long parallel wires carrying equal currents in opposite directions are placed at x = . a parallel to Y-axis with z = 0. Then: Column I Column II (A) Magnetic field B1 at origin O (P) 0 3 ia .. (B) Magnetic field B2 at P (2a, 0, 0) (Q) 0 4 ia ..
. (C) Magnetic field at M (a, 0, 0) (R) 0i a .. . (D) Magnetic field at N (�a, 0, 0) (S) Zero
MAGNETIC FIELD www.physicsashok.in 48 17. Two wires each carrying a steady current I are shown in four configurations in Column-I. Some of the resulting effects are deseribed in Column-II. Match the statements is Column-I with the statements in Column-II. Column I Column II [JEE 2007] (A) Point P is situated P (P) The magnetic fields (B) at P due to the currents midway between the wires. in the wires are in the same direction. (B) Point P is situated at the (Q) The magnetic fields (B) at P due to the currents in mid-point of the joining the the wires are in opposite directions. centers of the circular P wires, which have same radii. (C) Point P is situated at the (R) There is no magnetic field at P mid-point of the line joining the centres of the circular P wires, which have same radii. (D) Point P is situated at the (S) The wires repel each other. common center of the wires. P 18. Six point charges, each of the same magnitude q, are arranged in different manners as shon in Column-II. In each case, a point M and a line PQ passing through M are shown. Let E be the electric field and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let B be the magnetic field at M and µ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current. [JEE 2009] Column I Column II (A) E = 0 (P) Charges are at the corners of a regular hexagon. M is at the centre of the hexagon. PQ is + � + � � M + P Q perpendicular to the plane of the hexagon. (B) V . 0 (Q) � + � + � + Charges are on a line perpendicular to PQ at QPM equal intervals. M is the mid-point between the two innermost charges. (C) B = 0 (R) Charges are placed on two coplanar insulating rings at equal intervals. M is the common centre + � Q P M� � + + of the rings. PQ is rependicular to the plane of
the rings. (D) µ . 0 (S) Charges are placed at the corners of a rectangle of sides a and 2a and at the mid points of the P M Q +� + �� � longer sides. M is at the centre of the rectangle. PQ is parallel to the longer sides. (T) Charges are placed on two coplanar, insulating rings at equal intervals. M is the mid-point bet- P QM + + + � � � ween the centres of the rings. PQ is perpendicular to the line joining the centres and coplanar to the rings.
MAGNETIC FIELD www.physicsashok.in 49 PASSAGE TYPE PASSAGE = 1. A cyclotron is a device used by physicists to study the properties of subatomic particles. Small charged particles are deposited at high speed into a circular pipe. electric and magnetic field then accelerate the particle to an even higher speed. Finally, when the particle reaches the desired speed, a magnetic field keeps it moving in a circle at constant speed. In figure 1, amagnetic field pointing �into the page� keeps a charged particle travelling in a counterclockwise circle at constant speed inside the cyclotron. The magnetic force on the particle points towards the center of the circle, and has strength Fmag = qvB where q is the particle�s charge, v is its speed, and B is the magnetic field strength. As a result of this force, the particle moves in circles in the cyclotron at frequency 2 f qB. m . North West South East particle where m denotes the particle�s mass. The frequency (in hertz) is the number of revolutions completed by the particle per second. In the following questions, neglect gravity. 1. In figure 1, what is the direction (if any) of the particle�s acceleration? (A) North (B) East (C) West (D) It has no acceleration 2. If the magnetic field in figure were turned off, in which direction would the particle travel (until crashing)? (A) North (B) East (C) West (D) Now here; it would stop moving 3. An alpha particle has charge 2e and mass 4 amu. A proton has charge e and mass 1 amu. Let falpha and f proton denote the frequencies with which those particles circle a cyclotron. If both particles experience he same magnetic field in the same cyclotron, what is alpha proton f f ? (A) 2 (B) 1 (C) 1 2 (D) 1 4 4. In order for a cyclotron to work properly, the magnetic field must make the particle move in a circle. Which of the following particles would not work in a cyclotron? (A) lithium atom (Li) (B) positive lithium ion (Li+) (C) negative lithium ion (Li+) (D) all of the above particle would �work�. 5. Which of the following cannot create a magnetic field? (A) Electrical current flowing through a well-insulated straight metal wire. (B) A beam of electrons moving across a cathode ray tube. (C) Electric current flowing around a superconducting ring. (D) Static electricity (i.e., extra electrons) built up on a stationary door knob.
PASSAGE = 2. Magnetically levitated (MAGLEV) trains are considered to be important future travelling machines. The idea of MAGLEV transportation has been in existence since the early 1900s. The benefits of eliminating friction between the wheel and the rail to obtain higher speeds and lower maintenance costs has great appeal. The basic idea of a MAGLEV train is to levitate it with magnetic fields so that there is no physical contact between the train and the rails. For comparison, �bullet� trains in Japan have a maximum speed of about 250-300 km/h while a MAGLEV train under development has reached a speed of 411 km/hr. The MAGLEV train uses powerful on board superconducting electromagnets with zero electrical resistance to support the train above the rails.
(B) small size (D) zero electrical resistance. They have low maintenance cost but high running cost they have low maintenance cost but high energy efficiency they have low maintenance and running cost as well as low installation cost. (C) China (D) None of these (C) on the walls (D) on the station (B) small size (D) zero electrical resistance. They have low maintenance cost but high running cost they have low maintenance cost but high energy efficiency they have low maintenance and running cost as well as low installation cost. (C) China (D) None of these (C) on the walls (D) on the station another charged particle �2q of mass 2 m in a uniform magnetic field B as shown in figure. If the particles are projected towards each other with equal speeds v. 1. Find the maximum value of projection speed vm so that the two particles do not collide. 2. Find the time after which collision occurs between the particles if projection speed equals 2vm . 3. Assuming the collision to be perfectly inelastic find the radius of particle in subsequent motion. MAGNETIC FIELD The walls along the track contain a continuous series of vertical coils of ordinary wire. As the train passes each coil, the superconducting magnet on the train induces a current in these coils and makes them electromagnets. The electromagnets on the train and outside produce forces that levitate the train and keep it centered above the rails. In addition, electric current flowing through coils outside the train propels the train forward, as shown in Figure.
1. What is the major advantage of using super conductors in the electromagnets (A) very low resistance for electric current (C) low hysteresis loss 2. Choose the correct statement (A) MAGLEV trains have high maintenance cost (B) (C) (D) 3. (A) Japan 4. (A) MAGLE trains are operating in which country (B) USA Super conducting electromagnets are installed in (B) on the railroads Coils installed on the wall will carry currents only if train is moving very fast (B) all times whether train is standing or running only if train is moving irrespective of whether fast or slow. (D) None of these. PASSAGE = 3. A charged particle +q of mass m is placed at a distance d from xinside the train 5.
(A) (C) xxx
q,m -2q,2m VV xxxx
xxxx
d
xxxx
(Neglect the electric force between the charges) www.physicsashok.in 50
MAGNETIC FIELD www.physicsashok.in 51 Level # 2 1. A particle of charge q and mass m is projected from the origin with velocity v . = 0 . i�in a nonuniform magnetic field k�B . .B0x . . Here .0 and B0 are positive constants of proper dimensions. Find the maximum positive x coordinate of the particle during its motion. 2. A flat dielectric disc of radius R carries an excess charge on its surface. The surface charge density is . . The disc rotates about an axis perpendicular to its plane passing through the centre with angular velocity .. Find the torque on the disc if it is placed in a uniform magnetic field B directed perpendicular to the rotation axis. 3. A wire loop ABCDE carrying a current . is placed in the x-y plane as shown in figure. A particle of mass m and charge q is projected from origin with velocity . .j�i� 2 V V . 0 . m/s. Find the (a) instantaneous acceleration. (B) If an external magnetic field i�B . B0 is applied, find the x yO A B CD E 90° force and torque acting on the loop due to this field. r/2 4. A long wire of radius a is placed along z-axis and carries current i as indicated in the figure. y-axis is taken perpendicular to the plane of paper directed into the paper. An electron escapes from the surface of the wire with velocity .0 directed along x-axis. Determine the maximum distance from the wire along x-axis up to X Z e O V0 i which electron can move. 5. A long horizontal wire AB, which is free to move in a vertical plane and carries a steady current of 20 A, is in equilibrium at a height of 0.01 m over another parallel long wire CD which is fixed in a horizontal plane and carries a steady A B current of 30 A, as shown in figure. Show that when AB is slightly depressed, C
D it executes simple harmonic motion. Find the period of oscillations. 6. A current . flows along a thin wire shaped as a regular polygon with n sides which can be inscribed into a circle of radius R. Find the magnetic induction at the centre of the polygon.Analyse the obtained expression at n.. . 7. Find the magnetic induction at the centre of rectangular wire frame whose diagonal is equal to d = 16 cm and the angle between the diagonals is equal to . = 30°; the current flowing in the frame equals . = 5.0 A. 8. A thin conducting strip of width h = 2.0 cm is tightly wound in the shape of a very long coil with crosssection radius R = 2.5 cm to make a single-layer straight solenoid. a direct current . = 5.0 A flows through the strip. Find the magnetic induction inside and outside the solenoid as a function of the distance r from its axis. 9. Find the magnetic moment of a thin round loop with current if the radius of the loop is equal to R = 100 mm and the magnetic induction at its centre is equal to B = 6.0 .T. 10. Calculate the magnetic moment of a thin wire with a current . = 0.8 A, wound tightly on half a tore (Figure). The diameter of the cross-section of the tore is equal to d = 5.0 cm, the number of turns is N = 500.
MAGNETIC FIELD www.physicsashok.in 52 11. A non-conducting thin disc of radius R charged uniformly over one side with surface density . rotates about its axis with an angular velocity . . Find : (a) the magnetic induction at the centre of the disc ; (B) the magnetic moment of the disc. 12. A small coil C with N = 200 turns is mounted on one end of a balance beam and introduced between the poles of an electromagnet as shown in Figure. The cross-section area of the coil is S = 1.0 cm2, the length of the arm OA of the balance beam is . = 30 cm. When there is no current in the coil the balance is in equilibrium. On passing a current . = 22 mA through the coil the equilibrium is restored by putting the additional counterweight of mass .m = 60 mg on the balance pan. Find the magnetic induction at the spot where the coil is located. 13. A square frame carrying a current . = 0.90 A is located in the same plane as a long straight wire carrying a current .0 = 5.0 A. The frame side has a length a = 8.0 cm. The axis of the frame passing through the midpoints of opposite sides is parallel to the wire and is separated from it by the distance which is . = 1.5 times greater than the side of the frame. Find : (a) Ampere force acting on the frame ; (B) the mechanical work to be performed in order to turn the frame through 180° about its axis, with the currents maintained constant. 14. In an electromagnetic pump designed for transferring molten metals a pipe section with metal is located in a uniform magnetic field of induction B (Figure). A current . is made to flow across this pipe section in the direction perpendicular both to the vector B and to the axis of the pipe. a B I Find the gauge pressure produced by the pump if B = 0.10 T, . = 100 A, and a = 2.0 cm. 15. A proton accelerated by a potential difference V = 500 kV flies through a uniform transverse magnetic field with induction B = 0.51 T. The field occupies a region of space d = 10 cm in thickness (Figure). Find the d . + B angle . through which the proton deviates from the initial direction of its motion. 16. A slightly divergent beam of non-relativistic charged particles accelerated by a potential difference V propagates from a point A along the axis of a straigth solenoid. The beam is brought into focus at a distance . from the point A at two successive values of magnetic induction B1 and B2. Find the specific charge q/m of teh particles. 17. A non-relativistic proton beam passes without deviation through the region of space where there are uniform transverse mutually perpendicular electric and magnetic fields with E = 120 kV/m and B = 50 mT. Then the beam strikes a grounded target. Find the force with which the beam acts onthe target if the beam current is equal to . = 0.80 mA. 18. A beam of non-relativistic charged particles moves without deviation through
the region of space A (Figure), where there are transverse mutually perpendicular electric and magnetic fields with strength E and induction B. When the magnetic field is switched off, the trace of the beam on the screen S A S.. shifts by .x . Knowing the distances a and b, find the specific charge q/m a b of the particles.
MAGNETIC FIELD www.physicsashok.in 53 Level # 3 1. An electron in the ground state of hydrogen atom is revolving in anticlock-wise direction in a circular orbit of radius R. (JEE 1996) (i) Obtain an expression for the orbital magnetic dipole moment of the electron. � n 30° B (ii) The atom is placed in a uniform magnetic induction B . such that the plane-normal of the electron-orbit makes an angle of 30° with the magnetic induction. find the torque experienced by the orbiting electron. 2. A current of 10 A flows around a closed path in a circuit which is in the horizontal plane as shown in the figure. The circuit consists of eight alternating arcs of radii r1 = 0.08 m and r1 = 0.12 m. Each arc subtends the same angle at the center. (A) Find the magnetic field produced by this circuit at the center. (JEE 2001) A D C r1 r2 (B) An infinitely long straight wire carrying a current of 10 A is passing the center of the above circuit vertically with the direction of the current being into the plane of the circuit. What is three force acting on the wire at the center due to the current in the circuit? What is the force acting on the arc AC and the straight segment CD due to the current at the center? 3. A wheel of radius R having charge Q, uniformly distributed on the rim of the wheel is free to rotate about a light horizontal rod. The rod is suspended by light inextensible strings and a magnetic field B is applied as shown in the figure. The initial tensions in the strings are T0. If the breaking tension of the strings are 3T0/2, find the maximum .0 B T0 T0 d angular velocity 0 . with which the wheel can be rotated. (JEE 2003) 4. A pair of stationary infinitely long bent wires are placed in the x-y plane as shown in Figure. The wires carry currents of . = 10 A each as shown. The segments LR and MS are along the x- axis. the segments RP and SQ are parallel to the y-axis such that OS = OR = 0.02 m. Find the magnitude and . . . . L R O P S M x y
Q direction of the magnetic induction at the origin O. (JEE 1989) 5. A solenoid of length 0.4 m and having 500 turns of wire carries a current of 3 A. A thin coil having 10 turns of wire and of radius 0.01 m carries a current of 0.4 A. Calculate the torque required to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of the solenoid. (JEE 1990) 6. Three infinitely long thin wires, each carrying current in the same direction are in the x-y plane of a gravity free space. The central wire is along the y-axis while the other two are along x = . d. (JEE 1997) (a) Find the locus of points for which the magnetic field B is zero. (B) If the central wire is displaced along the z-direction by a small amount and released, show that it will execute simple harmonic motion. If the linear density of the wire is . , find the frequency of oscillations. 7. A proton and an alpha particle, after being accelerated through same potential difference, enter a uniform magnetic field the direction of which is perpendicular to their velocities. Find the ratio of radii of the circular paths of the two particles. (JEE 2004) 8. In a moving coil galvanometer, torque on the coil can be expressed as b = ki, whre �i� is current through the wire and �k� is constant. The rectangular coil of the galvanometer having numbers of turns N, area A and moment of inertia I is placed in magnetic field B. Find (a) �k� in terms of given parameters N, I, A and B. (b) the torsional constant of the spring, if a current i0 produces a deflection of ./2 in the coil in reaching equilibrium position. (c) the maximum angle through which coil is deflected, id charge Q is passed through the coil almost instantaneously. (Ignore the damping in mechanical oscillations) (JEE 2005)
MAGNETIC FIELD www.physicsashok.in 54 Answer Key ASSERTION&REASON Q. 1 2 3 4 5 6 7 8 9 Ans. A E A B A A D A B Q. 10 11 12 Ans. E B C Level # 1 Q. 1 2 3 4 5 6 7 8 9 10 Ans. C D C D ABCD C D B C C Q. 11 12 13 14 15 16 17 18 19 20 Ans. B A D D C A D A D C Q. 21 22 23 24 25 26 27 28 29 30 Ans. C C B BCD ACD ABC A BD C AD Q. 31 32 33 34 35 36 37 38 39 40 Ans. A A C B ACD B B ABD A A Q. 41 42 43 44 45 46 47 48 49 50 Ans. B C B D B C B A B B Q. 51 52 53 54 55 56 57 58 59 60 Ans. A B A ABD C C ABD B AC D Q. 61 62 63 Ans. AC A ACD FILL IN THE BLANKS / TRUE-FALSE / MATCH TABLE 1. D, B 2. 2 4iL. 3. 1.25 x 10�23 Am2 4. 0 1 2 1 1 4I R R . . . . . . . . 5. .lB ; +z direction 6. evB; ABCD 7. 14.0185 eV 8. True 9. True 10. False 11. False 12. [(A�PQ), (B�Q), (C�PQR), (D�QS)] 13. [(A�PT), (B�PQR), (C�P), (D�S)] 14. [(A�Q), (B�P), (C�R), (D�Q)] 15. [(A�Q), (B�R), (C�S), (D�P)] 16. [(A�R), (B�P), (C�Q), (D�Q)] 17. [(A�QR), (B�P), (C�QR), (D�QS)] 18. [(A�PRS), (B�RS), (C�PQT), (D�RS)] PASSAGE TYPES Passage = 1. 1. C 2. A 3. C 4. A 5. D Passage = 2. 1. D 2. C 3. D 4. A 5. C Passage = 3. 1. 2m V Bqd m . 2. Bq 6t m. . 3. R . 3d.
MAGNETIC FIELD www.physicsashok.in 55 Level # 2 1. B q 2mv 0 0 2. 4...BR4 3. (a) 0 . . . . 0 0 q � � a 4 2 B V j i 8 2 rm . . . . . . . (B) Zero, 0 2 B 21 4r ... .... . . 4. ei (2 mv ) ae 0 0 m .. . . 5. 0.2 Sec. 6. 0 nµ I tan 2R n. . 7. B 4 0 / d sin 0.10mT. . . . . . . 8. . . . . .. .. . . . . . . . . . . . . / 4 2 / r, r R. B /h (1 (h / 2 R) 0.3mT,r R, 0 2 0 9. 2 0 3 pm . 2.R B/. . 30mA.m 10. 2 2 pm . 1/ 2N.d . 0.5 A.m 11. (a) B 1/ 2 0 R . . .. (b) p 1/ 4 R4. m . ... 12. B . .mg. /N.S . 0.4 T. 13. (a) F 2 / .4 2 1. 0.40 N 0 0 . . .. . . . . . (b) A ( 0a 0 / ) ln [2 1) /(2 1)] 0.10 J. . . .. . . . . . . . 14. .p . .B/ a . 0.5 kPa. 15. 30 .
2mV arcsin dB q . . . . .. . . .. . . 16. 2 2 1 2 2 (B B ) q/m 8 V . . . . 17. F = mE . /qB = 20.N. 18. 2E xq/m a(a 2b)B2 . . . Level # 3 1. (i) M eh 4 m . . (ii) ehB 8 m . directed perpendicular to the both M ... and B.. . 2. (a) 6.54 x 10�5 Tesla (b) 0, Force on arc AC = 0, Force on segment CD = 8.1 × 10�6 N (inwards) 3. 0man 2 W DT BQr . 4. 1 x 10�4 T or Wb m�2. 5. . = 5.9 x 10�6 Nm. 6. (a) , z 0. 3 x . . d . (b) 1 2 0 d 2 . .. . . .. . .. . .. 7. p p p r m q 1 . r m q 2 . . . . .
8. (a) k = NAB, (b) 0 C . 2i NAB . ; (c) 0 Q NAB 21i . . �X�X�X�X�
ATOMIC PHYSICS
DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125
ATOMIC PHYSICS www.physicsashok.in 1 MASS OF A PHOTON Rest mass of photon is zero. Effectivemass : Ifwe assume thatmis the effectivemass of photons inmoving conditions, then according to Einstein�s theory. Energy of photon,E =mc2 m= 2 Ec but E = h. = pc, p =photon �momentum . m = 2 Ec = 2 hc. = pc C1: Drawthe graph of (a)momentumversus effectivemass of photon (b) Energyversus effectivemass, (c)wavelengthversusmomentum, (d) Effectivemass versuswavelength for a photon. Sol: . pO m . E O m (a) p = mc . straight line tan. = c (b) E = mc2 . straight line tan. = c2 . p . m (c)de-brogliewavelength of photon . = hp (d) Effectivemass of photon ..p = h . Rectangular hyperbola m = 2 Ec = 2 hc. = 2 hc .c m. = hc . Rectangular hyperbola Power of a light source Suppose P = Power of the light source, . = frequency of emitted photons, Energyof single photon= h. Let, n = numbers of photons emitted per sec by the source, then Thus, Energy emitted per sec. by the source =W= h. Intensity of light source at a point The amount ofenergyincident ona point per unit area at that point ina unit time is called the intensityoflight
at that point.
ATOMIC PHYSICS www.physicsashok.in 2 Thus, intensityat a point is energyincident per unit area unit time. Suppose, I = Intensityof light at a point . = Frequency of photon n0 = no. of photon incident / sec /Area . I = n0h. SI unit ofintensityis J/secondm2. Intensity of light at a distance r : Let, W= Power of a point source r Amount of energy received byspherical surface per second P = Amount of energy emitted by light source per second =W Hence, Intensityat distance r = W Area = 2 W 4.r Thus, I = 2 W 4.r Example 1: The intensityof sunlight on the surface of earth is 1400W/m2.Assuming themeanwavelength of sunlight to be 6000 Å, calculate the number of photons emitted fromthe sun per second assuming the average radius of earth�s orbit around sun is 1.49 × 1011m. Sol: Average radius, r = 1.49 × 1011m. intensity of sunlight received by earth = I = 1400W/m2, . =meanwavelength = 6000Å. Energy emitted per second by the sum= Power of the sun =W but, Power,W= nh.,where n is number of photons emitted bysun per sec. .....(i) Again, intensityat a distance �r�froma point source of powerW. I = 2 W 4.r Sun r Earth Power = P I = 2 nh 4 r. . [Fromequation (i)] n = I 4 r2 h . . . = I 4 r2 hc . . . [. h. = hc . ] . n = 4 r2 I hc . . =
2 22 7 34 8 4 (1.49) 10 6 10 1400 6.63 10 3 10. . .. . . . . . . . n = 1.18 × 1045 photons per sec. Photon Flux (i.e. photon/sec) Suppose,W= Power of a source. IfAis the area of themetal surface onwhichradiations are incident, thenthe power received bythe plate is W. = IA = 2 W 4 r . . .. . .. A If . is the frequency of radiation, then the energy is photon is given by h..
ATOMIC PHYSICS www.physicsashok.in 3 If n. is the number of photons incident on the plate per second, then, we have W. = n. h. 2 W 4 r . . .. . .. A=N. h. r P Area(A) . n. = Plate 2 W A 4 rh . . . . . . . . . . . . . n´ is called photon flux. Photon flow density (n0) : The number of photons incident on unit area of the plate in one second is called the photonflowdensity. i.e., Photon flowdensity, n0, = n ' A = 2 W 4.r h. Photon-concentration : The number ofphotons per unit volume of the space. If n is the number of photons incident per unit area per second, then Photon - concentration = nc , where c is the speed of light. Example 2: The power of light emitted bythe sun is 3.9× 1026W.Assuming themeanwavelengthof sunlight to be 6000 Å, calculate the photon flux arriving at 1m2 area on earth perpendicular to light radiations. The average radius of earth�s orbit around sun is 1.49 × 1011m. Sol: Power of light emitted by the sun,W= 3.9 × 1026W r = earth sunmean distance = 1.49 × 1011m . = meanwavelength of sun light = 6000 Å= 6 × 10�7m P Sun r Earth Intensityoflight an earth, I = 2 W 4.r Power received by areaAon earth,W. = I ×A . W. = 2 W 4.r × A but,W. = n. h., where n. is the number of photon incident per sec or photon-flux. . n. = W' h. = 2 W A 4 r h. . .
. . .
. . . . .
. .
. n. = 2 W A 4 r hc . . [..h. = hc/.] n´ = 26 7 2 22 34 8 (3.9 10 ) (6 10 ) 1 4 (1.49) 10 6.63 10 3 10 .. . . . . . . . . . . photon /sec = 4.22 × 1021 photon/sec
ATOMIC PHYSICS www.physicsashok.in 4 PRESSURE EXERTED BY PHOTONS OR RADIATIONS PRESSURE (a) Each photon carries energyandmomentum.Hencewhen photons of light is incident on a surface, the light is either absorbed or reflected or both is done by the surface. (b) change in momentumof light takes place, which causes impulse or force on the surface resulting in a pressure called radiation pressure. Let light of intensityI is incident on a surface. Each photon carries energy h. andmomentum= hc. . Energyincident onunit area in unit time = I(by the definition of intensity). Number of photons incident on unit area in unit time. i.e. N A.t = I h. , .....(i) Momentumdelivered to unit area inunit time, i.e., P A t .. = N A t . . . . . . . × (change inmomentumof each photon) .....(ii) Using equation (i)&(ii),we get P A t .. = I h . . . . . . . × (change inmomentumof each photon) but, P A t .. = Force exerted A , [. force exerted = Pt .. ] = Radiation pressure . Pr= Radiation pressure = I h. × (change inmomentumof each photon) .....(iii) Radiation pressure for perfectly absorbing surface : In this case, change ofmomentumsuffered by each photon= h
c. . Using the equation(iii),we get hv c Perfectly absorbing surface Radiation pressure = I h. × hc. = Ic . Radiation pressure = Ic for a perfectlyabsorbing surface. Radiation pressure for a perfectly reflecting surface : In this case, change inmomentumof each photon = hc. � hc. . . . . . . . = 2hc. hv c Perfectly reflecting surface hv Hence, using the equation (iii), we get c Radiation pressure = I h . . . . . . . × 2hc. . . . . . . = 2I c . Radiation pressure = 2I c for a perfectlyreflecting surface.
ATOMIC PHYSICS www.physicsashok.in 5 Radiation pressure for a surface of reflection coefficient (.) : hv c surface hv . c In this case,Momentumof incident photon= hc. Momentumof reflected photon= . hc. , . = reflection coefficient . Change inmomentum= hc. � hc. . . . .. . . . = (1 + .) hc. Thus, using equation(iii),we get, . Radiation pressure = (1 + .) Ic for photons falling normallyon a surface. NOTE : (i) For a perfectly absorbing surface . = 0 . Radiation pressure = Ic (ii) For a perfectly reflecting surface . = 1 . Radiation pressure = (1 + 1) Ic = 2I c C2: Alaser emits a light pulse of duration .= 0.13ms and energyE = 10J. Find themean pressure exerted by such a light pulsewhen it is focussed into a spot of diameter d = 10 .mon a surface perpendicular to the beamand possessing a reflection -coefficient . = 0.50. Sol: Laser energy, E = 10J Plate area = 4 Laser d d = 10 .m . d2 Pulse duration, . = 0.13 .s Pressure exerted bylight pulse i.e., Radiation pressure = (1 + .) Ic , by the definition of radiation pressure byphotons of light .....(i) Here, . = reflection � coefficient = 0.50 I = Intensityoflight c = speed of light = 3 × 108m/s
Bythe definitionof intensityof light, I = Laser energy (pulse duration).(Area of plate) i.e. I = 2 Ed 4. . . . . . . . .....(ii) . From(i)&(ii) Radiation pressure = (1 + .) 2 4E c. .d = (1 + 0.50) 8 3 12 4 10 3 10 (0.13 10. ) 100 10. . . . . . . = 4.9 × 107 Nm�2 = 490 atm
ATOMIC PHYSICS www.physicsashok.in 6 Example 3: 1 .Acurrent flows in an x-ray tube operated at 10,000 V. The target area is 10�4 m2. Find the pressure on the target, assuming that the electrons strike the target normallyand the photons leave the target normally.Consider the idealsituationwhere eachincident electron gives rise to a photon ofthe same energy. Sol: Energy of each electron = energy of each photon= E = 104 eV Momentumdelivered by each electron = p1 = 2mE (photon) p e� 1 p2 Plate Momentumtaken away byeach photon = p2 = Ec Change ofmomentumdue to each electron-photon pair = p1 � (�p2) = p1 + p2 Current incident on target = i= 10�6A . Number of electrons incident per second = i/e = no. of photons emitted per second. Totalmomentumchange per second = force = ie . . . . . . (p1 + p2) . Pr, Pressure = force / area = i Ae . . . . . . (p1 + p2) = i Ae . . . . . . 2mE Ec . . . . . . . . Pr = 6 4 19 10 10 1.6 10 . . . . . 4 19 31 4 19 1/ 2 8 (2 9.1 10 10 1.6 10 ) 10 1.6 10 3 10 . . . . . . . . . . . . . . . . . . = 3.7 × 10�6 Nm�2
FORCE EXERTED BY PHOTONS ON A SURFACE We know that,when photons of light is incident on a surface, the change inmomentumof photons takes place resulting in a force exertion by the photons on the surface. Let light ofpowerWis incident on a surface. Each photoncarries energy h. and,momentum= hc. .....(i) Energyincident in unit time =W(bythe definition of power) Power =W Surface But, power is given by W = N.t (h.), whereN= total no. of photons incident in time .t. N.t = Wh. .....(ii) Momentumdelivered inunit time. Pt .. = Nt . . . . . . . × (change inmomentumof each photon) = P h. (change inmomentumof each photon) but, Pt .. = rate ofchange ofmomentum
ATOMIC PHYSICS www.physicsashok.in 7 Hence, Pt .. = F = Force exerted by the photons. Thus, force = P h . . . . . . . × (change inmomentumof each photon) ...(iii) Force exerted by photons on a perfectly absorbing surface : In this case, change ofmomentumsuffered by each photon= hc. using equation(iii),we get hv c Perfectly absorbing surface force = Wh . . .. ... × hc. . . . . . . = Wc thus, F = Wc for a perfectly absorbing surface. Force exerted by photons on a perfectly reflecting surface: In this case, change ofmomentumsuffered by each photon = hc. � hc. . . . . . . . = 2hc. hv c surface hv . c using equation ( iii),we get Force = Wh . . .. ... × 2h
c. . . . . . . = 2Wc thus, F = 2P c , for a perfectlyreflecting surface. Force exerted by photons on a surface of reflection coefficient .: In this casemomentumof incident photon= hc. , hv c surface hv Momentumof reflected photon= . c . hc. . Change inmomentum = hc. � hc. . . . .. . . . = (1 + .) hc. Thus, using equation(iii),we get Force = h. . . . . . . . × (1 + .) hc. = (1 + .)Wc i.e. F = (1 + .)Wc for photons falling normally on a surface.
ATOMIC PHYSICS www.physicsashok.in 8 C3: If a point source of light of powerWis placed at the centre of curvature of a hemispherical surface,whose inner surface is completelyreflecting, then the force onthe hemisphere due to the light falling onit is given byF = W2c . W (Source) C4: If a perfectly reflecting solid sphere of radius r is kept in the path of a parallel beamof light of large aperture and having an intensityI, then the force exerted by the beamon the sphere is given F = pr2I c . I r Note that force is equal to the product of (I/c) and the projected area of the sphere. C5: If photon of light of powerWfalls at an angle . to a perfectly reflecting surface, then net force exerted on the plate is given by Fnet = 2Wc cos.. .. C6: Alaser emits a light pluse of durationT = 0.10ms and energyE = 10 J. Find themean pressure exerted by such a pulsewhen it is focused on a spot of diameter d =10 µmon a surface perpendicular to the beamand with a reflection coefficient . = 0.5. Sol. p,momentumof a photon = hv/c = E/c Momentumof reflected photon = .(E/c) . change ofmomentum= (E/c) � (�.E/c) = (1 + .)E/c Force exerted = [(1 + .) E/c]/T Pressure = 4 [(1 + .) E/c]/T.d2 = 2 8 4 10 4(1 )E 4(1 0.5) 10 cTd 3 10 10. 10. . . . . . . . . . . . = 6.37 × 106 Nm�2 C7: Aplane light wave of intensityI = 0.50Wcm�2 falls on a planemirror of reflection coefficient . = 0.8.The angle of incidence . = 45º. Find the normal presure exerted by light on that surface. Sol. If S is the area of the mirror onwhich light falls, the transverse section of the incident beamis S cos .. Momentumof the incident photons =(I/c)(S cos .).Normal component ofmomentum flux = (IS cos ./c) cos ..= IS cos 2./c. Momentumof reflected photons = .(IS cos2./c) in the opposite direction. . rate of change ofmomentum= force = (1 + .) IS cos2./c . normal pressure = force/area = (1 + .) I cos2./c . required pressure = (1 + 0.5) (0.5 × 104 cos2 45º)/3 × 108 = 1.25 × 10�5 Nm�2 Example 4:An isotropic point source of radiationpower Pis placed on the axis of an idealmirror.The distance between the source and themirror is n times the radius of themirro. Find the for
ce that light exerts on the mirror.
ATOMIC PHYSICS www.physicsashok.in 9 Sol. Energyfluxthrough the annular space between two cones ofhalf-angles . and . + d.= (P/4.) (2. sin . d.) . momentumof flux = P sin . d./2c d S nR BBRA . rate of change ofmomentumalong the normal. 2(P sin . d./2c) cos . = P sin . cos . d./c . force onmirror = 0 p sin cos d c . . . . . where . = half-angle of the cone subtened by themirror = tan�1(1/n) . F = P sin2 . / 2c = P/2(n2 + 1)c Example 5: Figure shows a small, plane strip ofmassmsuspended froma fixed support through a string.A continuous beamofmonochromatic light is incident horizontally on the strip and is completely absorbed. The energyfalling on the strip per unit time isW. find the deflection of the string fromthe vertical, ifthe strip stays inequilibrium. Sol: Force exerted by the photons of light = (number of incident photons per sec) × (change inmomentumof each photon) Let, Number of incident photons per sec =N Light change inmomentumof each photon= hc. hence force exerted by the photons of light, F = N hc. = (Nh ) c . but, W = Nhv, by the definition of power . F = Wc If the stringmakes an angle .withthe verticalwhen the strip is at equilibrium, thenfor the equilibriumofthe strip, Tsin. = F, in horizontaldirection, . T F m mg T = tension in the string and Tcos. =mg, in verticaldirection. . Dividing the above equations,we get
tan. = F mg = W/c mg = W cmg , . tan. = W cmg .... = tan�1 W cmg . . .. .. Example 6. Aplane light wave of intensity I = 0.80Wcm�2 illuminates a sphere of radiusR = 5.0 cm. Find the force that the light exerts on the sphere. Sol. p,momentumof the incident pulse = E/c. i p . (E / c) (sin .�i . cos.�j) . with respect to a frame of reference with the outward normal as the y-axis and a line perpendicular to it and lying in the plane of themirror as the x-axis.
ATOMIC PHYSICS www.physicsashok.in 10 pf = .E/c and f p . (.E/ c) (sin. �i . cos.�j) . f i . p . p . p . (..1)E / c sin .�i . (..1) cos.E / c �j . . . . | . p | . E / c (. .1)2 sin2 . . (..1)2 cos2 . . . E / c 1. .2 . 2.cos 2. | . p | . .10 / 3 .108 . 1. 0.82 . 2 . 0.8cos60º . = 5.2 × 10�8 Nm�2 Impulse applied by photon on a surface Let hn be the energy of photons of a light incident normallyon a surface, Momentumof anincident photon= hc. . Change inmomentumof the photon takes place due to impact ofthe photonswith the surface. This change inmomentumofthe photons causes impulse. Using impulse �momentumtheorem,we get Impulse = Total change inmomentumof photons = (Total number of photons) × (change inmomentumof each photon) . Impulse =N. × (change inmomentumof each photon) ......(i) Impulse on a totally absorbing surface: In this case, change inmomentumof each photon = hc. . using equation(i),we get Impulse = N'h c . = Ec , where E is the total energyof the light . Impulse = Ec for a perfectlyabsorbing surface. Impulse on a totally reflecting surface: Impulse = N. × (change inmomentumofeach photon) = N. × h h c c . . . . .. . . .. .. . . .. = 2(N'h ) c . = 2E c [. E = total energy=N.h.] . Impulse =
2E c for a perfectlyreflecting surface. Impulse on a surface of reflection-coefficient .: In this case, change inmomentumof each photon = hc. � hc. . . . .. . . . = (1 + .) hc. . Using equation(i),we get
ATOMIC PHYSICS www.physicsashok.in 11 Impulse = N.(1 + .) hc. = (1 + .) N'h c . . . . . . . = (1 + .) Ec , [E =total energy] . Impulse =(1 + .) Ec for photons falling normally on the surface. Example 7: A small perfectly reflecting mirror of mass m = 10 mg is suspended byaweightless thread oflength . = 10 cmas shown in the figure. Find the angle throughwhichthe threadwillbe deflectedwhen a short laserwith energyE =13J is shot in horizontaldirection at right angles to themirror. (g = 10m/s2). m m Laser . Sol: Impulse applies bylaser on themirror, Impulse= 2E c , asmirror is perfectly reflecting. Initialmomentumof themirror = 0 v = 0 v . . .(1 � cos.) = H Let, finalmomentumof themirror =mv change inmomentumof themirror due to impact =0 � (�mv) =mv,where v is the speed ofmirror just after impact. But, Impulsemomentumtheoremgives, Impulse =change inmomentum 2E c = mv or v = 2E mc .....(i) Totalmechanical energyof themirror willbe conserved after impact. Thus, Loss in kinetic energy=Gainin potential energy 12 mv2 = mgh ..... v2 = 2gH 2 2 2 4E m c = 2gH, [as v = 2E mc fromequation (i)] 2 2 2 2E m c
= gH . 2 2 2 2E m c = g.(1 � cos.), [. H = . (1 � cos.)] 1 � cos. = 2 2 2 2E m c g. 2sin2 2. . . . . . . = 2 2 2 2E m c g. [. 1 � cos. = 2sin2 2. . . . . . . ] sin 2. = E2 mc g. Here E = 13J, m= 10 × 10�6 kg, . = 0.1 m . sin 2. = 5 8 13 10. .3.10 10.0.1 = 0.0043 or . = 0.5º PHOTONS UNDER GRAVITY
ATOMIC PHYSICS www.physicsashok.in 12 Photons can be considered as a particle ofmassm. If m = mass of photon v = frequencyof photon E = energy ofphoton, then mc2 = E = hv . m = 2 Ec = 2 hc. Thus, a photonof frequency v acts gravitationally like a particle ofmass 2 hc. . C8: If a photon frequency v falls on the surface of earth froma height h, thenwhat will be its frequency on the surface of earth. Sol: Change in frequencyof the photontakes place. v, E = hv + mgH v,. E = hv Ground H Let v. be the frequency of photon on the surface of earth. Mass of photon = 2 hc. Mass of the photon depends onits frequency, but wewill consider themass to be constant as difference in v´ and v is very small. Energyconservation gives Initial energy= final energy . hv + mgH = hv. . hv + 2 hc. gH = hv. . v. = v 2 1 gH c . . . . . . . Example 8:Aplanet ofmassMand radius R emits a photon of frequency v.What will be the frequency of photonwhen it covers an infinite distance. Sol: Let v..be the frequencyphoton at a infinite distance. v. R at . M G=Gravitational constant v Mass of the photon, m= 2 hc. Energy of photon on the surface = hv �G Mm
R = hv � 2 GM h R c. .. .. .. Energyof photonat infinity= hv. Energymust be conserved, . hv � 2 GM h R c. .. .. .. = hv´ . v. = v 2 1 GM Rc . . . .. .. NOTE: .. . = 2 GM Rc is called frequency shift. PHOTOELECTRIC EFFECT
ATOMIC PHYSICS www.physicsashok.in 13 Ejectionofelectrons fromametalplatewhenilluminated bylight or anyother radiationofsuitablewavelength (or frequency) is called photoelectric effect. This phenomenonwas first discovered byHeinrichHertz in 1887.One year later,Hallwachsmade the important observation thatwhen a negatively-charged zinc plate is irradiatedwith ultra-violet light, it loses its negative charge.Afterwards, itwas discoveredthat alkalimetals like lithium, sodium, potassium, rubidium and caesiumeject electronswhen visible light falls on them.Ten years later, J.J. Thomson and P.Lenard showed that the action of light was to cause the emission of free electrons frommetal surface.Although these electrons are no different fromallother electrons, it is customaryto refer to themas photoelectrons. Experimental Study of Photoelectric Effect Quartz C A � + E V Photoelectric phenomenoncanbe studiedwith the help ofa simple apparatus showninfigure. It consists oftwo photosensitive surfaces E andC enclosed in an evacuated quartz bulb. In the absence of light, there is no flowof current in the circuit and the ammeterA reads zero. When plate E is exposed to some source of monochromatic light, current starts flowing.However, no current is found to flowwhen light falls on plate C. The explanation of the above behaviour lies in the fact that whenE is irradiated with light, the incident photons eject electrons by collisionwith its atoms. These photoelectrons are immediatelyattracted bythe collector plateC thereby starting current flowas indicated bythe ammeter. WhenC is irradiated, even then photoelectrons are produced but theyare not allowed to leave plate C (i) firstly, because of the pulling effect of positive potentialofCand (ii) secondly, due to repulsion fromthe negative plate E. Hence, no current is found to flowin the circuit. Einstein�s Photoelectric Equation: Following Planck�s idea that light consists of photons, Einstein proposed an explanation of photoelectric effect as early as 1905.According to this explanationwhena single photon is incident on ametalsurface, it is completely absorbed an imparts its energyhf to a single electron. The photon energy is utilised for two purposes: (i) Partlyfor getting the electronfree fromthe atomand awayfromthemetal surface. This energyis known as the photoelectricwork function of themetal and is represented byW0. (ii) the balance of the photon energyis used up in giving the electron a kinetic energyof 1/2mv2. . hf =W0 + 1/2mv2 .....(i) It is knownasEinstein�s photoelectric equation. In case, the photon energy is just sufficient to liberate the electrononly then no energywould be available for imparting kinetic energy to the electron.Hence, the above equationwould redu
ce to hf0 = W0 .....(ii) where f0 is calledthe thresholdfrequency. It is definedas theminimumfrequencywhichcancause photoelectric emission. For frequencies lower than f0, therewould be no emission of electronswhereas for frequencies greater than f0, electronswould be ejectedwith a certain definite velocity(and hence kinetic energy). Substituting this value ofW0 in equation (i) above, the Einstein�s photoelectric equation becomes hf = hf0 + 12 mv2 or hf = hf0 + K.E. Long Wavelength Limit (.0)
ATOMIC PHYSICS www.physicsashok.in 14 It is thewavelength corresponding to the threshold frequency f0. Its physical significance is that radiations havingwavelength longer than .0would not be able to eject electrons froma givenmaterialwhereas those having . < .0will. Inotherwords, it represents the upper limit ofwavelengthfor photoelectric phenomenon. By analogy, it is also referred to as thresholdwavelength. Now, c = f0 .0 ...0 = 0 c f Also, W0 = hf0 . 0 1f = 0 h W . .0 = 0 ch W (i) WhenW0 is in joules .0 = 8 34 0 3 10 6.625 10 W . . . . = 26 0 19.875 10 W . . metre (ii)WhenW0 is electron volts (eV) .0 = 8 34 19 0 3 10 6.625 10 1.602 10 W . . . . . . = 7 0 12.4 10 W . . metre = 0 12, 400
W Å Kinetic energy of Photoelectrons Einstein�s photoelectric equationcan be used to find the velocityand hence the kinetic energyof an ejected photo-electron. hf = W0 + 12 mv2 = hf0 + K.E. . K.E. = hf � hf0 = h(f � f0) Now, f = c. and f0 = 0 c . . K.E. = ch 0 . 1 1 . . . . . . . . = 3 × 108 × 6.625 × 10�34 0 . 1 1 . . . . . . . . = 19.875 × 10�26 0 . 1 1 . . . . . . . . joules . and .0 inmetres = 19.875 × 10�26 0 . 1 1 . . . . . . . . joules . and .0 in Å
ATOMIC PHYSICS www.physicsashok.in 15 = 16 19 19.875 10 1.602 10 . . . . 0 . 1 1 . . . . . . . . eV . and .0 in Å . K.E. = 12,400 0 . 1 1 . . . . . . . . eV . and .0 in Å Incidentally, itmaybe noted that this also represents themaximumvalue ofthe kinetic energya photoelectron can have. . Emass = h(f � f0) = h..f joules = 12,400 0 . 1 1 . . . . . . . . eV . and .0 in Å Photoelectric Work Function As explained above it is defined as the energywhich is just sufficient to liberate electrons froma bodywith zero velocity. Its value is given by W0 = hf0 = 0 ch . = 8 34 0 3.10 .6.625.10. . = 26 0 19.875.10. . joules .0 inmetres = 16 0 19.875.10. . joules .0 in Å = 16 19 0 19.875 10 1.602 10 . . .
. . eV .0 in Å
. W0 = 0 12, 400 . eV .0 in Å Laws of Photoelectric Emission R B V Quartz C A The apparatus shown in figurewas used byMillikan to studythe photoelectric effect and the various laws governing it. S is a source of radiations of variable but known frequency f and intensity I. E is the emitting electrodemade of the photosensitivematerialand C is the collecting electrode. Both these electrodes are enclosed in an evacuated glass envelopewith a quartzwindowthat permits the passage of ultraviolet and
ATOMIC PHYSICS www.physicsashok.in 16 visible light.As shown, anypotentialdifference can be established between the two electrodes.Areversing switch RS enables the polarities of the two electrodes to be reversed. If the tube is in the dark, then no photoelectrons are emitted and themicroammeterAread zero. However, if ultraviolet or visible light is allowed to fall on the emitting electrode, electrons are liberated and circuit current is set up. Fromthe experimentaldata collected byRichardson and Compton in 1912, photoelectric emissionwas found to be governed bythe following laws: (i) Photoelectric current (i.e., number of electrons emitted per second) is directly proportional to the intensityofthe incident light (or radiation). Light Intensity I Photoelectric current Frequency constant This can be verified by increasing the intensity of light and measuring the corresponding photoelectric current while holding the frequency of the incident light frequency of the incident light and the accelerating potential V of the collecting electrode C constant. The graph is similar to one shown in figure. Increase in intensity means more photons and hence greater ejection of electrons. (ii) For each photosensitive surface, there is aminimumfrequency of radiation (called threshold frequency) at whichemission begins. f0 Frequency Photoelectric current Intensity constant Light oflower frequency(or longerwavelength), however strong, willnot be able to produce any electron emission. This fact can be verified by keeping the light intensity constant while varying the frequency.The graph so obtained is shownin figure.The current is found to increase (though non-linearly) with the frequency of the incident light. Moreover, it is seen that there is a limiting or critical frequency below which no photoelectons are emitted. It is called threshold frequency and its value depends on the nature of thematerial irradiated because for eachmaterial there is a certain minimumenergynessecaryto liberate anelectron. This energy is known as photoelectricwork function or threshold energyW0.As seen fromf0 =W0/h. Thewavelengthcorrespondingto be thresholdfrequencyf0 is called thresholdwavelengthor longwavelength limit.No photo-electrons are emitted for wavelength greater than .0, nomatter howlong the light falls on the surface or howgreater is its intensity.The photoelectric or quantumyield (which is definedas the photoelectric current in amperes per watt of incident light) depends on the frequency (and not intensity) of the incident light. (iii) Themaximumvelocityofelectron emission (and hence kinetic energy) varies
linearlywith the frequencyo the incident light but is independent of its intensity. As seen fromEinstein�s photoelectric equation of 2max 1 mv 2 = h(f � f0) or Emax . f f0 Frequency Emax . Intensity constant Hence, increase in the frequency ofthe incident light increase the velocitywithwhich photoelectrons are ejected.The same fact is illustrated byfigure. Incidentally, itmay be noted that the slope of the curve gives the value ofPlanck�s constant h. If the intensityof the incident light is increased (keeping frequencyconstant),more photonswillbe incident
ATOMIC PHYSICS www.physicsashok.in 17 on themetal surface, each photon having the same energy.Hence,more photo electronswill be ejected. Since an electron can absorb only one photon, each photoelectronwillhave the same energyandwill be ejectedwiththe same velocity.Obviously, increase inintensityonlyincreases the number of photo-electrons ejected but not their kinetic energy. (iv) The velocities of emitted electrons have values betweenzero and a definitemaximum. The proportion of the electrons having a particular velocityis independent of the intensityof radiation. (v) Electrons are emitted almost instantaneously evenwhen the intensityof incident radiations is very low. The time lag between the incidence of radiation and emission of first electrons is less than 10�8 second. (vi) For a givenmetalsurface, stopping potentialV0 isdirectlyproportional to frequencybut is independent of the intensityof the incident light. v i v0 O a b IH Suppose in figure, the frequencyand intensityofincident light are held constant but the potential difference V between the two electrodes E and C is increased. Up to some stage as this p.d. is increased photoelectric current is also increased.However, soon some value ofVis reachedwhen the current reaches a limiting or saturation value when al the photoelectrons emitted by E are immediately collected byC. Further increase inVhardlyproduces anyappreciable increase in current as shown by the flat portion of curve I in figure. IfVis reversedwith the help of the reversing switchRS i.e. E ismade positive and Cnegative, the current i does not immediatelydrop to zero proving that electrons are emitted fromE with some definite velocity. This velocity is such that it gives enough kinetic energy to the electrons so as to surmount the retarding electric field between the two electrodes.Hence, some electrons do succeed in reachingCdespite the fact that the electric field opposes theirmotion. When reversedVismade large enough, a valueV0 (called stopping or inhibiting potential) is reachedwhen current is reduced to zero. Stopping potentialis that value ofthe retarding potentialdifference betweenthe two electrodeswhich is just sufficient to half themost energetic photoelectrons emitted. As seen fromcurve II of figure, doubling the intensity of the incident light merely doubles the current but does not affect the value ofV0. Now, if vmax is themaximumvelocityofemission of a photoelectron andV0 the stopping potential, then 2max 1 mv 2 = eV0 or vmax = 0 2.eV m
or vmax = 11 2.1.76.10 .V0 = 5.93 × 105 0 V m/s Obviously Emax = eV0 joules = V0 electron-volt If however, the experiment is repeated by varying the frequency of the light, it is found that the stopping potential varies linearlywith frequency as shown in figure. Below threshold frequency, no electrons are emitted, hence stoppingpotential is zero for that reason. But as frequencyis increased above f0, the stopping potentialvaries linearlywith the frequencyofincident light.
ATOMIC PHYSICS www.physicsashok.in 18 Einstein�s photoelectric equationmay be expressed interms of stopping potentialas given below: hf = W0 + 2max 1 mv 2 f0 Frequency Stopping potential, V0 Now, W0 = hf0 and 2max 1 mv 2 = eV0 . hf = hf0 + eV0 or V0 = 0 h(f f ) e. Now, f = c. and f0 = 0 c . V0 = ch e 0 . 1 1 . . . . . . . . = 12.4 × 10�7 0 . 1 1 . . . . . . . . volt . and .0 inmetres . V0 = 12,400 0 . 1 1 . . . . . . . . volt . and .0 in Å C9: Photoelectrons with a maximumspeed of 7 × 105m/sec are emitted froma metal surfacewhen light of frequency 8 × 1014Hz falls on it. Calyculate the threshold frequency of the surface. Sol: Emax = h(f � f0) . 2max 1 mv 2 = h(f � f0) or 12 × 9.1 × 10�31 × (7 × 105)2 = 6.625 × 10�34 (8 × 1014 � f0) . f0 = 4.635 × 1014 Hz C10:Atungsten cathodewhose thresholdwavelength 2300Åis irradiated by ultraviolet light ofwavelength 1800Å. Calculate (i) the maximumenergy of the photoelectrons emitted and (ii) thework function for tungsten, bothin electron-volts. Sol: (i) W0 = 0
12, 400 . = 12, 400 2300 = 5.4 eV (ii) Emax = 12,400 0 . 1 1 . . . . . . . . eV = 12,400 1 1 1800 2300 . . . . . . .= 1.5 eV C11: If light of . = 6000Åfalls on ametalsurface and emits photoelectronswith a velocityof 4 × 105m/s,what is photoelectric thresholdwavelength ?
ATOMIC PHYSICS www.physicsashok.in 19 Sol: K.E. of photoelectrons = 12 × 9.1 × 10�31 × 5 2 19 (4 10 ) 1.6 10. .. = 0.445 eV Energy content of photon of . = 6000Å= 12, 000 6000 = 2.07 eV . W0 = 2.07 � 0.445 = 1.625 eV ..0 = 12, 000 1.625 = 7631 Å C12: Calculate the threshold frequency for gold having photoelectricwork functionequal to 4.8 eV. If light of wavelength 2220Åfalls on gold,what will bemaximumkinetic energyof the photoelectrons coming out? Sol: Energy of the light photon = 12, 000 2220 = 5.58 eV. Out of this, 4.8 eV would be used for dislodging the electron and the balancewould represent its kinetic energy. Emax = 5.58 � 4.8 = 0.78 eV Alternatively, .0 = 12, 400 4.8 = 2583 Å. Hence, wemay use . Emax = 12,400 0 . 1 1 . . . . . . . . eV. C13:When violet light of . = 4000 Au strikes the cathode of a photocell a retarding potential of 0.4 V is required to stop emission of electrons.Calculate (i) light frequency(ii) photon energy(iii)work function (iv) threshold frequencyand (v) net energyafter the electron leaves the surface. Sol: (i) f = c. = 8 10 3 10 4000 10. .. = 7.5 × 1014 Hz (ii) E = hf = 6.625 × 10�34 × 7.5 × 1014 = 4.95 × 10�19 J = 3.1 eV (iii) W0= hf � K.E. = hf � V0 = 3.1 � 0.4 = 2.7 eV (iv) f0 = 0 Wh = 19 34 2.7 1.6 10 6.625 10
.
.
. .
. = 6.5 × 1014 Hz (v) Net energy hf � W0 = 3.1 � 2.7 = 0.4 eV = 6.4 × 10�20 J Example 9:Aphoton ofwavelength 3310Åfalling on a photo cathode ejects an electron of energy 3 × 10�19 J and one of wavelength 5000 Å ejects an electron of energy 0.972 × 10�19 J. Calculate the value of Planck�s constant and the thresholdwavelength for the photo cathode. Sol: hf = W0 + K.E. or hc . = W0 + K.E.
ATOMIC PHYSICS www.physicsashok.in 20 In the first case, 810 h 3 10 3310 10. . . . = W0 + 3 × 10�19 In the second case, 810 h 3 10 5000 10. . . . = W0 + 0.972 × 10�19 Subtracting one fromthe other, h = 6.62 × 10�34 J-s Substituting this value of h,W0 = 3 × 10�19J . .0 = 0 ch W = 8 34 19 3 10 6.62 10 3 10 . . . . . . = 6620 × 10�10 m. vmax = 7.12 × 105 m/s Example 10:Acertainmetalhas a thresholdwavelength of 6525Å. Find the stopping potentialwhen themetal is irradiatedwith (a) monochromatic light having awavelengthof 4000Å. (b) light having twice the frequencyand three times the intensity of that in (a) above. (c) If a material having double the work function were used, what would be the answer to (a) and (b) above? Sol: (a) V0 = 12,400 1 1 4000 6525 . . . . . . . = 1.2 volt (b) Stopping potential is independent ofthe intensity of the incident light but varies directlyas frequencyf provided it ismore than f0. Since frequencyis twice, thewavelength of the light is half i.e. 2000Å V0 = 12,400 1 1 2000 6525 . . . . . . . = 4.3 V (c) Ifwork function is double, then .0 is reduced to half i.e. .0 = 6525 2 =3262.5Å. Since the incident light
has . = 4000 Å, it would not be able to produce photoemission. In the second case, . = 2000 Å. . V0 = 12,400 1 1 2000 3262.5 . . . . . . . = 2.4 volt C14:Acertainmetallic surface is illuminated bymonochromatic light of variablewavelength.No photoelectrons are emitted above awavelength of 5000Å.With an unknownwavelength, a stopping potential of 3.1Vis necessary to stop photoelectric current. Find the unknownwavelength. Sol: Here, .0 = 5000 Å, V0 = 3.1 V, . = ? Now, V0 = 12,400 0 . 1 1 . . . . . . . . . 3.1 = 12,400 1 1 5000 . . . . . . . . . . = 2,222 Å
ATOMIC PHYSICS www.physicsashok.in 21 C15: Light ofwavelength 2000Åfalls on analuminiumsurface. In aluminium, 4.2 eVare required to remove an electron. Determine (i) KE of the fastest emitted photo-electron (ii) KE of the slowest emitted photoelectron (iii) stopping potentialand (iv) cut-offwavelength for aluminium. Sol: Photon energyofthe incident light is = 12400 2000 = 6.2 eV (i) Emax = (6.2 � 4.2) eV = 2 eV = 2 × 1.6 × 10�19 = 3.2 × 10�19 J (ii) Emin = 0 (iii) Ve = Emax = 2 V (iv) .0 = 0 12400 W = 12400 4.2 = 2952.4 Å C16: The stopping potential is 4.6Vfor light offrequency 2 × 1015Hz.When light of frequency 4 × 1015Hz is used, the stopping potential is 12.9V. Calculate the value ofPlanck�s constant. Sol: eV0 = h(f � f0) Substituting the two given values,we get 4.6e = h (2 × 1015 � f0) 12.9e = h(4 × 1015 � f0) Subtracting one fromthe other,we have 8.3e = 2h × 1015 8.3 × 1.6 × 10�19 = 2h × 1015 . h = 19 15 8.3 1.6 10 2 10 . . . . = 6.44 × 10�34 Js Example 11. 10�3Wof 5000Ålight is directed on a photoelectric cell. If the current in the cell is 0.16 µA, the percentage of incident photonswhich produce photoelectrons, is (A) 0.4% (B) .04% (C) 20% (D) 10% Sol. The percentage of incident photonswhichproduce photoelectrons is = n / t 100 N/ t . . . ...(1) . I net . . 16 19 n I 0.16 10 t e 1.6 10 .
.
.
. .
. .
. n 1012 t . . ...(2) and W N hc t . . . . N W 1016 t hc 4 . . . . ...(3) Persentage = 12 16 10 4 100 10 . . [fromeq. (1), (2) and (3)] Persentage . 0.04% Hence (B) is correct.
ATOMIC PHYSICS www.physicsashok.in 22 Example 12. In a photo-emissive cell, with excitingwavelength ., themaximumkinetic energyofelectron isK. If the excitingwavelengthis changed to 34. the the kinetic energyof the fastest emitted electronwillbe : (A) 3K/4 (B) 4K/3 (C) less than 4K/3 (D) greater than 4K/3 Sol. hc . . .K . ...(1) hc K´ 3 / 4 . . . . ...(2) Substracte eqn. (1) fromeq. (2) 4hc hc K´ K 3 . . . . . 4hc 3hc K´ K 3. . . . hc K´ K 3 . . . K K´ K 3 . . K´ K K 3 . . 4K K´ 3 . Hence (D) is correct. Example 13. Let K1 be themaximumkinetic energy of photoelectrons emitted by a light ofwavelength .1and K2 corresponding to .2. If ..1= 2.2, then : (A) 2K1 = K2 (B) K1 = 2K2 (C) K1 < 2 K2 (D) K1 > 2K2 Sol. K.E. = P2 2m P . 2mK P . K h . K . 1K . . 1 2 2 1 K
K . . .
ATOMIC PHYSICS www.physicsashok.in 23 . .1 = 2.2 . 2 2 2 1 2 KK . . . 21 K 4 K . 2 1 K K4 . . 2 1 K K2 . Hence (C) is correct. Example 14. Radiation of two photon energies twice and five times thework function ofmetal are incident sucessively on themetal surface. The ratio of themaximumvelocity of photoelectrons emitted is the two caseswill be (A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 4 : 1 Sol. E1 = 2 . E2 = 5 . E1 = . + K1 2 . = . + K1 K1 = . and E2 = . + K2 5 . = . + K2 K2 = 4 . . 12 K 1 K 4 .2max 1 2max 2 1 mv 1 21 mv 4 2 . 2max1 2max 2 v 1 v 4
. max 1 max 2 v 1 v 2 . Hence (A) is correct. Example 15.When photons of energy 4.25 eVstrike the surface ofametalA, the ejected photoelectrons have maximumkinetic energyTa eVand de-Brogliewavelength.a. themaximumkinetic energyofphotoelectrons liberated from another metal B by photones of energy 4.7 eV is Tb = (Ta � 1.5) eV. If the De-Broglie wavelength of these photoelectrons is .b= 2.a, then find (a) Thework function of a (b) Thework function of b is (c) Ta and Tb
ATOMIC PHYSICS www.physicsashok.in 24 Sol. 2 A Amax h 1 mv 2 . . . . and a A h P . . . A a P . h . . 2 2 A A max 1 P mv 2 2m . wheremismass of electron. . 2A A P h 2m . . . . or 4.25 = 2 A 2a h 2m . . . ...(1) For B, 4.7 = .B + Tb 2 B 2b h 4.7 2m . . . . ...(2) But .b = 2.a and Tb = Ta � 1.5 After solving (a) 2.25 eV (b) 4.2 eV (c) 2 eV and 0.5 eV C17: An isolatedmetal body is illuminatedwithmonochromatic light and is observed to become charged to a steadypositive potential1.0Vwithrespect to the surrounding.Thework function ofthemetal is 3.0 eV. the frequency of the incident light is __________. Sol. h . = . + eV or h . = 3 + 1 = 4 eV .
19 34 4 eV 4 1.6 10 h 6.63 10 . . . . . . . . . = 0.96 × 1015 Hz Example 16. 663mWof light froma 540 nmsource is incident on the surface of ametal. If only 1 of each 5 × 109 incident photons is absorbed and causes an electronto be ejected fromthe surface, the totalphotocurrent in the circuit is _______. Sol. N.t = no. of photon incident per second. . 663 10 3 N hc t . . . . .
ATOMIC PHYSICS www.physicsashok.in 25 . N 663 10 3 663 10 3 t hc 1242 nmeV 540 . . . . . . . . . 9 n / t 1 N/ t 5 10 . . . . . 3 9 9 n 1 N 1 663 10 540 t 5 10 t 5 10 1242 nmeV . . . . . . . . . . . . I ne 5.76 1011 A t . . . . Example 17. Light ofwavelength 330 nmfallingon a piece ofmetal ejects electronswithsufficient energywhich requires voltageV0 to prevent a electron fromreading collector. In the same setup, light ofwavelength 220 nm, ejects electronswhichrequire twich the voltageV0 stop theminreaching a collector. Find the numerical value of voltageV0. (take plank�s constant, h = 6.6 × 10�34 Js and 1 eV= 1.6 × 10�19 J) Sol. 0 1 hc . . . eV . and 0 2 hc . . . 2eV . Here .1 = 330 nm and .2 = 220 nm After solving,0 V 15 volt 8 . Example 18. Asmall 10Wsource of ultraviolet light ofwavelength 99 nmis held at a distance 0.1 mfroma metal surface. The radius of an atomof themetal is approximately0.05 nm. Find (i) the average number of photons striking an atomper second. (ii) thenumber ofphotoelectrons emitted per unit area per secondifthe efficiencyofliberationofphotoelectrons is 1%. Sol. (i) I = intensity = 2 10
4.(0.1) . .w= power incident on atom = . .2 9 2 2 I r 10 0.05 10 4 10 . . . . . . . . . w n hc t . . . . . n w 5 t hc / 16 . . . . .
ATOMIC PHYSICS www.physicsashok.in 26 (ii) no. of photons incident per unit area per second I hc / . . . no. of ejected electrons per unit area per second I 1 1020 hc / 100 80 . . . . . Example 19. The surface ofcesiumis illuminatedwithmonochromatic light of variouswavelengths and the stopping potentials for the wavelengths are measured. The results of this experiment is plotted as shownin the figure. estimate the value ofwork functionof the cesiumand Planck�s constant. 120 �1 �2 0.49 0.5 1.0 1.5 v × 1015Hz supporting potential (volt) Sol. h . = . + eVs or s V he e . . . . Fromgraph 15 h 2 e 0.49 10 . . . 19 15 h 2 1.6 10 0.49 10 . . . . . h = 6.53 × 10�34 Js But 2 e. . . . . . = 2 eV Example 20. In a photoelectric effect set-up a point source of light of power 3.2 × 10�3Wemitsmonoenergetic photons of energy 5.0 eV. The source is located at a distance of 0.8 mfrom the centre of a stationary metallic sphere of work function 3.0 eV and of radius 8.0 × 10�3 m. The efficiency of photoelectron emission is one for every 106 incident photons.Assume that the sphere is isolated and initially neutral and that photoelectrons are instantlyswept awayafter emission. (a) Calculate the number of photoelectrons emitted per second. (b) Find the ratio of thewavelengthof incident light to the de-Brogliewavelength
of the fastest photoelectrons emitted. (c) It is observed that the photoelectronemission stops at a certain time t after the light source is switched on why? (d) Evaluate the time t. Sol. (a) Energyof emitted photons E1 = 5.0 eV = 5.0 × 1.6 × 10�19 J E1 = 8.0 × 10�19 J Power of the point source is 3.2 × 10�3 watt or 3.2 × 10�3 J/s Therefore, energyemitted per second, E2 = 3.2 × 10�3 J. s 0.8 m r = 8.0×10�3m Hence number of photons emitted per second
ATOMIC PHYSICS www.physicsashok.in 27 2 1 1 E n E . or 3 1 19 3.2 10 n 8.0 10 . . . . . n1 = 4.0 × 1015 photons/sec. Number of photons incident on unit area at a distance of 8.0mfromthe source S will be n2 = 1 2 n 4.(0.8) = 4.0 1015 4 (0.64) . . . 5.0 × 1014 photon/sec � m2. The area ofmetallic sphere overwhich photonswill fallis : A = .r2 = .(8 × 10�3)2 m2 . 2.01 × 10�4 m2 Therefore, number of photons incident on the sphere per second are n3 = n2 A= (5.0 × 1014 × 2.01 × 10�4) . 1011 per second But since one photoelectron is emitted for every 106 photons hence number of photoelectrons emitted per second, n = 36 n 10 = 11 6 10 10 = 105 per second or n = 105 per second (b)Maximumkinetic energyof photoelectrons Kmax = Energy of incident photones �work function Kmax = (5.0 � 3.0) eV = 2.0 eV = 2.0 × 1.6 × 10�19 J Kmax = 3.2 × 10�19 J The de-Brogliewavelength of these photoelectronswillbe 1 max h h p 2 K m . . . (Here h = Planck�s constant andm=mass of electron) .
34 1 19 31 6.63 10 2 3.2 10 9.1 10 . . . . . . . . . . .1 = 8.68×10�10 m = 8.68 Å Wavelength of incident light .2 (inÅ) = 112375 E (in eV) or .2 = 12375 5 Å = 2476 Å Therefore, the desired ratio is 21 2475 285.1 8.68 . . . . (c)As soonas electrons are emitted fromthemetal sphere, it gets positively charged and acquires positive potential. The positive potential graduallyincreases asmore andmore photoelectrons are emitted fromits surface. Emissionof photoelectrons is stoppedwhen its potential is equal to the stopping potential required for fastestmoving electrons. (b)As discussed in part (c), emission of photoelectrons is stoppedwhen potential on themetal sphere is equal to the stoppeing potentialof fastestmoving electrons. Since Kmax = 2.0 eV
ATOMIC PHYSICS www.physicsashok.in 28 Therefore, stopping potentialV0 =2 volt.Let q be the charge required for the potentialon the sphere to be equal to stopping potentialor 2 volt. Then . 9 . 3 0 2 1 . q 9.0 10 9 4 r 8.0 10. . . . .. . . q = 1.78 × 10�12 C Photoelectrons emitted per second = 105 [part a] or charge emitted per second = (1.6 × 10�19) × 105 C = (1.6 × 10�14) C Therefore, time required to acquire to charge qwill be 2 14 q 1.728 10 t sec sec 1.6 10. 1.6. . . . or t . 111 second Example 21.Monochromatic radiation ofwavelength .1= 3000Åfalls on a photocelloperating in saturation mode.The corresponding spectral sensitivityof photocellis J = 4.8mA/W.When anothermonochromatic radiation ofwavelength .2 =1650Åand power P = 5mWis incident. It is found thatmaximumvelocityof photoelectrons increases to n = 2 times.Assuming efficiency of photo-electron generation per incident photon to be same for both the cases, calculate (i) thresholdwavelength for the cell and (ii) saturation current in second case. [Given, h = 6.6 × 10�34 Js, c = 3 × 108 ms�1 and e = 1.6 × 10�19 coul.] Sol. Maximumkinetic energy of photoelectrons is given byEk = hc . �W0,where . iswavelength of incident radiation andW0 iswork function of the surface onwhich radiation is incident. . Maximumkinetic energyof photoelectrons emitted byradiation ofwavelength .1 is given by 12 mv12 = 1 hc . �W0 or 21 0 1 mv 2 hc W . . . . . . . . . ...(1) wheremismass of an electron and v1 ismaximumvelocityof photoelectrons. Similarly, for radiation ofwavelength .1, 220
2 1 mv hc W 2 . . . ...(2) But v2 = 2v1, therefore fromequation (2), 21 0 2 2mv . hc .W . ...(3) Fromequations (1) and (3), 0 0 1 2 4 hc W hc W . . . . . . . . . . . or W0 = 3 eV
ATOMIC PHYSICS www.physicsashok.in 29 But work-functionW0 = 0 hc . where .0 is thresholdwavelength. . .0 = 0 hc W = 4125 Å Ans. (i) In saturationmode, spectalsensitivitywithwavelength .1 =3000Åis J = 4.8mA/Wor 4.8mC/J. Itmeans when 1 joule radiation ofwavelength .1= 3000Åis incident, a charge of 4.8mCflows in saturationmode or 4.8 mC e electrons are ejected. Energy of each photon ofwavelength .1 is E1 = 1 hc . . Number of photons in 1 joule radiation ofwavelength .1 1 1 1 E hc . . . No. of electrons ejected by these photons = Je = 3 19 4.8 10 1.6 10 . . . . = 3 × 1016 . Efficiencyof photo-electrongeneration per incident photon, 16 1 3 10 0.0198 ( / hc) . . . . . . Energy of eachphoton ofwavelength .2, E2 = 2 hc . . Rate of incidence of photons ofwavelength .2 in a radiation of power P 2 2 P P E hc .
. . per second Since, efficiency . of photo-electron generation is same for both the case, therefore, rate of ejection of electrons in later case2 . Phc . . . per second . Rate of flow of charge in saturationmode = 2 Phc . . e Cs�1= 13.2 µ Cs�1 But rate of flowof charge is current. Hence, saturation current is second case = 13.2 µA. Ans. (ii) Example 22. A monochromatic point source S radiating wavelength . = 6000Åwith power P = 2watt, an apertureAof radius R = 1 cm and a large screen are placed as shownin fig.Aphotoemissive detector D of surface area S = 0.5 cm2 is placed at centre of the screen. Efficiency of detector for photoelectric emissionper incident photon is . = 0.9. D A L S60cm 6 m (i) calculate photonflux at centre of screen and photo current in the detector. (ii) If a convex lens L of focal length f = 30 cmis inserted in the aperture as shown, calculate new value of photon flux and photo current assuming a uniformaverage transmission of 80%fromthe lens. (iii) Ifwork function of photo-emissive surface isW0 = 1 eV, calculate value ofstopping potential intwo cases (without andwiththe lens in aperture). Given, h = 6.625 × 10�34 J-S, c = 3 × 108 ms�1.
ATOMIC PHYSICS www.physicsashok.in 30 Sol. Photon flux is rate of incidence ofphotons per unit are ofdetector. Therefore, to calculate photon flux, first rate of emission of electrons fromthe source should be calculated. Energy of each photon is E . hc . Rate of emission of energy fromthe source is P = 2 watt = 2 Js�1 . Rate of emission of photons fromthe source is n P P E hc. . . or n = 6.04 × 1018 photons per second (i) Distance of detector fromsource is r1 = 6m . Photon flux at detector, 16 1 2 1 n 1.33 10 4 r . . . . . photons/m2s Rate of incidence of photons on detector = .1 . S Rate of emission of electrons fromdetector = ..1S per second Since, current is charge flowing per second, therefore photo current = (..1S)e = 9.6 × 10�8 amp (ii) When a concave lens is inserted in the aperture, it refracts incident rays. Therefore, photon flux and hence photo-current changes. Distance of lens fromsource is r2 = 0.60m . Photon flux at lens is .´= 2 2 n 4.r = 1.33 × 1018 photons/m2s Considering a very small areaAof the lens, Rate of incidence of photons on this area of lens =A.´ Nowconsidering refractionthrough the lens, u = � 60 cm, f = + 30 cm v = ? Using lens formule, 1 1 1 v u f . . , v = + 60 cm Since, average transmissionfromlens is 80%, therefore, rate of transmissionofphotons fromareaAof lens = 0.8 A.´ But these photons are transmitted in a solid angle subtended by the areaAat P as shown in fig. This solid angle, 2 2 A A v (0.6) . . . . Rate ofphotons transmitted per unit solid angle is T (0.8 A ´) . . . = (0.8 .´) (0.6)2 = 0.288 .´
D S P v 4.80m
ATOMIC PHYSICS www.physicsashok.in 31 Solid angle subtended by unit area of detector at P, .´= 2 1 (4.80) . Photon flux at D, .2 = T.´ = 1.67 × 1016 photons/m2s Rate of incidence of photons on detector = .2S Rate of emission of electrons fromdetector = ..2S . Photo current = (..2S)e = 1.20 × 10�7 amp (iii) Since, stopping potentialV0 is given bye.V0 = 0 . hc .W . .. . .. and . andW0 bothremainunchanged, therefore, stopping potential is same for boththe cases. . 0 0 V 1 hc W 1.07 volt e . . . . . .. . .. PHOTOELECTRIC EFFECT AND WAVE THEORY OF LIGHT According towave theory, when light falls on ametal surface, energy is continuouslydistributed over the surface, energyis continuouslydistributed over the surface.All the free electrons at the surface receive light energy.An electronmaybe ejected onlywhen it acquired energymore than thework function. Ifwe use a low-intensity source, itmaytake hours before an electron acquires thismuch energy fromthe light. In this period, there will be many collisions and any extra energy accumulated so far will be shared with the remainingmetal. Thiswill result in no photoelectron. This is contrary to experimental observations. No matter howsmall is the intensity, photoelectrons are ejected and that too without any appreciable time delay. In the photon theory, lowintensitymeans less number ofphotons and hence less number of electrons get a change to absorb energy.But anyfortunate electrononwhich a photonfalls, gets the fullenergyof the photon andmaycome out immediately. In figure,we illustrate an analogyto thewave the particle behaviour of light. In part (a), water is sprayed froma distance on an area containing several plants. Each plant receiveswater at nearly the same rate. It takes time for a particular plant to receive a certain amount ofwater. In part (b) of the figure,water is filled in identical, loosely-tiedwater bags and a particle physicist throws the bags randomly at the plants.When a bag collideswith a plant, it sprays all itswater on that plant ina very short time. In the sameway,whole of the energyassociatedwith a photon is absorbed by a free electronwhen the photon hits it. (a) (b) Themaximumkinetic energyof a photoelectrondoes not depend on the intensityof the incident light. This fact is also not understood bythewave theory.According to this theory,more intensitymeansmore energy and themaximumkinetic energymust increasewiththe increase inintensitywhichis not true.The dependence ofmaximumkinetic energyonwavelength is also against thewave theory.There should not be anythreshold wavelength according to thewave theory. According to this theory, byusing sufficientlyintense light of any
wavelength, an electronmaybe given the required amount of energyto come out. Experiments, however, showthe existence of thresholdwavelength.
ATOMIC PHYSICS www.physicsashok.in 32 DUAL NATURE OF LIGHT (a) Wave nature: Wave nature of light can be explained on the basis of reflection, refraction, interference, diffraction and polarization. (b) Particle Nature: Energy is transported byenergy particles, photons. It could be explained byphotoelectric effect, Zeeman effect, Comptoneffect etc. MATTER WAVE THEORY OR DE-BROGLIE�S THEORY (a) This theorywas given on the basis of duelnature of light. (b) According to de-Broglie theoryeach and everymoving particle has somewave nature associatedwithitself which is calledmatter waves. (c) Thus,moving particles like e�, proton, neutron, .-particle etc, also behave likewaves. (d) Thesewaves arewaves of probability. (e) The wavelength associatedwith a moving particles is given by . = h/p, where p is themomentumof the particle. (f) Thiswavelength is known as the de-Brogliewavelength of the particle. DE BROGLIE WAVELENGTHS For Photon For moving particle Rest mass Zero m Effectivemass m = 2 Ec = 2 hc. m= 02 2 m 1. v / c Energy(kinetic) E = hv = hc . E = 12 mv2, E = p2 2m Momentum p = Ec = hc. = h. p = mv = 2mE Wavelength . = hp = hc E . = h
p = h 2mE Wavelength for charged particle . = h 2mE accelerated byVvolts K.E. = qV Speed c = 3 × 108 m/s v = 2E /m = 2qV/m De-Broglie�s explanation for stable Bohr�s orbits : (a) De-Broglie suggested that non-radiation ofenergybythe electrons circling in aBohr�s orbit can be explained on the basis on the basis ofthe formation of stationarywaves bythe electrons in circularmotion inBohr�s orbits. . r . O Fourth Bohr-orbit (b) Comparing this to the vibrations ofawire loop suchstationarywaveswould be formed ifeachwave joins smoothlywith the next.
ATOMIC PHYSICS www.physicsashok.in 33 (c) In otherwords the number ofwavelengthsmust be an integer. (d) Condition for stable orbits : An election cancircle around an atomic nucleuswithout radiation energyif the circumference ofits orbit is an integralmultiple ofthe electronswavelength. i.e. 2.r = n... condition for stable orbits. 2.r = n hp . pr = nh 2. . mvr = nh 2. [. p = mv] mvr = nh 2. Asmvr is the angularmomentumof the circling electron. Bohr�s postulate is justified. C18: Calculate the de-Broglie wavelength associated with themotion of earth (mass = 6 × 1024kg) orbiting around the sun at a speed of 3 × 106ms�1. Sol: . = h mv = 34 24 6 1 6.63 10 (Js) (6 10 ) (3 10 )(kg ms ) . . . . . . . . . = 3.68 × 10�65 m NOTE: The wavelengths associated with the motion of macroscopic objects like earth, train etc, are negligible compared to their sizes. This is why the wave-like character of these objects is not observable in our daily life. C19: Calculate the de-Broglie wavelength of an .-particle ofmass 6.576 × 10�27 kg and charge 3.2 × 10�19 coulomb, accelerated though 2000V. Sol: E = kinetic � energy of a-particle = qV . E = 3.2 × 10�19 × 2000 J = 6.4 × 10�16 J deBrogliewavelength, . = h 2mE , [. 2mE =momentumof photon] = 34 27 16 6.63 10 (J s) 2 6.576 10 6.4 10 (J kg) . . . . .
. . . . . = 2.28 × 10�13 m NOTE: The wavelength associated with .-particles is of the order of size of the .-particle. That is why the wave like character of a-particle is observable. C20:Aparticle ofmassmand charge q is accelerated through a potentialdifferenceV. Find (a) its kinetic energy (b)momentum, and (c) de-Brogliewavelength associatedwith itsmotion. Sol:When the particle is accelerated througha potentialdifferenceV, gain in kinetic energyis given byK= qV. (a) Thus, kinetic energy,K= qV. (b) Momentumof the particle (p) is given byK= p2 2m p = 2mK = 2mqv . Momentum= 2mqv
ATOMIC PHYSICS www.physicsashok.in 34 (c) De- Brogliewavelengthis given by, . = hp = h 2mqV Thus, wavelength= h 2mqV Example 23. Assume that the de-Broglie was associatedwith an electron can forma standing wave between the atoms arranged in a one dimensional arraywith nodes at each of the atomic sites. It is found that one such standingwave is formed if the distance d between the atoms of the array is 2 Å.Asimilar standing wave is again formed if d is increased to 2.5 Åbut not for any intermediate value of d. Find the energy of the electron in eV and the least value of d for which the standing wave of the type described above can form. Sol. Fromthe figure it is clear that p . (./2) = 2 Å (p + 1) . ./2 = 2.5 Å . ./2 = (2.5 � 2.0) Å = 0.5 Å or ..= 1 Å = 10�10 m (i) deBrogliewavelengthis given by 2 Å2.5 Å N N p-loops (p + 1) loops /2 h h p 2 km . . . K= kinetic energy of electron 2 34 2 17 2 31 10 2 K h (6.63 10 ) 2.415 10 J 2m 2(9.1 10 )(10 ) . . . . . . . . . . . 17 19 K 2.415 10 eV 1.6 10 . . . . . . . . . . . . K = 150.8 eV (ii)N N The least value of d will bewhen only one loop is formed
. dmin = ./2 or dmin = 0.5 Å C21: Find the de-brogliewavelength associatedwith an electron accelerated through a potential difference of 30 kV. Sol: Kinetic energyof the electron K = qV = e(30 kV) = 30 keV = 30 × 103 × 1.6 × 10�19 J = 4.8 × 10�15 J Now,momentumof the electron = 2mK = 2.9.1.10.34 . 4.8.10.15 kg.J = 2.96 × 10�24 kg.m/s . De-brogliewavelength, . = h Momentum = 34 24 6.63 10 J s 2.96 10 Kgm/ s . . . . . = 2.24 × 10�10 m = 2.24 A
ATOMIC PHYSICS www.physicsashok.in 35 ATOMIC STRUCTURE ATOMIC MODEL By now, it iswell-known thatmatter, electricity and radiation etc. are all atomic incharacter.Although, no one has so far seen individualatoms, there is no doubt that they reallyexist. In 1895, it was discovered by J.Perrin in Paris that the cathode rays consist of negatively-charged practices called electrons. In 1897, J.J. Thomson measured the e/mratio for an electronwhereas its chargewasmeasured byMillikan in 1906 byhis famous oil-drop experiment.Mass of the electronwas found by dividing charge e bythe ratio e/m.Discoveryofpositive rays during the latter part of 19th centuryindicated that a normalatomconsisted of both negative and positive charges. But howthese charges are distributed in an atomwas not known at that time. Consinuous efforts have beenmade since then to studythe physical structure of anatomsuch as its extra-nuclear electronic structure chieflywiththe help of spectralproperties of atoms.To account for the spectroscopic dataobtained experimentallyover the years, several theories regarding atomic structure have been proposed fromtime to timewhich are called the atomicmodels.Various atomicmodels proposed by scientists over the last fewdecades are: (i)Thomson�sPlumpuddingmodel, (ii) Rutherford�sNuclearmodel, (iii) Bohr�smodel(iv) Sommerfeld�s Relativisticmodel (v)Vectormodel and finally(vi)Wave-mechanicalmodel. These differentmodels have been suggested one after the other inaneffort to get a satisfactoryinterpretation of the experimentaldatawhich, it is hoped, willultimately lead to a perfect and complete understanding of the physical structure of an atom. Thomson�s Plum PuddingModel According to thismodel, the atomis regarded as a heavy sphere of positive charge seasonedwith enough electron �plums� to make it electricallyneutral. Thomson visualized the positive charge ofanatombeing spread out uniformly throughout a sphere of about 10�10 metre radius with electrons as smaller particles distributed in circular shells as shown infigure.Whereas the net force + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + ++ + + + + + +
+ + + + + + + + + + + + + + + + + + + � � � � � � exerted bythe positively-charged sphere on each electron is towards the centre of the sphere, the different electrons experiencemutual repulsion and get arranged in the formof circular shells. This atomicmodelwas given up after some time because it could not provide any satisfactorymechanism for explaining the large deflection suffered by.-particles inRutherford�s experiment. Rutherford�s Experiment on .-particle Scattering As showninfigure high-speed a-particle (i.e. heliumnucleieachwitha charge of+2e) fromsome radioactive material like radiumor radon, confined to a narrowbeamby a hole in a lead block, weremade to strike a verythin goldfoilG.Whilemost ofthe .-particleswent straight throughthe foilas iftherewerenothing there (and produced scintillations on a fluorescent screen), some of them�collided�with the atoms ofthe foil and were scattered around at various angles -a fewbeing turned back towards the source itself. 150º 45º 60º . .. . . . Radium Lead block + Beam of .-particles
ATOMIC PHYSICS www.physicsashok.in 36 Althoughsome smallangle scattering could be expected fromThomson�smodel, large-angle scatteringwas absolutelynot expectedat all. Further detailedexperiments inRutherford�s laboratorybyGeiger andMarsden showed conslusivelythat large-angle scattering canbe expected onlyifone assumes that amassive positive point charge exists at the centre of each gold atomas shown in figure. According to thismodelproposed byRutherford in 1911, the positivemassive part of anatomis assumed to be concentrated in a very smallvolume at its centre. This central core, nowcalled nucleus, is surrounded by a cloud of electronswhichmakes the entire atomelectrically neutral. . . . P+ZeNucleus Asymptote of trajectory Trajectory of .-particle The large-angle scattering of positively-charged .-particles could be easily explained on this atomic model as shown in figure. This scattering is due to themutual repulsion (as per Coulomb�s law) between the .-particles and the concentrated positive charge on the nucleus. The .-particle approaches the positively-charged nucleus alongAO. If there were no repulsion fromthe nucleus, it would have passed at a distance of p fromit.However, due to coulombic force ofrepulsion, the .-particle follows a hyperbolawith nucleus as its focus.The linesAOand A.Oare the asymptotes of the hyperbola and present approximatelythe initial and finaldirections of the .- particlewhen it has passed out of the effective range of the nuclear electric field. As seen, the .-particle is deflected through an angle .. The perpendicular distance fromthe nucleus to the lineAOis called the impact parameter and is denoted by p.The Rutherford�s scattering formula is tan 2. = 1 2 Q Q 2pE where Q1 = Charge of the incoming .-particle Q2 = Charge ofthe scattering nucleus E = Kinetic energy of the incident .-particle p = Impact parameter Distance of Closest Approach Suppose that an .-particle approaches a positively-charged nucleus for a head on collisionwith a kinetic energy ofK.As shown in figure at pointA, the repulsive force ofthe nucleus is so strong as to stop the .- particlemomentarily.At this point, all the kinetic energyof the .-particle is converted into potential energy. Let Dbe the distance of closest approach of the .-particle. The potential at pointAdue to nuclear charge Ze is
= 0 Ze 4.. D + D A . Potential energy of the .-particlewhen at pointAis = 0 Ze.2e 4.. D = 2 0 2Ze 4.. D . K = 2 0 2Ze 4.. D or D = 2 0 2Ze 4.. K Since .-particles are generally obtained fromnatural radioactive substances, their kinetic energyK is known. Hence, value ofDcan be found easily fromthe above equation.
ATOMIC PHYSICS www.physicsashok.in 37 C22: InGeiger-Marsden experiment on .-particle scattering fromgold foil, the kinetic energy of .-particles usedwas 7.68MeV. Calculate the distance of closest approach of .-particle if atomic number of gold in 79. Sol: D = 2 0 2Ze 4.. K Here, Z = 79, e = 1.6 × 10�19 C; .0 = 8.854 × 10�12 F/m K = 7.68 MeV = 7.68 × 1.6 × 10�13 J D = �19 2 12 13 2 79 (1.6 10 ) 4 8.854 10. 7.68 1.6 10. . . . .. . . . . = 2.96 × 10�14 m Major Deficiencies in Rutherford�s Nuclear Model It was found later on that Rutherford�smodel had two serious drawbacks concerning (i) distribution of electrons outside the nucleus and (ii) the stabilityof the atomas awhole. It canbe shown that electrostatic forces between the positive nucleus and the static negative extra-nuclear electrons are not enough to produce equilibriumin such a nuclear atom. For example, consider the case of an atom* having two electrons and a nucleuswith a charge of+2e. Suppose the electrons are symmetrically placed at a distance of r fromthe nucleus and are stationary.The force of attraction between the nucleus and each of the electrons is F = e. 2 2e 4. .r = 2 2 2e 4..r while the force of repulsion between the two electrons is 2 2 e 4...4r = 2 2 e 16..r . Since the force of attractionis eight times the force of repulsion, the electronswill fall into the nucleus therebydestroying the stable structure of the atom. �� + + + + 2e + + + + � 2e �
v v r me + � Photon (a) (b) (c) 2e (hv) To overcome this difficulty, Rutherford suggested that stability can be achieved (as in a solar system) by assuming that electrons, instead ofbeing static, revolve round the nucleuswithsucha speedthat the centrifugal force balances the attractive force exerted bythe nucleus onthe electrons.As seen fromfigure (b), condition for stabilityis achievedwhen mv2 r = 2 0 2e.e 4.. r or mv2r = 2 0 2e 4. . In general, ifZ is the atomic number, then nuclear charge isZe, so that the above relation becomes mv2 r = 2 0 Ze.e 4.. r or mv2r = 2 0 Ze 4. .
ATOMIC PHYSICS www.physicsashok.in 38 Incidentally, itmaybe noted that according to the above relation, it is possible to have an infinite number of orbits inwhich electrons can rotate. But this assumptionofrevolving electrons ledto serious difficultyfromthe point ofviewo the electromagnetic theoryaccording towhich an accelerated chargemust continuouslyemit electromagnetic radiation or energy. Since an electronrevolving in a circular orbit has centripetal acceleration (= v2/r), itmust radiate energy as per the laws of classical electrodynamics. Due to this continuous loss of energy, the electronswillgraduallyapproachthe nucleus bya spiral path and finally fall into it as shown infigure (c). Hence, it is seen that the orbitalmotion of the electron destroys the verypurpose forwhich itwas postulatedi.e. the stabilityofthe atom.Obviously, eitherRutherford�s nuclear atomicmodelwithrevolving electrons isdefective or the classicalelectromagnetic theoryfails inthis particular case. This dilemmawas solved in 1913 byNeils Bohrwho proposed an improved version ofRutherford�s atomicmodel. Bohr�s Atomic model Thismodel (first proposed for hydrogen atombut later applied to other atoms as well) retains the two essential features ofRutherford�s planetarymodel i.e. (i) the atomhas amassive positively-charged nucleus and (ii) the electrons revolve round their nucleus in circular orbits the centrifugal force being balanced, as before, by the electrostatic pull between the nucleus and electrons. However, he extended thismodel further byutilizing Planck�sQuantumTheory. Hemade the following three assumptions: (iii) an electron cannot revolve round the nucleus in any arbitrary orbit but in just certain definite and discrete orbits.Onlythose orbits are possible (or permitted) forwhichthe orbital angularmomentum(i.e.moment of momentum) ofthe electron is equal to anintegralmultiple of h 2. i.e. orbitalangularmomentum= nh 2. where n is an integer and h is Planck�s constant. Such orbits are also known as stationary orbits. (iv) while revolving in these permitted stationary (or stable) orbits, the electron does not radiate out any electromagnetic energy. Inother words, the permissible orbits are non-radiating paths of the electron. (v) theatom radiatesout energy only when an electron jumps from oneorbit to another. If E2 and E1 are the energies corresponding to two orbits before and after the jump, the frequencyof the emitted photon is given bythe relation E2 � E1 = hv or .E = hv where v is the frequency of the emitted radiations. C 23: If I is the moment of inertia of an electron and . its angular velocity, then as per assumption (iii) given above
.I = nh 2. or (mr2).= nh 2. or (mr2 )v r = n.h 2. or mvr = n.h 2. Alternatively, sincethemomentumoftherevolvingelectronismv, itsmoment about the nucleus is =mvr Hence mvr = n.h 2. .....(i) e� m v +ze r
ATOMIC PHYSICS www.physicsashok.in 39 where, n= 1, 2, 3 for the first second and third orbits respectively. It is called the principalquantumnumber and because it can takewhole number values only, it fixes the sizes ofthe allowed orbits (also calledBohr�s circular orbits). r1 r2 r3 n = 1 n = 2 n = 3 +ZeE1 E2 E3 Permitted orbits (a) KLM +ZeE1 Electron Jump (b) E2 .E E3 Let the different permitted orbit have energies ofE1, E2, E3 etc. as shown in figure (a). The electron can be raised from n = 1 orbit to any other higher orbit if it is given proper amount of energy.When it drops back to n =1 orbit after a short intervalof time, it gives out the energydifference .E in the formof a radiation as shown in figure (b). The relationbetween the energy released and frequency of the emitted radiation is E2 � E1 = hv or .E = hv Expressions for velocity, radius, energy of electron and orbital frequency in Bohr�s orbit Here it should be kept inmind that Bohr�smodelis valid onlyfor hydrogen atomand hydrogen-like ions. In otherwords,we can saythat is applicable to hydrogen atomand ions having just one electron. Examples of such ions are He+, Li++, Be+++ etc. According to Bohr�s + ze r �e, m First postulate, v 2 2 0 Ze 4.. r = mv2 r where 0 1 4. . = 9 × 109 Nm2 C�2 v = velocityof electron and .0 = 8.854 × 10�12 C2/Nm2 m = mass of an electron = Absolute permittivity Z = atomic number
of vacuumor free space r = radius ofthe orbit e = magnitude of charge on an electron n = principalquantumnumber i.e. orbit number . r = 2 2 0 Ze 4.. mv .....(i) Fromfourth postulate ofBohr, we have mvr = nh 2. .....(ii) where, n = 1, 2, 3, 4, ........ i.e., n is a positive integer. Fromequations (i) and (ii),we get mv × 2 2 0 Ze 4.. mv = nh 2. . v = 2 0 ze 2.. nh Now, let us substitute the value of v in equation (ii),
ATOMIC PHYSICS www.physicsashok.in 40 m × 20 ze r 2 nh . . = nh 2. 2 2 0 2 r n h mZe . . . Note that for fixed n, v . Z, r . 1Z For fixed z, v . 1n , r . n2 IfV1 is the speed of the electron in the 1st orbit Then, Vn = V1 Zn , where V1 = c 137 , where c is speed of light. If a0 = first Bohr radius = 0.53Å, then rn = a0 n2 Z K.E. of the electron = 12 mv2 = 2 4 2 2 2 0 mZ e 8. h n = K = kZe2 2r , Potential energy of the atom= � 20 Ze 4.. r = U = � kZe2 r [k = 1/4...0] U = � 2 4 2 2 2 0 mZ e
4. h n . Total energy of the atom, E = K + U E = � 2 4 2 2 2 0 mZ e 8. h n In general, En = �13.6 22 Zn in eV. Orbital frequency for the electron, v = V 2.r = 2 2 2 2 0 0 Ze mZe 2 2 hn h n . . .. . .. . v = 2 4 2 3 3 0 mZ e 4. n h Time period of revolution (T) is given by T = T1 32 nZ where T1 = time period of revolution in the 1st orbit = 1.52 × 10�16 s = 0 1 2 a v. NOTE : It is assumed that the acceleration of the nucleus is negligible on account of its large mass. Some important results for H-atoms when n = 1 1. Bohr radius, a0 = 0.53 Å rn = a0n2/Z 2. v1 = 2.18 × 106 ms�1 . c/137 3. E1 = �13.6 eV = 1 U2
ATOMIC PHYSICS www.physicsashok.in 41 4. K1= 13.6 eV = � 1 U2 5. U1= �27.2 eV 6. v1 = 6.6 × 1015 Hz 7. 1 rydberg = �13.6 eV 8. B1 = 12.5 tesla, where B1 is themagnetic field at the centre ofBohr atomdue to the current generated by themotion ofelectron in 1st orbit. Ground state and excited states The state ofan atomwith the lowest energyis called its ground state or normal state. For ground state, n= 1. The stateswith higher energies are called excited states. For the first excited state, n= 2; for the 2nd excited state, n = 3 and so on. Formth excited state, n =m+ 1 Ionization energy and ionization potential Theminimumenergyneeded to ionize an atomis called ionization energy. The potentialdifference throughwhich anelectron should be accelerated to acquire the value of ionization energyis called ionization potential. The value of ionization energyofH-atomin ground state is 13.6 eVthat of ionization potential is 13.6 eV. Binding energy Binding energyof a systemis the energyneeded to separate its constituents to large distances or itmay be defined as the energy releasedwhenits constituents are brought frominfinityto formthe system. The value ofbinding energyofH-atomis 13.6 eV, identical to its ionization energy. Excitation energy and excitation potential The energyneeded to take the atomfromits ground state to an excited state is called the excitation energy of that excited state. The potentialdifference throughwhich an electron should be accelerated to acquire the value of excitation energyis called excitation potential. NOTE : (A) I.E. = E. � E1 = �E1 = Binding energy of theH-atom. (B) Movement of electron in circular orbits in aBohr atomcauses electric current inthe orbit. This current will lead to self-generatedmagnetic field in the atomand alsomagnetic current (m) (a)Magnetic field(B) IfBis themagnetic field generated at the centre of atom, then H-atom +e a0 �e i B = 00i 2a . , where i is the current due to motion of the electron and a0 is the 1st Bohr-radius. Hence, i = ev
hence B = 0 0 e 2a . .
ATOMIC PHYSICS www.physicsashok.in 42 Putting the values of v and a0, we get B = 12.5 Tesla (b) Magneticmovement vector ( .. ) .. = iA . . . = iA = .ev 20a (c) Relation between .. & L . (angularmomentumvector) µ = iA = .ev 20a , [i = ev, A= . 20a ] L = mV1a0 = m2.v 20a [V1 = 2.va0] . L. = 20 20 e a m2 a . ... = e 2m . L. = e 2m Vectorially, .. = e 2m . L . Example 24: The quantumnumber ofBohr orbit inH-atomwhose radius is 0.01mmis (a) 223 (b) 435 (c) 891 (d) none of these Sol: (b)We know that rn = a0n2 n2 = n0 ra n = 310 0.01 10 0.529 10.. .. . n = 435 Example 25:The quantumnumber n in theBohr�smodel ofH-atomspecifies: (a) radius of the electron (b) energyof the electron
(c) angularmomentumof the electron (d) allof these Sol: (d) rn = 2 0 2 h me .. . . . . . . × n2 En = � 4 2 2 2 0me 8. h n , Ln= n h 2 . . . . . . . Example 26: The radius Bohr�s orbit in ground state for H-atomis (Take .0 = 8.86 × 10�12 C2/Nm2, h = 6.6 × 10�34 J-s) (a) 0.528 Å (b) 0.0528 Å (c) 5.28 × 10�10 m (d) 5.28 × 10�10 cm Sol: (a) . a0 = 2 0 2 h me .. (in this case, Z = 1, n = 1) a0 = 34 2 12 31 19 2 (6.6 10 ) (8.86 10 ) 3.14 (9.1 10 ) (1.6 10 ) . . . . . . . . . . . = 0.528 Å
ATOMIC PHYSICS www.physicsashok.in 43 Example 27: The speed of the electron inthe first Bohr orbit ofH-atomis (Take c = speed of light in vacuum) (a) c (b) c 13.6 (c) c 137 (d) 137 c Sol: (c) For H-atom,we knowthat v = 20 e 2. h [since n this case, Z = 1 and n = 1] = 20 e c 2. hc But 2 0 e 2. hc = 1 137 = fine structure constant . v = c 137 Example 28: In a H-atom, binding energy of the electron in the ground state is E1. Then the frequency of revolution of the electron in the nth orbit is (a) 1 3 2E n h (b) 3 1 2E n h (c) 1 3 2mE n h (d) none of these Sol: (b) The frequencyof revolution of the electron in the nth orbit is given by v = 2 4 2 3 3 0 mZ e 4. h n = 13 2E hn where E1 = 2 4 2 2 0
mZ e 8. h C24: Calculate the energy of aHe+ ion in its first excited state. Sol: . En =� 2 2 (13.6eV)Z n Here, Z = 2, n = 2 . En = 2 2 13.6 2 n . . eV = �13.6 eV C25. An electron in a hydrogen like atomis in an excited state. It has a total energy of �3.4 eV. Calculate : (i) the kinetic energy (ii) the de-Brogliewavelength of the electron Sol. (i)Kinetic energyof electron in the orbits of hydrogen and hydrogen like atoms = | Total energy | . Kinetic energy = 3.4 eV (ii) The deBrogliewavelength is given by h h P 2 Km . . . (K= kinetic energyof electron) Substituting thevalues,we have 34 19 31 (6.6 10 J s) 2(3.4 1.6 10 J)(9.1 10 kg) . . . . . . . . . . = 6.63 × 10�10m or . = 6.63 Å
ATOMIC PHYSICS www.physicsashok.in 44 Example 30. The electron in a hydrogen atommakes transition fromMshell to L. The ratio ofmagnitudes of initial to final centripetal acceleration ofthe electron is (A) 9 : 4 (B) 81 : 16 (C) 4 : 9 (D) 16 : 81 Sol. . rn × n2 and vn × 1n2n n n v a r . n 2 2 a 1 n n . . n 4 a 1 n . n 4 a Kn . M 3 a a K81 . . {. M = 3} L 2 a a K16 . . {. L = 2} ML a 16 a 81 . Hence (D) is correct. Example 31. The angularmomentumopf an electron in the hydrogenatomis 3h 2. .Here h is Planck�s constant. The kinetic energyof this electronis : (A) 4.53 eV (B) 1.51 eV (C) 3.4 eV (D) 6.8 eV Sol. L nh 3h 2 2 . . . . . n = 3 In the electronic third orbit, the energy of electron 22 Z E 13.6
n . E 13.6 19 . . E 13.6 1.51 eV 9 . . Hence (B) is correct. Example 32. A particle of mass m moves along a circular orbit in a centrosymmetrical potential field U(r) = kr2 2 .Using theBohr�s quantization condition, find the permissible orbital radii and energy levels of that particle.
ATOMIC PHYSICS www.physicsashok.in 45 Sol. F = � dU dr = � kr Negative signimplies that force is acting towards centre. The necessary centripetal force to the particle is being provided bythis force F. Hence mv2 r = kr ...(1) and mvr = n h where h h 2 . . . .. . .. ...(2) solving equations (1) and (2) we get r = rn = n h m. where . = km and total energy E = U + K= kr2 2 + 12 mv2 Substituting the values,we get E = n h . = En Example 33. In a hypothetical systema particle ofmass mand charge �3q is moving around a very heavy particle having charge q.AssumingBohr�smodel to be true to this system, the orbital velocity ofmassm when it is nearest to heavyparticle is (A) 2 0 3q 2. h (B) 2 0 3q 4. h (C) 0 3q 2. h (D) 0 3q 4. h Sol. 2 e mv F r . 2 2 0
q 3q mv 4 r r . . .. . . 20 3 q mvr v 4 . .. . mvr h 2 . . . . . 20 3 q h v 4 2 . .. .2 0 v 3 q 2 h . . Hence (A) is correct.
ATOMIC PHYSICS www.physicsashok.in 46 SPECTRUM Dispersed light arranging itself ina patternofdifferent wavelength is referred to as a spectrum. Light coming froma sourcemay be dispersed bya prismor by any other dispersingmedium. Whenwhite light falls on a prismand the transmitted light is collected on awhitewallorwhite paper then a spectrumis obtainedwhich consists ofdifferent colours fromred to violet. Kinds of spectra : (A) Emission spectra:When a light beamemitted by certain source is dispersed to get the spectrum, it is called an emission spectrum. An emission spectrumcan be three types : (a) Continuous spectrum: That emission spectrumwhich is obtained by continuouslyvaryingwavelength, is called continuous emissionspectrum. In this case,when light is dispersed, a bright spectrumcontinuously distributed ona dark background is obtained. Light emitted froman electric bulb, a candle or a red hot iron piece comes under this category. (b) Line spectrum: The atoms and molecules can have certain fixed energies.An atomormolecule, in an excited state, can emit light to lower its energy. Light emitted in such a process has certain fixedwavelengths.When such a light is dispersed, certain sharp bright lines on a dark background is obtained. Such a spectrumis called line emission spectrum. Atomic energy levels Energy Line spectrum . For example, when electric discharge is passed through sodiumvapour, they vapour emits light of the wavelength 589.0 nmand 589.6 nm.Whenthis light is dispersed bya high resolution grating , one obtains two bright yellowlines on a dark background. (c) Band spectrum: Thewavelengths emitted bythemolecular energylevelswhich are generallygrouped into severalbunches, are also grouped; eachgroup beingwell separated fromthe other.The spectrumlooks like separate bands of varying colours. Such a spectrumis called band emission spectrum. Molecular energy levels Energy Band spectrum . (B) Absorption spectrum:Whenwhite light is passed throughan absorbingmaterial, thematerialmayabsorb certainwavelengths selectively.When the transmitted light is dispersed, dark lines or bands at the positions of themissing (absorbed) wavelengths are obtained. Such type of spectrumis called absorption spectrum. White lightAbsorbing material Absorption spectrum
ATOMIC PHYSICS www.physicsashok.in 47 An absorption spectrum may be of two types : (a) LineAbsorption spectrum : Light may be absorbed by atoms to take themfromlower energy states to higher energystates. In the similarwaywhenwhite light is passed through a gas, the gas is found to absorb light of certainwavelength. The absorption spectrumconsists of dark lines on bright background. Such a spectrumis calleda line absorption spectrum.When light coming fromthe sun is dispersed, it shows certain sharply defined dark lines. This shows that certainwavelengths are absent. There missing lines are called Fraunhofer lines. (b) Band Absorption spectra : If absorbing media is polyatomic such as H2, CO2 or KMnO4 solution, instead of dark lines we get few characteristic dark bands (against coloured background) called band absorption spectra. Hydrogen Spectra : If hydrogen gas enclosed in a sealed tube is heated to high temperature, it emits radiation. This radiation consists of components of different wavelengths which deviate by different amounts. The radiationwith different amounts ofdeviation formsH-spectrum. Explanation of hydrogen spectra by Bohr (a) The electron in aH-atomif not disturbed remains in the ground state (i.e. n = 1 state).When the electron receives energyfromoutside, it is elevated to anyone of the higher permitted states (excited state). (b) The electronremainsonlyfor a short intervaloftime (generallyin theorder of10�8 s) in the excited state and comes back to the ground state finally. (c) The electroncan reach the ground state fromanyone of the excited states inmanyways.As a result,many electron transitions take place. (d) According toBohr, all electron transitions terminating at a particular state give rise to a particular spectral series. Series limit Energy Lyman series Balmer series Paschen series Brackett series n =1 n = 2 n = 3 n = 4 n = 5 n = . E = 0 (e) Lyman: nf = 1: 1. = � 1 Ech 2 2 1 1 1 n
. . . . .
. . n = 2, 3, 4,.......... Balmer: nf = 2: 1. = � 1 Ech 2 2 1 1 2 n . . . . . . . n = 3, 4, 5, .......... Paschen: nf = 3: 1. = � 1 Ech 2 2 1 1 3 n . . . . . . . n = 4, 5, 6, .......... Brackett: nf = 4: 1. = � 1 Ech 2 2 1 1 4 n . . . . . . . n = 5, 6, 7, ......... Pfund: nf = 5: 1. = � 1 Ech 2 2 1 1 5 n . . . . . . . n = 6, 7 , 8,......
ATOMIC PHYSICS www.physicsashok.in 48 (f) If an electronmakes a jump fromthe nith to nfth orbit (ni> nf), the extra energyEi � Ef is emitted as a photon of electromagnetic radiation. The correspondingwavelength is given by 1. = i f E E hc . where c = speed of light in vacuum. According to Bohr,we canwrite 1. = 2 4 2 3 0 mZ e 8. ch 2 2 f i 1 1 n n . . . . . . . 1. = RZ2 2 2 f i 1 1 n n . . . . . . . where R = 4 2 3 0 me 8. ch is called theRydberg constant. (g) The value of R is 1.0973 × 107m�1 NOTE: � En = 2 2 RhcZ n . , 1 rydberg = �13.6 eV, Rhc = 13.6 eV (h) 1. is calledwave number ( .. ) of the line and 2. . is called angular wave number of the line. (i) Photon energy= Ep = hv (j) Momentumof a photon = p = Photon Energy speed of light . p = p Ec Important points regarding H-spectra
(a) The sharplydefined, discretewavelengths exist in the emitted radiation in theH-spectrum. (b) Ahydrogen sample emits radiationwithwavelengths less than those inthe visible range (i.e. uv light) and alsowithwavelengthsmore than those in the visible range (i.e. infrared). (c) the linesmay be grouped in separate series. (d) In each series, the separation between the consecutive wavelengths decreases as wemove fromhigher wavelength to lowerwavelength. (e) Aparticularminimumwavelength in each series approach a limiting value knownas series limit. (f) The series corresponding to uv region, visible region and infrared region are known asLyman, Balmer and Paschen series respectively. (g) In the Balmer series of hydrogen, theH. line (3 . 2) is red, theH. line (4 . 2) is blue, theH. ( 5 . 2) and H.(6 . 2) lines are violet, and the other lines are in the near ultraviolet (uv). (h) Theoretically possible no. of emission spectral lines = n(n 1) 2. . (i) Theoretically possible no. of absorption spectral lines = (n � 1)
ATOMIC PHYSICS www.physicsashok.in 49 (j) Approximate range ofwavelength for different colours of visible light Colour Wavelength Range Violet + Indigo 3800 Å to 4500 Å Blue 4500 Å to 5000 Å Green 5000 Å to 5500 Å Yellow 5500 Å to 6000 Å Orange 6000 Å to 6500 Å Red 6500 Å to 7200 Å Infrared rays: 720 nmto 50 mm Ultraviolet light: 10 Å to 3800 Å Limitations of Bohr�s Model (a) It is valid onlyfor one electron atomand hydrogen-like ions e.g. : H, He+, Li+2, Na+10 etc. (b) Orbitswere taken as circular but according to SOMMER field these are elliptical. (c) Intensity of spectral lines could not be explained. (d) Nucleuswas taken as stationary but it also rotates onits own axis. (e) It could not be explained theminute structure in spectrumline. (f) This does not explain the ZEEMANeffect (splitting up of spectral lines inmagnetic field)&Stark effect (splitting up in electric field). (g) This does not explain the doublets in the spectrumof some of the atoms like sodium(5890Åto 5896Å). C26. Total number of emission lines fromsome excited state n1 to another energy state n2(< n1) is given by 1 2 1 2 (n n )(n n 1) 2 . . . . For example total number of lines fromn1 = n to n2 = 1 are n(n 1) 2. . C27. As the principalquantumnumber n is increased in hydrogenand hydrogen like atoms, some quantities are decreased and some are increased. The table given belowshowswhich quantities are increased and which are decreased. Table Increased Decreased Radius Speed Potentialenergy Kinetic energy Totalenergy Angular speed Time period Angularmomentum C28. Whenever the force obeys inverse square law 2 F 1r . . . .. .. , and potential energy is inversely proportional to r, kinetic energy(K), potential energy(U) and total energy (E) have the following relationships. K = | U | 2 and E = �K = U2 . If force is not proportional to 2
1r or potential energyis not proportional to 1r , the above relations do not
ATOMIC PHYSICS www.physicsashok.in 50 hold good. In JEE problems, this situation arises at two places, in an atom(betweennucleus and electron) and in solar system(between sun and planet). C29. Total energy of a closed systemis always negative and the modulus of this is the binding energy of the system. For instance, suppose a systemhas a total energyof �100 J. Itmeans that this systemwill separate if 100 J of energyis supplied to this.Hence, binding energy of this systemis 100 J. Thus, totalenergyof an open systemis either zero or greater than zero. C30. Kinetic energyofa particle can�t be negative,while the potentialenergycan be zero, positive or negative. It basically depends on the reference point where we have taken it zero. It is customary to take zero potential energywhenthe electron is at infinite distance fromthe nucleus. In some problemsupposewe take zero potential energyin first orbit (U1 = 0), then themodulus of actual potential energyin first orbit (when reference point was at infinity) is added inUand E inall energy states, whileKremains unchanged. Example 33:The value of series limit inLyman series is: (a) 121.6 nm (b) 91.2 nm (c) 656.3 nm (d) 365.0 nm Sol: min 1 . = R.1. 1 . .. ... min 1 . = R[1 � 0] .min = 7 1 1.097.10 = 91.2 nm C31: Find the longest wavelength present in theBalmer series of hydrogen: Sol: In the Balmer series, nf = 2. The longest wavelength in this series corresponds to the smallest energy difference between energy levels. Hence the initial statemust be ni = 3. . 1. = R 2 2 f i 1 1 n n . . . . . . . = R 2 2 1 1 2 3 . . . . . . . = R 1 1 4 9 . . . . . . . 1. =
5R 36 . = 7 36 5.1.097.10 = 6.56 × 10�7 m . = 656 nm(near the red end of the visible spectrum) C32:Ahydrogen atomemits uv radiation of102.5 nm. Calculate the quantumnumbers of the states involved in the transition. Sol: The uv radiation 102.5 nmlies inthe lyman region of spectrum. Thus nf=1 . 1. = R 2i 1 1 n . . . . . . . . 2i 1 n = 1 � 1.R . 2i 1 n = 1 � 9 7 1 102.5.10. .1.097.10 = 1 � 1 1.124
ATOMIC PHYSICS www.physicsashok.in 51 2i 1 n = 0.1 2i n = 10 ni . 3 Hence transition is from3 . 1. C33:What is the unit of reciprocalofRydberg constant in S.I. units ? Sol:We know that 1. = R 2 2 f i 1 1 n n . . . . . . . the terminthe bracket is unitless. Now, we canwrite 1. = R . 1R = . Hence, the unit of 1R willbemetre i.e.m. C34: Howmanydifferent wavelengthsmay be observed in the spectrumfroma hydrogen sample if the atoms are excited to stateswith principal quantumnumber n ? Sol: The totalnumber of possible transitions is (n � 1) + (n � 2) + (n � 3) + ............. + 2 + 1 = n(n 1) 2. C35: Consider the following two statements: (A) Line spectra contain information about atoms only (B) Band spectra contain information aboutmolecules (a) BothAand B are wrong (b)Ais correct but B iswrong (c) B is correct butAis wrong (d) BothAand B are correct Sol: (c) Line spectra contain information about atoms andmolecules both. C36. Ultraviolet light ofwavelength 800Åand 700Åwhen allowed to fallon hydrogen atoms in their ground state is found to liberate electronswith kinetic energy 1.8 eVand 4.0 eVrespectively. Find the value of the Planck constant. Sol. hv = E0 + T where E0 = ground level energy and T = kinetic energy of electron 0 hc . E . T . . 10 hc 800 .10. = E0 + 1.8 × 1.6 × 10�19 and 10 hc 700 .10. = E0 + 4.0 × 1.6 × 10�19
Subtracting 8 hc 1 1 10. 7 8 . . . .. .. = 2.2 × 1.6 × 10�19 or h = 27 8 2.2 1.6 10 56 3 10 . . . . . = 6.57 × 10�34 Js
ATOMIC PHYSICS www.physicsashok.in 52 C37: The excitation energy of a hydrogenlike ion in its first excited state is 40.8 eV. Find the energy needed to remove the electronfromthe ion. Sol: The excitationenergy in the first excited state is E = 13.6Z2 2 2 1 1 1 2 . . . . . . . 40.8 = (13.6 eV) × Z2 × 34 . Z = 2 Now, ionization energy= 2 2 13.6Z 1 = �4 × (13.6 eV) Eion = �54.4 eV C38: The first ionization potentialof some hydrogen likeBohr atomis xV.Then the value of the first excitation potential for this atomwillbe: (a) xV (b) x2 V (c) 34 xV (d) 20 xV Sol: The value of first excitation potentialis given by x. = x 1 14 . . . . . . . V x. = 3x 4 V Example 34:Adoublyionised lithiumatomis hydrogen likewith atomic number 3. Find thewavelength of the radiation required to excite the electron inLi++ fromthe first to the thirdBohr orbit. (Take ionization energy ofH-atomequal 13.6 eV). Sol: For E1, Z = 3, n = 1 . E1 = � 2 2 13.6Z n = �13.6 9 1. = 122.4 eV E n = 3 3 E n = 2 2 E n = 1 1
.E For E3, Z = 3, n = 3 . E3 = � 2 2 13.6 3 3. = �13.6eV We have .E + E1 = E3 . .E = E3 � E1 = 108.8 eV . . = hc.E = 12400 108.8 Å = 114 Å Example 35: InBohr�smodelof hydrogen atomwhenthe electron ismoving in one of the stationaryorbits then: (a) velocityof the electron is fixed and no emission of energy takes place (b) velocitychanges continuouslybut no emission of energytakes place (c) energyis emitted but the velocitydoes not change (d) energyis emitted and the velocity also changes. �e �e �e �e v1 v2 v3 v4 . Sol: (b) According to the second postulate ofBohr. 1 | v | . = 2 | v | . = 3 | v | . = 4 | v | . = v
ATOMIC PHYSICS www.physicsashok.in 53 Example 36. An electron in an unexcited hydrogenatomacquired an energyof 12.1 eV. Towhat energy level did it jump?How many spectral linesmay be emitted in the course of transition to lower energy levels? Calculate the shortest wavelength. Sol. E = � E0 22 Zn where E0 = 2.18 × 10�18 J = 1 rydberg .E (energygap between unexcited state (n = 1) and an excited state (n= m)) 2 0 2 2 E Z 1 1 1 m . . . . .. .. . 12.1 × 1.6 × 10�19 = 2.18 × 10�18 × 12(1 � 1/m2) n = 3 n = 2 or 1 � n = 1 2 1 m = 0.888 or m= 3 Three lines are emitted. The shortestwavelength corresponds to the greatest energygap. . min hc . = E3 � E1 = E0 2 2 1 1 1 3 . . . .. .. or .min = 34 8 18 6.6 10 3 10 9 2.18 10 8 . . . . . . . . = 1.02 × 10�7 m = 1020 Å Example 37. Ahydrogen like atom(atomic number Z) is in a higher excited state of quantumnumber n.The excited atomcanmake a transition to the first excited state bysuccessivelyemitting two photons of energy 10.2 and 17.0 eVrespectively.Alternatively, the atomfromthe same excited state canmake a transition to the second excited state bysuccessivelyemitting two photons ofenergies 4.25 eVand 5.95 eVrespectively. Determine the values of n and Z. (lonization energyofHatom= 13.6 eV) Sol. Fromthe given conditions En � E2 = (10.2 + 17) eV = 27.2 eV ...(1) and En � E3 = (4.25 + 5.95) eV = 10.2 eV ...(1) Equation (1) and (2) gives E3 � E2 = 17.0 eV or Z2 (13.6)(1/4 � 1/9) = 17.0
. Z2 (13.6) (5/36) = 17.0 . Z2 = 9 . Z = 3 Fromequation (1) Z2 (13.6) (1.4 � 1/n2) = 27.2 or (3)2(13.6) (1/4 � 1/n2) = 27.2 or 1/4 � 1/n2 = 0.222 or 1/n2 = 0.0278 or n2 = 36 . n = 6 ATOMIC EXCITATION WITH THE HELP OF COLLISION (a) An atomcanbe excited to an energyabove its ground state bya collisionwith another particle inwhich part of their joint kinetic energy is absorbed by the atom. (b) An excited atomreturns to its ground state in an average of 10�8 s by emitting one ormore photons.
ATOMIC PHYSICS www.physicsashok.in 54 (c) Energytransfer is amaximumwhenthe colliding particles have the samemass. (d) The energy used in this process will be of discrete nature. (.E = E2 � E1 = hv = hc . ). NOTE: If the joint kinetic energy of colliding particles is less than 20.4 eV (considering particles as Hatoms or one neutron and one H-(atom) then the nature of collision will be necessarily elastic. Example 38: Aneutronmovingwith speed vmakes a head-on collisionwith a hydrogen atomin ground state kept at rest. Find the minimumkinetic energy of the neutron for which inelastic (completely or partially) collisionmay take place. The mass of neutron= mass of hydrogen = 1.67 × 10�27 kg. Sol: Let us suppose that neutron andH-atommove at speeds v1 and v2 after the collision. Suppose an energy .E is used in thisway. On the basis of conservation of linearmomentumand energy,we canwrite mv = mv1 +mv2 .....(i) 12 mv2 = 21 1 mv 2 + 221 mv 2 + .E .....(ii) Fromequation (i) we have v2 = 21 v + 22v + 2v1 v2 .....(iii) Now, fromequation (ii) v2 = 21 v + 22v + 2 E m. .....(iv) . Fromequation(iii) and (iv) 2v1v2 = 2 E m. Hence, (v1 � v2)2 = (v1 + v2)2 � 4v1v2 = v2 � 4 E m. Since v1 � v2 is always real, v2 � 4 E m. . 0 mv2 . 4.E . 12 mv2 . 2.E
Theminimumkinetic energyof the neutron needed for aninelastic collision corresponds to transition from n = 1 to n = 2 . Kmin = 2min 1 mv 2 = 2 × 10.2 eV . Kmin = 20.4 eV Example 39. Consider anexcited hydrogen atominstate nmovingwith a velocityv (v<< c). It emits a photon inthedirectionofitsmotionand changes its state to a lowr statem.Applymomentumandenergyconservation principle to calculate the frequencyv of the emitted radiation. Compare thiswiththe frequencyv0 emitted if the atomwere at rest. Sol. Let En and Em be the energies of electron in nth andmth states. Then
ATOMIC PHYSICS www.physicsashok.in 55 En � Em = hv0 ...(1) In the second casewhen the atomismovingwith a velocity v. Let v´ be the velocityof atomafter emitting the photon.Applyingconservationof linearmomentum, v v´ m m v mv = mv´ + hv c (m=mass of hydrogen atom) or v´ = v hv mc . . . .. .. ...(2) Applying conservationof energy En + 12 mv2 = Em + 12 mv´2 + hv or hv = (En � Em) + 12 m(v2 � v´2) hv = hv0 + 12 m 2 v2 v hmc . . . . . . . . . . . .. . . .. hv = hv0 + 12 m 2 2 2 2 2 2 v v h 2h v m c mc . . . . . . . . . . . hv = hv0 + 2 22 h v h c 2mc . . . Here the termis 2 22 h2mc . is very small. So, can be neglected . hv = hv0 + h v h v
c c . . . or v 1 vc . . . .. .. v0 or v = v0 1 1 vc . . . . .. .. ; v . v0 1 vc . . . .. .. as v < < c Example 40: Thewavelength ofD1 andD2 lines of sodiumare 5890Åand 5896Årespectively, if theirmean wavelength is 6000Åthen find the difference of excited energy states. Sol: E = hc . ....E = 2 hc . .. .E = 34 8 10 20 6.62 10 3 10 6 10 6000 6000 10 . . . . . . . . . . . .E = 3.31 × 10�22 J . .E = 22 19 3.31 10 1.6 10 . . . . , ........2 × 10�3 eV C39:Alithiumatomhas three electrons.Assume the following simple picture of the atom.Two electronsmove
ATOMIC PHYSICS www.physicsashok.in 56 close to the nucleusmaking up a spherical cloud aroundit and the thirdmoves outside this cloud in a circular orbit. Bohr�smodel can be used for themotion of this third electron but n = 1 states are not available to it. Calculate the ionization energyof lithiumin ground state using the above picture. Sol: In this picture, the third electronmoves in the field of a total charge +3e �2e = +e. Thus, the energy are the same as that of hydrogen atoms. The lowest energy is E2 = 1 E4 = 13.6eV 4 . = �3.4 eV Thus , the ionization energy of the atomin this picture is 3.4 eV. Example 41: Find the wavelengths in a hydrogen spectrumbetween the range 500 nmto 700 nm. Sol: The energy of a photon ofwavelength 500 nmis hc . = 1242eV nm 500nm. = 2.44 eV The energy of a photon ofwavelength 700 nmis hc . = 1242eV nm 700nm. = 1.77 eV The energydifference between the states involved in the transition should, therefore, be between 1.77 eV and 2.44 eV. Figure shows some of the energies of hydrogen states. It is clear that onlythose transitionswhich and at n = 2mayemit photons of energybetween 1.77 eVand 2.44 eV. Out of these only n = 3 to n = 2 falls in the proper range. The energyof the photon emitted in the transition n = 3 to n= 2 is E . . = (3.4 � 1.5)eV= 1.9 eV.Thewavelength is . = hc .E = 1242eV nm 1.9eV. = 654 nm. Example 42: Calculate the (a) velocity, (b) energy, and (c) frequency of the electron in first Bohr orbit of hydrogen atom. Sol. (a)We have, vn = 2
0 Ze 2. nh ; but here Z = 1 and n = 1 . v1 = 2 0 e 2. nh = 19 2 9 34 (1.6 10 ) 1 36 10 2 1 6.62 10 . . . . . .. . . . = 2.18 × 106 m/sec (b) We have, En = � 4 2 2 2 2 0 me Z 8. n h Again here, z = 1 and n = 1
ATOMIC PHYSICS www.physicsashok.in 57 . E1 = � 31 19 4 2 9 2 2 34 2 9.1 10 (1.6 10 ) 1 (4 9 10 ) 8 1 (6.62 10 ) . . . . . . . . .. . . . . = �21.758 × 10�19 joule = � 19 19 21.758 10 1.6 10 . . . . = �13.6 eV (c) We have, v = 2 4 2 3 3 0 Z me 4. n h ; here also n = 1 and Z = 1 = 4 2 3 0 me 4. h = 6.57 × 1015 Hz. C40: Find out the radius ofthe hydrogen atomin ground state. Sol. We have, rn = 2 2 0 2 h n mZe. . ; here Z = 1 and n = 1 r1 = 34 2 9 19 2 31 (6.62 10 ) 1 4 9 10 1 (1.6 10 ) (9.1 10 ) . . . . . .. . . .. . . . . r1 = 0.53 Å Example 43: If thewavelength of the firstmember of the Balmer series of hydrogen spectrumis 6562Å, then calculate thewavelength of firstmember ofLymen series in the same spectrum. Sol. We have, for the first member of the Balmer series v1 = R 2 2
1 1 2 3 . . . . . . . = 5 36 R and for the first member ofLyman series, v2 = R 2 2 1 1 1 2 . . . . . . . = 3R4 . 12 .. = 12 .. = 5R 36 × 4 3R = 5 27 . .2 = 1 527 . = 5 6552 27 . = 1215.18 Å Example 44.Thehydrogenatominits groundstate is excitedbymeans ofmonochromatic radiationsofwavelength 975Å.Howmanydifferent lines are possible in the resulting spectrum?Calculate the longest wavgelength among them.Youmayassume the ionizationenergy for hydrogen atomto be 13.6 eV, the Planck constant = 6.63 × 10�34 Js. Sol. En= � E0Z2/n2. The energyrequired to take the electronfromn = 1 to infinityis the ionization energyof the hydrogen atom. . 13.6 = E0(1/12 � 1/.) or E0 = 13.6 eV
ATOMIC PHYSICS www.physicsashok.in 58 Therefore, for hydrogen En = �13.6/n2 eV The energyof the photon incident on hydrogen is 34 8 18 10 hc 6.63 10 3 10 E h 2.04 10 J 975 10 . . . . . . . . . . . . . Let the electron jump fromn= 1 to n=mafter absorbing the incident photon. Then .E = Em � E1 = 13.6(1/12 � 1/m2)eV = 13.6(1 � 1/m2) × 1.6 × 10�19 J . 13.6(1 � 1/m2) × 1.6 × 10�19 = 2.04 × 10�18 . (1 � 1/m2) = 0.9375 or m= 4 The resulting transitions are shown in the figure. So there are six possible lines. The longest wavelength corresponds to theminimumenergygap.Hence longest wavelengthcorresponds to transitionfromm= 4 to m= 3 . 0 2 2 h E 1 1 3 4 . . . . . .. .. . 0 hc E 1 1 9 16 . . . . . .. .. E0 m = 4 m = 3 m = 2 m = 1 . 0 hc 144 7E . . . . 34 8 19 6.63 10 3 10 144 7 13.6 1.6 10 . . . . . . . . . . . . = 1.88 × 10�6 m = 18800 Å. Example 45: The energy of an excited hydrogen atomis �3.4 eV. Calculate the angular momentumof the electron according to Bohr�s theory. Ans: 2.11 × 10�2 joule × sec. Since the energy of an electron in nth level in hydrogen atomis
En = � 2 RCh n or, �3.4 = � 2 13.6 n . RCh = 13.6 eV . n = 2 FromBohr�s theory, L= n h 2. . L = 2 × 0.6 10 34 2 3.14. . . = 2.11 × 10�3 joule sec. REDUCED MASS .In our earlier discussionwe have assumed that the nucleus (a protonin case of hydrogenatom) remains at rest.With this assumption the values of the Rydberg constant R and the ionization energy of hydrogen predicted byBohr�s analysis arewithin 0.1%ofthemeasured values. Rather the proton and electron both revolve in circular orbits about their common centre ofmass.We can
ATOMIC PHYSICS www.physicsashok.in 59 take themotion of the nucleus into account simplybyreplacing themass ofelectronmbythe reducedmass µ of the electron and nucleus. Here . = Mm M.m .....(i) where M= mass of nucleus. The reduced mass can also be written as, . = m1 mM . Now, whenM> > m, mM . 0 or . . m For ordinary hydrogen we let M = 1836.2 m. Substituting in equation (i),we get µ = 0.99946mwhen this value is used instead of the electronmassmin the Bohr equations, the predicted values arewellwithin 0.1%of themeasured values. + � cm m m v v Separation r Applying the Bohr model to positronium. The electron and the positron revolve about their common centre of mass, which is located midway between them because they have equal mass The concept of reduced mass has other applications.Apositron has the same rest mass as an electron but a charge +e. Apositroniumatomconsists of an electron and a positron, each withmassm, in orbit around their common centre ofmass. This structure lasts only about 10�6 s before two particles annihilate (combine) one another and disappear, but this is enough time to studythe positroniumspectrum. The reducedmass ism/2, so the energy levels and photon frequencies have exactlyhalf the values for the simpleBohrmodelwithinfinite protonmass. CM M m r2 r1 Now, let us provewhymis replaced bythe reducedmass .when motion of nucleus (proton) is also to be considered. In figure both the nucleus (mass =M, charge = e) and electron (mass =m, charge =e ) revolve about their centre ofmass (CM) with same angular velocity (.) but different linear speeds. Let r1 and r2 be the distance ofCMfromproton and electron. Let r be the distance between the proton and the electron. Then, Mr1 = mr2 .....(ii) r1 + r2 = r .....(iii) . r1 = mr M.m and r2 = Mr M.m .....(iv) Centripetal force to the electron is provided by the electrostatic force. So, mr2.2 =
22 0 1 e 4.. r or m Mr M r . . . . . . . .2 = 22 0 1 . e 4.. r or Mm M m . . . . . . . r3.2 = 2 0 e 4..
ATOMIC PHYSICS www.physicsashok.in 60 or .r3.2 = 2 0 e 4.. .....(v) where Mm M.m = . Moment of inertia of atomabout CM, I = 2 2 1 2 Mr .mr = Mm M M . . . . . . . r2 = .r2 .....(vi) According toBohr�s theory, nh 2. = I... µr2..= nh 2. .....(vii) Solving equations (v) and (vii) for r,we get r = 2 2 0 2 n h e .. . .....(viii) Electrical potential energyof the system, U = 20 e 4 r ... and kinetic energy,K = 12 I.2 = 12 µr2.2 Fromequation (v),.2 = 2 3 0 e 4.. .r , K = 20 e 8.. r .Total energyof the system, E = K + U = � 20 e 8.. r
Substituting value of r fromequation(viii),we have E = � 4 2 2 2 0 e 8 n h . . .....(ix) The expression for Enwithout considering themotion of proton is En = � 4 2 2 2 0me 8. n h , i.e.,mis replaced byµ while considering themotionof proton. NOTE :(i) Variation of rn, vn and Enwithmass of election is as under, rn . 1m , vn = independent ofmand En .m Sometimes the electron is replaced bysome another particlewhich has a charge �e but mass different fromthemass of electron. Here, two cases are possible. Case 1: Let saymass of the replaced particle is x times themass of the electron and nucleus is still very heavycompared to the replaced particle, i.e., themotion of the nucleus is not to be considered. Inthis case
ATOMIC PHYSICS www.physicsashok.in 61 rnwill become 1x times, vn will remain unchanged and En becomes x times. Case 2: In this case motion of nucleus is also to be considered, i.e., mass of the replaced particle is comparable to themass of the nucleus. In this case themass of the electron is replaced bythe reducedmass of the nucleus and the replaced particle. Let say the reducedmass is ytime themass of the electron. Then, rn will become 1y times, vn remains unchanged and En becomes y-times. (ii)Reduced massm= 1 2 1 2 m m m . m ofm1 and m2 is less than both the masses. C41:Apositroniumatomis a systemthat consist of a positron and an electron that orbit each other. Compare thewavelengths of the spectral lines of positroniumwith those of ordinaryhydrogen. Sol: In this case reducedmasswillbe given by m. = mM m.M = m2 2m = m2 wherem=mass of the electron. Hence, the energy levels of a positroniumatomare E.n = m' m . . . . . . 12 En = 12 E 2n It means that the Rydberg constant for positroniumis half as large as it is for H-atom. As a result the wavelength inthe positroniumspectrallines are all twice thoseof the corresponding lines in theH-spectrum. Example 46: Bohr�s theoryassumes that nucleus is of infinitemass and so electron rotates round the stationary nucleus. Assuming the nucleus to be of finite mass MH, the value of correct Rydberg constant will be (consider hydrogen atom) (a) 4 H 2 3
H 0 mM e M m 8 ch . . . . . . . . (b) 4 H 2 2 3 H 0 M m e mM 8 c h . . . . . . . . (c) 4 H 2 3 H 0 mM e M m 8 ch . . . . . . . . (d) 4 2 3 0e 8. ch Sol: (c) In this case both electronand nucleuswill rotate about a common centre ofmass, sayO. Suppose that the radiiof the electron and nucleus orbits are re and rn respectivelythenbydefinition of centre ofmass; the electron, the centre ofmassOand the nucleus are always in a straight line. . MHrn =mre whenmis themass of electron . n e n r r . r = H m m.M Let r = re + rn then rn = H mr m.M and re = H H M r m.M FromBohr�s quantizationrule,we have
ATOMIC PHYSICS www.physicsashok.in 62 2 2 m.re .MH.rn = nh 2. [.is the angular velocity about the centre ofmass] Putting thevaluesof rn and re,we get HH mM m M . . . . . . . .r2 = nh 2. The above equation can be comparedwithm.r2 = nh 2. . TheRydberg constantwill be given by R = HH mM m M . . . . . . . 4 2 3 0e 8. ch Obviously the reducedmass of the electron is HH mM m M . . . . . . . . Example 47.Apositromiumatomis a bound systemofan electron (e�) and its antiparticle positron (e+) revolving about their centre ofmass. Find thewavelength of the radiationwhen the systemde-excites fromits first excited state to the ground state. Sol. This problemcan be solved byBohr�s theoryof the hydrogen atombyreplacing themass of the electron by its reducedmass. . . e e e e e m m m µ m m 2 . . . . . 2 4 e n 2 2 2 0
Z e mE 2 8 h n . . . .. .. . .4 2 e n 2 2 2 0 e m Z E 8 h 2n . . . . . .. . . . . . . . . . En = � 2.18 × 10�18(12/2n2) (. e4me/8.02h2 = 2.18 × 10�18) When n = 1, E1 = � 2.18 × 10�18/2 When n = 2, E1 = � 2.18 × 10�18/8 . .E = E1 � E2 = 2.18 × 10�18(1/2 � 1/8) = 2.18 × 10�18 × 3/8 . .E = hc/. = 2.18 × 10�18 × 0.375 . . . . . 34 8 18 6.63 10 3 10 2.18 10 0.375 .. . . . . . . . , . = 1.2165 × 10�7 m = 2433 Å Example 48:Themass ofmuon (. ) is 207 times that of the electron and charge = �1.6 × 10�19 C.Amuon can be captured by a nucleus to formamuonic atom. Calculate the value of ionization energyof themuonic atom.
ATOMIC PHYSICS www.physicsashok.in 63 Sol: The ionization energyof themuonic atomis obtained by replacingme inH-atomformula bythe reduced massmof the proton-muon system. This reducedmass is m = p p m m m m. . . = e e e e 1836m 207m 1836m 207m .. . 186me Thus the ground state energy is (n = 1, Z= 1). 2 2 4 1 2 2 mk e E h . . . 2 2 4 e 1 2 e m 2 m k e E m h . . . . . . . . . E1 = � 186 × 13.6 eV The ionizationenergy = �E1 = 186 × 13.6 eV = 2.53 KeV PRODUCTION OF X-RAYS X-rayswere accidentallydiscovered byWilhelmRontgenin 1895 during the course of some experiments with a discharge tube.At present, it is well known that these rays are produced whenever fast moving electrons strike a high atomicweight solid like tungsten kept in vacuum. (a) X-ray Tube: The essentialelements of amodernCoolidgeX-rayvacuumtubewhich iswidely used for commercial and medical purposes are shown in figure. Electrons are produced thermionically froma tungsten filamentary cathode F which is heated to incandescence either by a storage battery or by a low-voltage alternating current froma stepdown transformer T2. These electrons are focussed on the target T with the help of a cylindrical shield Swhichsurrounds Fand ismaintained at a negative potential. The electrons are accelerated to veryhigh speeds (upto 10%of velocity of light) by the d.c. potentialdifference (of about 50 kV- 100 kV) applied between F and the anode (also called anticathode). This high d.c. potential is obtained froma step-up transformer T1whose output is converted into direct current by full-wave
rectifier and a suitable filter. � + 50 kV ElectronsT A Cooling Tube Fins X-rays R F S B The targetTusuallyemployed inX-raytubes is amassive block oftungstenor inmanycases, amolybdenum
ATOMIC PHYSICS ATOMIC PHYSICS plug embedded in the face of a solid copper anode. The face of the copper anode is sloped at about 45º to the electron beam. Being very good conductor of heat, copper helps to conduct heat efficiently to the externalcooling finsorthewater-coolingsystem.Undertheterrificbombardmentofthetargetbysomany electrons,mostmetalswillmelt. Thatiswhymetalsliketungsten,platinumandmolybdenumetc.areused whichhave highmelting points and also have a highatomicweight (whichis essentialfor abundant production of X-rays).
When the electrons strike the tungsten target, they give up their kinetic energyand thereby produce X-rays. T A Cooling Water Tube X-rays R F ST1 T2 Rectifier (b) Control of Intensity and Quality The intensity of X-rays depends on the number of electrons striking the target. This number is determined by the temperature of the electron-emitting filament which itself is proportionalto the heater current. Hence by controlling the filament current with the help of a rheostat R, thermionic emission and hence intensity of X-rays can be controlled. The quality of X-rays is measured in terms of their penetrating power which is dependent on the potential difference between filamentary cathode and the anode. Greater this accelerating voltage, higher the speed of the striking electrons and consequently, more penetrating the X-rays produced. It is customaryto refer to highly penetrating X-rays (i.e. those possessing high frequency) as hard X-rays and to those less penetrating (i.e. of low frequency) as soft X-rays. Obviously, the quality or penetrating power of X-rayscan be controlled by varying the potential difference between the cathode and anode. It will be noticed from above explanation that in coolidge X-ray tube, it is possible to achieve separate control ofthe intensity and quality of X-rays independent of each other. It has beenfound that apart fromthe intensity and quality ofX-rays, their abundance depends on the atomic weight ofthe target material. Target materials of higher atomic weights yield a greater abundance of X-rays than those oflower atomic weights.
ORIGIN OF X-RAYS X-rays are produced when high-speed electrons strike some material object. However, majority of the electronsthatstrikeasolidtarget,donothingspectacular atall.Most ofthemundergoglancingcollisions
with the matter particles, lose their energy a little bit at a time and thus merely increase the average kinetic energyofthe particles of the target material. It is found that nearly99.8 percent of the energyof the electron beam goes into heating the target. But a smallnumber ofthe bombardingelectrons produce X-rays bylosing their kinetic energyinthe following two ways:
(i) Someofthehigh-velocityelectronspenetratetheinterioroftheatomsofthetargetmaterialandareattracted www.physicsashok.in 64
ATOMIC PHYSICS www.physicsashok.in 65 bythe positive charge of their nuclei.As an electronpasses close to the positive nucleus, it is deflected from its pathas shownin figure. The electronexperiences deacceleration duringits deflectionin the strong field of the nucleus. The energy lost during this de-acceleration is given off in the formofX-rays of continuously varyingwavelength(and hence frequency). TheseX-rays produce continuous spectrumwhen analysed by Bragg spectrometer.This spectrumhas a sharplydefined short-wavelength limit .min(or high-frequencylimit fmax)which corresponds to themaximumenergyofthe incident electron. X-RayX-Ray + + + + X-Ray 1 v. 2 mv.2 Continuous spectrum12 mv2v If, as shownin figure, the striking electronhas its velocityreduced fromvto v. during its passage through the atomofthe targetmaterial, then its loss of energyis = ( 12 mv2 � 12 mv.2). Thismust equal the energyof the X-rayphotons emitted. . 12 m(v2 � v.2) = hv The highest ormaximumfrequency of the emitted X-rays corresponds to the case when the electron is completely stopped i.e. when v. = 0. In that case 12 mv2 = hvmax .....(i) If the electron is accelerated through a potential ofVvolts, then 12 mv2 = eV .....(ii) From(i) and (ii), we get hvmax= eV; vmax = eV/h Now, hvmax= hc/.min [. c = v ...] . hc/.min = eV or .min = hc/eV Substituting the values of e = 1.602 × 10�19C h = 6.62 × 10�34 J-s and c = 3 × 108 m/s, we get .min = 34 8 19 6.62 10 3 10 1.602 10 V . . . . .
. = 1.24 10 6 V . . m or .min = 12400 V Å [1Å = 10�10 m] SuchX-rays are veryaptly called �braking� radiations because they are due to braking or slowing down of high-velocity electrons is the positive field of a nucleus. These radiations constitute, as said earlier, the continuous spectrumof theX-rays because they consist of a series of uninterruptedwavelengths having a sharplydefined short-wavelength limit .min. TheseX-rays are independent ofthe nature of the targetmaterial but are determined by the potentialdifference between the cathode and anode of theX-ray tube. (ii) Some ofthe high-velocityelectronswhile penetrating the interior ofthe atoms of the targetmaterial, knock
ATOMIC PHYSICS www.physicsashok.in 66 off the tightlybound electrons in the innermost shells (likeK, L-shells etc) of the atoms.When electrons fromouter orbits jump to fillup the vacancyso produce, the energydifference is given out in the formofXrays of definite wavelength (and frequency). These wavelengths constitute the line spectrumwhich is characteristic of thematerial of the target. KL M e e e (a) KL M (b) X-Ray K -line . Figure (a) shows the case when the high-velocity incident electron knocks off one electron fromtheKshell.As shown in figure (b), this vacancyinK-shellis filled by a nearby electron in the L-shell. During the jump anX-rayradiation is emittedwhose frequencyis given by Ek � E. = hv where Ek is the energyrequired to dislodge an electron formtheK-shell and El is that required for L-shell. Since this energy difference is comparativelyvery large, theX-rays emitted have verylarge energycontent and hence are highly penetrating. If, however, this vacancy inK-shell is filled up by an electron jumping fromM-shell, theX-rays emitted would be stillmore energetic andwould consequentlypossess stillhigher frequencybecause .E =(Ek �Em) ismore than .E = (Ek � EL). SuchX-rays arising frommillions of atoms produce the K-lines as shown in figure. Usually, two linesK. andK. of this series are detected although there aremanymore. Similarly, when the incident electron carries somewhat lesser amount of energy, it dislodges an electron fromthe L-orbit and an electron either fromM-orbit or other outer orbits takes its place so that X-rays of frequency lower than that of theK-series are produced. This gives the L-series of theX-ray spectrumas shown byK., L. and L. lines in figure (a). KLMN O K. K. L. L. L. M (a) E = 0 �2 �20
�200 �2000 �20,000 K. K.L. L. L. M. M. KLMNO .SHELL Energy in eV (b) M Spectral lines ofM-series are produced ina similarway as shownin the energy-leveldiagramof figure (b). As stated earlier, theseK, LandMseries constitute the line spectra of theX-rayswhich are characteristic of thematerialused as target in theX-ray tube. Hence, theX-rays produced byanX-ray tube consist of two parts: (i) one part consists ofa series of uninterruptedwavelengths having a short cut-offwavelength .min.This
ATOMIC PHYSICS www.physicsashok.in 67 constitute the continuous spectrumand (ii) the other part consists of a number of distinct and discrets wavelengths which constitute the line or discontinues spectrumof theX-rays. X-ray Spectrum As explained inX-ray spectrumconsists of (i) continues spectrumand (ii) line spectrum. These two are shownin figure. (a) Continuous Spectrum (i)It is produced due to the de-acceleration of high-velocityelectronswhen theyare deflectedwhile passing near the positively, charged nucleus of an atomof the targetmaterial. K. K. L. L. L. Intensity Continuous Spectrum .min . f Continuous Spectrum K. K. L. L. L. (a) (ii) It has a sharply-defined short wavelength limit given by .min = 12.400 V × 10�10 m or 12, 400 V Å (iii)The cut-offwavelength.min is independent ofthenature ofthe targetmaterialbut is inverselyproportional to the potentialdifference between the cathode and anode of anX-ray tube. The value of .min decreases as this potentialdifference is increased. (iv) The intensity of the continuous spectrum(given by the area enclosed bythe curve of figure (b) is found verynearlyproportional to the square of the applied voltage for a given target and to the atomic number of the target materialwhen a constant potential difference is applied. 0 1.0 2.0 3.0 4.0 5 kV 10 kV 15 kV 20 kV K 25 kV . K. L. L. L. Wave length(A.U.) X-ray Intensity Tungsten Target (b) (v) There is a shift of themaximumintensity position towards the short wavelength side as voltage is increased. (b) Line Spectrum
(i) It is producedwhen electrons are dislodged fromthe innermost orbits ofthe atoms ofthe targetmaterialfollowed byelectronjumps fromouter orbits. (ii) It consists of discrete spectral lineswhichconstituteK-series, L-series andM-series etc.K-series consists of those lines for which electron jumps end at K-level. (iii) K-series beingmost energetic constitute the hardX-rayswhereasL- andM-series formthe �soft�Xrays. (iv) Line spectrumis characteristic of the targetmaterial used. In
ATOMIC PHYSICS www.physicsashok.in 68 fact, X-rays constituting the line spectrum are known as characteristicX-rays. The number of lines present in the spectrum depends both on the nature of target material and the excitation voltage. K. K. 90 79 50 42 28 Cr 24 Cu Mo Sn Au Cf . Mass No. (Z) (c) (v) There is a regular shift towards shorter wavelength in theKspectrumas the atomic number ofthe target is increased figure (c). The exact relationship, as found byMoseley, is 21 .. = 12 .. = 2 1 2 2 (Z 1) (Z 1) .. where v1 is the frequency of theK. line for a target material having an atomic number ofZ1 and v2 and Z2 are similar quantities for some different targetmaterial. C42:AnX-raytubeworks on 60,000V.What will be the wavelength ofX-rayemitted in it. Sol: .min = 12, 400 V Å Here, V = 60,000 V . .min= 12, 400 60,000 = 0.2 Å C43: If the potential difference applied across anX-ray tube is 12.4 kV and the current through it is 2 mA, calculate: (i) the number of electrons striking the target per second (ii) the speedwithwhich theystrike it (iii) the shortest wavelength emitted Take e = 1.6 × 10�19 C and m = 9.1 × 10�31 kg
Sol: (i) If n is the number of electrons striking the anode per second, then I = ne . n = Ie = 319 2 10 1.6 10.. .. = 1.25 × 1016 s�1 (ii). v 2eV 5.93 105 V m . . . v = 5.93 × 105 v = 5.93 × 105 12, 400 = 6.6 × 107 m/s (iii) .min = 12, 400 V = 12400 12400 = 1Å C44: Calculate theminimumapplied potentialrequired to produceX-rays of 1Åwavelength. Sol: .min = 12400 V Å . V = min 12400 . = 12400 1 = 12.4 kV C45:An X-ray tube passes 5 mAat a potential difference of 100 kV. Calculate the maximumspeed of the
ATOMIC PHYSICS www.physicsashok.in 69 electrons striking the target and the rate of productionofheat at the target if only0.1 percent of the incident energy is converted into X-radiations. Take e/m= 1.76 × 1011 C/kg and J = 4.18 joules/cal. Sol: . v 2eV 5.93 105 V m . . . v = 5.93 × 105 100,000 = 1.88 × 108 m/s � + 100 kv 5 mA Incident power = 100,000 × 5 × 10�3 = 500W Power converted into heat = 99.9%of 500 = 499.5W Heat produced / second = 499.5 4.18 = 119 cal/s. C46:AnX-raytube operated at 30 kVemits a continuousX-rayspectrumwith a short wavelength limit .min = 0.414 Å. Calculate Planck�s constant h if e = 1.602 × 10�19 C and c = 3 × 108m/s. Sol: .min = ch eV . h = min eVc. = 19 3 19 8 1.602 10 30 10 0.414 10 3 10 . . . . . . . . = 6.63 × 10�34 J-s MOSELEY�S LAW In 1913-14,Moseleycarried out a systematic study of the characteristicX-ray spectra ofvarious elements used as targets in anX-raytube.By usingBragg�s spectrometer for the purpose, the remarkably similar to each other in the sense that each consists of K-L and M-series. However, there is one very important difference. The frequency of lines (in every series) produced froman element of higher atomic number is greater thanthat produced byanelement of lower atomic number. It is due to the fact that binding energyof electrons increases aswe go fromone element to another ofhigher atomic number. Because there is greater positive charge onthe nucleus ofanelement of higher atomic number, larger amount ofenergyis required to liberte an electron fromtheK, L andMshells of that element. Consider theK. line of the characteristicX-rayspectrumof any element. It is found that higher the atomic number ofthe targetmaterial, higher is the frequencyof theK. line produced byit. The exactmathematical relationship betweenfrequency and atomic number is given by v . (Z � b)2 or . . (Z � b) or . = a(Z � b)
where Z is the atomic number of the element and a and b are constants for a particular series but varyfrom one series to another i.e. their values for K-series are different fromthose for L-series etc. The constant b is known as nuclear screening constant. For lines ofK-series, b = 1. Its values for lines ofL-series ismore. The above relation is known asMoseleylawfor the characteristic or lineX-ray spectrum. Itmaybe stated as follows: The frequency of a spectral line in the characteristic X-ray spectrumvaries directly as the square of the atomic number ofthe element emitting it. Figure showsMoseleydiagramfor K. andK. lineswhich is
ATOMIC PHYSICS www.physicsashok.in 70 obtained by plotting . versus atomic number of different elements of the periodic table.As expected, the graph is linear. 5 10 15 20 25 × 108 Al-13 10 Ca-20 Zn-30 Ze-40 Sn-50 K. K. Atomic number Frequency An exact formofMoseley�s lawis 1. = R(Z � .)2 2 2 1 2 1 1 n n . . . . . . . whereR is Rydberg�s constant, Z the atomic number, . a correction factor and n1 and n2 the principal quantum numbers of the energy levels between which the transition occurs. Importance of Moseley law The great significance ofMoseley law lies in the fact that it proves for the first time that it is the atomic number and not the atomicweight of an elementwhichdetermines its characteristic properties (bothphysical and chemical). It provides the proper guideline that elementsmust be arranged intheperiodic table according to their atomic numbers and not their atomicweights. Accordingly,Moseleylawhas been used to place elements in their proper sequence in the periodic table in certain questionable cases. For example, ifwe go by the atomic weight, potassium(19K39) should come before argon (18A40) and similarly, nickel (28Ni58.7) should precede cobalt (27Co58.9). ButMoselylawdictates that as per their atomic numbers, their order should be just opposite of the above. This fact is further supported bythe chemical properties of these elements. Moseleylawhas led to thediscoveryof newelements like hafnium(72), promethium(61), technetium(43) andrhenium(75) etc. bytheindicationofgaps inMoselydiagram.This lawhas beenalso helpfulindetermining the atomic number of rare earths therebyfixing their position in the periodic table. It can be shown thatMoseleylawis in accordancewithBohr�s theory of spectral emissionfromatoms.As shown inwhenan electron jumps froman orbit n2 to the orbit n1, the frequencyof the radiationgiven out is, v = 4 2 3 0
me 8. h .Z2 2 2 1 2 1 1 n n . . . . . . . Thismaybe put asv = 4 2 3 2 2 0 1 2 me 1 1 8 h n n .. .. .. .. .. . .. .. . .. .. Z2 or v . Z2 or . . Z Bohr did not take into account the screening effect of electrons whereasMoseley did. That is why the expression becomes . . (Z � b) Formulae for K- and L-series of X-ray Spectrum The frequencies of the various lines in the K- and L-series of the X-ray line spectrumare given by the following empiricalformulae. K-series: The general formula is 1. = R(Z � 1)2 2 1 1 n . . . . . . . where n = 2, 3 etc Here, the nuclear screening constant is unity.
ATOMIC PHYSICS www.physicsashok.in 71 (i) For K. line, n = 2. . 1. = R(Z � 1)2 1 14 . . . .. .. = 3R4 (Z � 1)2 (ii)For K. line, n = 3 . 1. = R(Z � 1)2 1 19 . . . .. .. = 8R9 (Z � 1)2 L-Series: The general formula is 1. = R(Z � 7.4)2 2 1 1 4 n . . . . . . . where n = 3, 4 etc. Here, screening constant is 7.4. (i) For H. line, n = 3 . 1. = R(Z � 7.4)2 1 1 4 9 . . . . . . . = 5R 36 (Z � 7.4)2 (ii)For H. line, n = 4 1. = R(Z � 7.4)2 1 1 4 16 . . . . . . . = 3R 16 (Z � 7.4)2 C47: Find the nuclear screening constant for the L-series ofX-rays if it is known that X-rayswith awavelength of . = 1.43Åare emittedwhen an electron in a tungsten atom(Z = 74) is transferred fromtheM-level to L-level. Take Rydberg constant = 10.97 × 106m�1. Sol:When electron jumps fromM to L-level, the first member of the L-series i.e. L. line is given out. Its wavelength as given byMoseley�s lawis 1. = R(Z � b)2 2 2 1 1 2 3
. . . . .
. . = 5R 36 (Z � b)2 Substituting the givenvalues,we have 1010 1.43 = 10.97 × 106 × 5 36 (74 � b)2 . (74 � b)2 = 4589.8; (74 � b) = 67.75 ; b = 6.25. Example 49: The K. and L. absorption edges of copper occur at wavelengths 1.380 Å and 11.288 Å respectively. Calculate the atomic number of copper. Sol: It shouldbe remembered that absorptionages are found inthe absorptionspectrumofX-rays.The absorption edge ofeachseries (ofline spectrum) represents the limit ofthat series. Inotherwords, the short-wavelength limit of each series is called its absorption edge and iswritten as ... It maybe obtained by putting n= . in the formulae.Moreover, forK-series, the value of screening constant forK.member is 3.3 (instead of1 for othermembers). Corresponding value for L-series is 11(instead of 7.4). K-Series: 1. . = R(Z � 3.3)2 2 .1. 1 . . . . . . = R(Z � 3.3)2 . 1010 1.38 = (Z � 3.3)2 L-Series: 1. . = R(Z � 11)2 2 1 1 4 . . . . . . . . = R4 (Z � 11)2
ATOMIC PHYSICS www.physicsashok.in 72 . 1010 11.288 = R4 (Z � 11)2 Dividing one bythe other, we get 2 2 (Z 3.3) (Z 11) .. = 11.288 4.1.38 = 2.045 or (Z 3.3) (Z 11) .. = 1.43 or Z = 29 Example 50:An impure tungsten target emits a strongK. line of . = 0.21Åand aweakK. line of . = 1.537 Å. Can you identify the impuritytaking the nuclear screening constant as unity. Given for tungsten,Z= 74, for Ni = 28. Sol: Thewavelength for K. line is given bythe relation 1. = R(Z � 1)2 1 14 . . . . . . . = 34 R(Z � 1)2 For tungsten 10 1 0.21.10. = 34 R(74 � 1)2 = 15,987R 4 For impurity 10 1 1.537.10. = 34 R(Z � 1)2 Dividing one bythe other, we get 1.537 0.21 = 2 5329 (Z.1) ; Z = 28 Obviously, impurity is nickelwhose atomicmass number is 28. Absorption of X-rays When a narrowandmonochromatic beamofX-rays passes throughmatter, part of it is a
bsorbed and the remaining part is transmitted.Absorption of x-rays can be studiedwith the help of the apparatus shown in figure. TheX-rays produced by anX-raytube are first made into awell defined narrowbeamby passing themthrough two fine slits S1 and S2 in the two lead plates. The beamis thenmonochromatised byBragg reflectionfroma crystal(not showninthe figure) and allowedto enter the ionizationchamberwhichmeasures the ionization current.The strength of the ionization current is ameasure of the intensityof theX-ray. Next, a sheet of the absorbing material is interposed in the path of the X-ray beambefore it enters the ionization chamber. It is found that the ionization current and hence the intensity ofX-rays is reduced by their passage through the absorber sheet. The ratio 0 I I of the two ionization currents can be used to measure the absorptioncoefficient of thematerial.
ATOMIC PHYSICS www.physicsashok.in 73 L1 L2 S1 S2 Absorber Ionization Chamber Let I0 be the initial intensityof a homogeneousX-raybeamincident normallyon an absorber sheet and I the intensity after the beamhas travelled a thickness x of the absorber. If dI is the further decrease in intensityover a thickness dx of the absorber figure, then dI dx gives the rate of decrease of intensity with thickness. Assuming that this rate is proportionalto the intensityI,we have I0 I (I � dI) x dx � dI dx . I or dI dx = �.I where . is a constant of proportionality and is called the linear absorption coefficient of the absorber (it is also known asmacroscopic absorption coefficient or linear attenuation coefficient). Now, dI I = �..dx .....(i) Integrating both sides of the above equation, we get . dI / I = �. . dx . logeI = �.x + K .....(ii) The value ofthe integration constant Kcan be found fromthe known initial conditionswhichare that when z = 0, I = I0. Substituting these values in equation (ii) above, we have I0 x I O logeI0 = K Hence, equation (ii) becomes logeI = �.x + logeI0 or e 0 log I I = �.x or 0 I I = e�.x or I = I0e�.x .....(iii) It is seen that intensity of the X-ray beamdecreases exponentiallywith the thickness of the absorbing material as shown in figure. BRAGG�S LAW . .
A P S Q E F M B G H N R Figure gives a 3-dimensional view of how a beam of monochromaticX-rays undergoesBragg�s reflection from different planes in a NaCl crystal. Figure gives a
ATOMIC PHYSICS www.physicsashok.in 74 2-dimensional viewof the same diagram. It shows a beam ofmonochromaticX-rays incident at a glancing angle . on a set of parallelplanes ofNaCl crystal. The beamis partically reflected at the successive layers rich in atoms Ray no.1 is reflected fromatomAinplane1whereas rayno. 2 is reflected from atom B lying plane 2 immediately below atom A. Whether two reflected rays will be in phase or antiphase with each other will depend on their path difference. This pathdifference can be found bydrawing perpendicularsAMandANon rayNo. 2. Since the two rays travel the same distance frompointsAandNonwards, it is obvious that ray no. 2 travels an extra distance = MB + BN Hence, the path difference between the two reflected beams is = MB + BN M. . N . A . 1 2 3 1 2dd Plane 1 Plane 2 Plane 3 B = d sin. + d sin. = 2d sin. where d is the interplanar spacing i.e. vertical distance between two adjacent planes belonging to the same set. The two reflected beamswillbe in phasewith eachother if this pathdifference equals an integralmultiple of la dnwill be antiphase if it equals an oddmultiple of ./2. Hence, the conditionfor producingmaxima becomes 2d sin. = n. where n= 1, 2, 3 etc., for the first order, second order and third ordermaxima respectively.This equation is known as Bragg�s Law. PROPERTIES OF X-RAYS Main properties ofX-raysmay be summarised as under: (i) Likevisible light,X-rays consist ofelectromagneticwavesofveryshortwavelength(or ofveryhighfrequency) and showreflection, refraction, interference, diffraction and polarisation etc. (ii) They are not deflected by electric andmagnetic field. (iii) They posses high penetrating power and can pass throughmanysolidswhich are opaque to visible light. The transparency depends on the density of the material. Higher the density of the substance, the less transparent it is to theX-rays. For example, sheet of lead 1 cmthick can absorbX-rayswhereas aluminium sheet ofsame thickness cannot. The penetrating power ofX-rays depends upon (a) the voltage applied across the cathode and anode of theX-raytube and (b) the atomic number of thematerialof the cathode.Greater the accelerating potential of the X-ray tube and higher the atomic number of its target material, the more penetrating the X-rays
produced. (iv) They ionize a gas and also eject electrons frommetals onwhich they fall. (v) Theycause fluorescence inmanysubstances like barium, cadmium, tungstate and zinc sulphide etc. (vi) Theysuffer compton scattering. (vii) They have a destructive effect on living tissue. Exposure of humanbodyofX-rays causes the reddening of skin and surface sores. Practical Applications of X-rays On account of their diverse and distinctive properties,X-rays have beenput tomanyuses in different fields of our dailylife i.e. in industry,medicine and research. (a) Industrial applications: Some of these applications are as under:
ATOMIC PHYSICS ATOMIC PHYSICS (i) to detect and photograph defects within a body i.e. in its internal structure such as metals, machine parts intended for withstanding highpressures, cracks in wood, porcelainand other insulators, defectsindiamonds and other precious stones, in moulds, forgings and castings etc. (ii) toanalysesthestructureofalloysandotherrcompositebodiesbydeterminingthecrystalforminaningot with the help of diffraction of X-rays. In this way, alloys like cobalt-nickel, steel, bronze, duraluminium, artificial pearls and old paintings have been analysed. (iii) to study the structure of rubber, cellulose and plastics. The diffraction of X-rays bythese substances leads to valuable information about their molecular grouping. (b) Applications for pure scientific research: (i) for investigating the structure of the atom. (ii) for studying the structure of the crytalline solids and alloys (X-ray crystallography). (iii) for identification ofchemical elements including determination oftheir atomic numbers. (iv) for analysing the structure of coraplex organic molecules by examining their X-ray diffraction patterns. (c) Medical Applications: These can be broadly divided into two classes; one for diagnosis purposes (radiography) and the other for curative purposes (X-ray therapy). (i) Radiography: X-rays are being widely used for detecting fractures, tumours, the presence of foreign matter like bullets etc. in the human body as well as diseased organs of the boy. It is due to differential absorption of X-rays between bones, tissues and metals. Radiographs or X-ray photos are used for this purpose. Since bones are more dense and hence more opaque to X-rays then flesh, a contrasting radiograph ofhuman body can be obtained for leisurely study by interposing it between the X-ray tube and photo film. Where the organs do not provide contrast as, for example, the intestines or other fleshy parts of the human body, a artificial means are adopted to create sufficient contrast between them. In such cases, before taking radiograph, barium or bismuth meal is given to the patient. This meal consists of milk to which some amount of barium sulphate or bismuth carbonate has been added.Afew hours after the meal has been taken, these powders settle in the gastrointestinal tract and if a radiograph is taken at that item, the intestines stand out in sharp contrast to the surrounding tissues due to the fact that absorptioncoefficient ofbarium is greater on account ofitshigh atomic number. In this way, pepticulcers and ruptures etc. in the internal organs ofthe human body can be accuratelylocated. Similarly, radiographs are routinely used for the diagnosis of tuberculosis, stones in kidneys and gall-bladders etc. (ii) X-ray or Rontgen Therapy Many types of skin diseases, malignant sores, cancers and tumours have been cured bycontrolled exposure to X-rays of suitable quality. This curative power of X-rays is due to the fortu
nate fact that diseased tissue is more susceptible to destructionthan the surrounding healthytissue. Unnecessarylong exposure of human body to X-rays produces many injurious effects including the loss of white cells in the blood, sterility and harmfulgenetic changes. X-rays have been used for the identification of different types of cells and tissues and for bringing about genetic mutations.
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MODERN PHYSICS www.physicsashok.in 123 THINKING PROBLEMS X-rays 1. What leads you do believe that X-rays are electromagneticwaves ? 2. The occurrence ofa lower bound ofwavelengths ofX-rays produced inanX-raytube lends support to the quantumconcept ofradiation. Explain how. 3. IfYoung�s experiment is repeatedwith electron beams interference is observed. Does thismean that an electron gets divided into twowhile passing through the slits ? 4. Water irradiatedwithX-rays is unsafe for drinking. Is this true or false ? 5. An electronmoves through a gas-filled region in the presence of a transversemagnetic field. Describe its motion. 6. Fluerescence is produced by ultraviolet rays but never by infrared rays. Explainwhy. 7. Aneutron, a proton, an electron and an alpha particle enter a region of constant magnetic fieldwith equal velocities. Themagnetic field is along the inward normal to the place of the paper.The tracks of the particles are labelled in the figure. Which tracks do the electron and alpha particle follow? 8. Achargedandanunchargedparticlehavethe samemomentum.Willtheyhave the samedeBrogliewavelength? 9. The electricalconductivityofa gas increaseswhenX-rays or .-rays pass throughit.Explainthis phenomenon. 10. Why are tungstenor platinumwidely used as the target theX-raytubes ? 11. Auniformelectric field acts normally on amoving charge. Can the charge be deflected through 90º ? 12. Auniformelectric field acts normally on amoving charge. Iswork done by the field on the charge (a) as it enters the field, (b) later ? 13. Does the speed of a charged particle changewhen (a) a magnetic field, (b) an electric field, acts on it for some time ? 14. Parallel electric and magnetic fields act on on a charged particle moving perpendicular to these fields. Describe its subsequentmotion. 15. Howdo you conclude that cathode rays are fast moving negativelycharged particles ? 16. �Magneticmirror�is a termfor the region of amagnetic field inwhich there is an intense concentration oflines ofinductionas shown inthe figure. Suppose a charged particle approaches amagneticmirror.What will happen to it ? 17. Howis amonoenergetic, slightlydiverging beamof charged particles focussed byamagnetic field ? 18. X-rays are producedwhen a fast electron hits a proper target.What happens to the electron ? 19. Why does the target in anX-ray tube become hot ? 20. X-rays can be produced in cathode ray tubes and also in Coolidge tubes.Why are the latter preferred in actual use ? 21. Why is thewave nature ofmatter not apparent in our daily lives ? 22. Aneutralpion decays into two gamma photons. .0 .... Why cannot a single photon be born?What conservation lawis in contradictionwith it ?
23. Quarks inside protons and neutrons are thought to carryfractional charges 2 e, 1 e 3 3 . ..... . .. ....Whyare theynot evidenced inMillikan�s oildrop experiment ?
MODERN PHYSICS www.physicsashok.in 124 Atomic Structure, Radioactivity, Nuclear Fassion and Fusion 1. Distinguish between�excitation� and �ionization� bycollision. 2. According to the theory of electron transitions, the spectral lines froma glowing gas should be �sharp� i.e., of one particularwavelength each. In practice, theyare found to be somewhat �diffuse�, i.e., spread over a small range ofwavelengths. Suggest a reason for this. 3. Can it be concluded from.-decay that electrons exist inside the nucleus ? 4. Why are .-rays emitted only in nuclear processes and not in orbital electron transitions ? 5. What are the principles that are obeyed in filling the orbits of an atom? 6. Howis the radioactivityof an element affectedwhen it forms chemical compounds ? 7. How can Becquerel ray, i.e., the combination of .-, .- and .-rays, be separated ? 8. When a nucleus undergoes .-dcay, is the product atomelectrically neutral ? In .-decay? 9. Do .-decay and .-decay cause a change of element, called transmutation ? 10. Experimental results inradioactivityshowsmallvariations fromthe results predicted bytheory. Explain this. 11. Does the relationE =mc2 suggest that mass can be converted to energy onlywhen it is inmotion ? 12. What is a �thermalneutron� ? 13. Does a nucleus have to be bombardedwith fast or showneutrons in order for it to undergo fission ? 14. Why has it not been possible so far to control the fusion process and obtain usable energy fromit ? 15. An atomhas a continuous distribution ofmass in..... (Thomsonmodel, Rutherfordmodel) but has a highly non-uniformdistributionin ..... (Thomsonmodel,Rutherfordmodel). 16. Which level of the doubly-ionized lithium(Li++)ion has the same energy as the ground state energyof the hydrogen atom? 17. If the a-decayofU238 is allowed fromthe point of viewof energy (the decayproducts have a totalmass less than themass ofU238) what preventsU238 fromdecaying all at once ?Why is its half-life so large ? 18. Can a spectral line belong to both the Lyman andBalmer series ? 19. Although theLymanseries involves transitions to the ground level, andtheBalmer series to the second orbit, the latter was discovered earlier.Why ? 20. Bohr�s principle of quantization of angularmomentumis not a postulate but an essentialcondition. Explain how. 21. Whyare .-particle tracks as observed in a cloud chambermuch shorter than .-particle tracks though they emerge froma radioactive samplewith almost the same speed ? 22. Cathode rays and .-particles are streams of electrons. Inwhat respect do theythen differ fromeach other? 23. When a radioactive substance emits an .-particle its position in the periodic table is lowered bytwo places. Is this true or false ? 24. When Rutherford bombarded a thin foil of gold by .-particles, he found that 1 in 2500 are deflected through verylarge angles.What inference did he drawfromthis result ? 25. Auraniumnucleus (atomic number 92, mass number 238) emits an .-particle and the resultant nucleus
emits a .-particle.What are the atomic andmass numbers of the final nucleus ? 26. Among .-particles, .-particles, protons and neutronswhich have the greatest penetrating power through matter andwhy ? 27. The isotope of hydrogen 31 H (tritium) is radioactive.What would be its decay process and the product? 28. What ismeant by the disposal of radioactivewaste in a nuclear reactor ? 29. .- and .-particles suffer equal and opposite deflections in an external electric field. Is this true or false? 30. It requires infinite time for all the atoms in a radioactive sample to decay,whatever be the half-life of the material. Is this true or false ? 31. What ismean by �enrichment of uranium� ? 32. Which yields greater energy per atom�fission or fussion ? .... per unitmass ? 33. Explain the statement �Themoderator ina nuclear reactor thermalizes the neutrons.�
MODERN PHYSICS www.physicsashok.in 125 34. When is a chain reaction said to be critical ? 35. If a nucleus emits only a .-ray photon, does itsmass number change?Does itsmass change ? 36. Aclassical atombased on ..... (Thomsonmodel, Rutherfordmodel) is doomed to collapse.Why ? 37. What is Bohr�s correspondence principle ? 38. Bohr�s quantization principle, i.e., angularmomentum= nh/2. is a basic lawin nature.Why dowe never speak of quantizationof the angularmomentumof a planet, around the sun ? 39. Aphoton is emitted by a dense star. Scientists say there is a change in the frequency of the photon as it moves awayfromthe star and call the difference infrequcney gravitational shift. Can you explain this? SOLUTION OF THINKING PROBLEMS X-rays 1. X-rays cannot be deflected by electric andmagnetic fields. They are reflected, refracted, diffracted like ordinarylight waves.All these facts lead us to believe that X-rays are electromagneticwaves. 2. The quantumtheory predicts a lower bound (hv =Ve) ofwavelengths ofX-rays produced in anX-rays tube. This is found to be in good agreement withexperimentalobservations. So this supports the quantum theoryof light. 3. No. Infact electrons have associatedwaves.At the accelerating voltage of the experiment thewave behaviour of electrons becomes quite prominent and interference fringes are observed. 4. False, X-rays damage living cells and hence kill the bacteria present inwater. Hence, the water actually becomes safer for drinking. 5. The electronloses energydue to collisionswith the gas atoms. It therefore describes a circlewith decreasing radius, i.e., it spirals inwards. 6. The phenomenon offluorescence consists of absorptionof higher energyphotons and re-radiation of lower energy visible light. This is possiblewith ultraviolet rays as these have greater photon energy than visible light. Infrared rays have lower photon energy, so fluorescence cannot occur. 7. D, B. Alpha particles are heavy and positively charged, and so they are deflected the least to the left according to Fleming�s left-hand rule. The electrons are deflected to the right. 8. Yes. The de Brogliewavelength does not depend on charge, onlyonmomentum. 9. X-rays or .-rays cause ionization bycollision inthe gas atoms. The free electron and ionpairs produced can nowmove and conduct electricity. Thus, the gas becomesmore conducting. 10. Theyhave large atomic numbers and highmelting points. 11. No.The electric fieldwillnot affect the initialvelocity. Itwillonlyproduce an additional acceleratedmotion perpendicular to the initial velociyt.A90º deflection requires that the initial velocity be reduced to zero. 12. (a) No work is done, since the displacement due to the motion is perpendicular to the field. (b)Work iis done, as the displacement howhas a component parallel to the force. 13. (a) The speed does not change since the force is perpendicular to the displacement at everypoint. The field does no work on the particle. The energy and speed of the particle remain unchanged.
(b)When an electric field acts for some time, the particlewill acquire a displacement parallel to the force. Work is done by the field on the particle. The energyand speed of the particlewill change. 14. Due to themagnetic field, the particlewillmove in a circlewith the field as axis.Due to the electric field, a forcewill act along this axis, producing an acceleratedmotion perpendicular to the plane of the circle. The result willbe a helicalpath ofgradually increasing pitch. 15. Their negativelycharged character is shownbythe directionofdeflection inan electric field. Their streaming character is shown by deflection in amagnetic field because a magnetic field can produce deflection only when charged particles are inmotion.
MODERN PHYSICS www.physicsashok.in 126 16. The charged particlewillmove along a helixwinding around the lines of induction. Let us resolve the velocityalong the field and perpendicular to it. The resolved part along the field vII is called orbital velocity. Since the magneticmoment due to the orbitalmotion is opposite to themagnetic field is tends to push the charged particle out of the field, i.e., the charged particle is strongly decelerated and so its drift velocitydecreases, becoming zero inthe case of a sufficientlyhigh field gradient. Fromthis place it begings tomove in the opposite direction. 17. The charged particles followa helicalpath of periodT = 2.m/Bqwhichis independent of the velocityof the particles and pitch p = (2.m/Bq) v cos.. = (2.mv)/Bq when . small. Thus after a path of length p all the particles come down to the same point whatever be the angle of inclination of their initialmotionwith the field. 18. It is absorbed bythe target,which is also the anode of theX-raytube. Subsequently, the electronreturns to the cathode via the external voltage circuit. 19. Less than 1%of the incident electronenergy is actuallyconverted toX-rays. The balance is lost in inelastic collisions be between the electrons and the target atoms. This energy appears as heat in the target. 20. In aCoolide tube, the hardness and intensity of theX-rays can be controlled independently. The hardness is controlled by the applied voltage. The intensity is controlled by the filament temperature, i.e., by the filament current. Such independent control is not possible in the cathode ray tube. 21. Thewavelengthof amatterwave is given by. =h/p.Themomentumofordinarymaterialbodies at ordinary speeds is verylarge and so the associatedwavelength is extremelysmallbecause of the verysmallvalue of h = 6.6 × 10�34 Js. This iswhy thewave nature ofmatter is not apparent in our daily lives. 22. Asingle photoncannot be bornbecause the principle ofconservation ofmomentumwould then be violated. The meson is at rest and so if a single photon is created, that photon must also be at rest to conserve momentum. But it is not possible for a photon to be at rest. This iswhy two photons are bornwhichmove in opposite directions after creation, in conformitywith the principle of conservation ofmomentum. 23. Because they are held together by a strong force, they are not exhibited separately. Atomic Structure, Radioactivity, Nuclear Fission and Fusion 1. When amoving electron collideswith an atom, one orbital electron in the atomabsorbs part or allof the kinetic energyof the incident electron.As a result, the orbital electronmaymove to an outer orbit, of higher energy (excitation), or become completelyfree fromthe attractive field ofthe nucleus (ionization). 2. The gas atoms emitting light due to electron transitions are inmotion. They behave like fixed frequency sources inmotion.Due toDoppler effect, the observed frequency(and hencewavelength) becomes different fromthe emitted frequency.This difference depends onthe velocityofthe atom, caus
ing the small spread of the spectral line. 3. No. The .-particle, although an electron, is actuallycreated at the instant of .-decay and ejected at once. It cannot exist inside the nucleus as its de Broglie wavelength ismuch larger than the dimensions of the nucleus. 4. The energyof a .-rayphotonis of the order ofMeV. Energies of thismagnitude occur innuclear processes but not inorbital electron transitions. 5. The universalprinciple of stability ofa system, that is, �a systemlies instable equilibriumwhen its energy is at the lowest possibelvalue�and Pauli�s exclusion principle, i.e., �no two electrons canhave all their quantum numbers identical�. 6. In no way. Chemical bonds involve onlyorbital electrons, whereas radioactivity is a nuclear process. 7. Bypassing themthrough transverse electric ormagnetic fields. 8. No. In .-decay, the atomic number decreases by2, hence the atomis left with two extra orbital electrons. It therefore has a double negative charge. In .-decay, the atomis left with a net single positive charge.
MODERN PHYSICS www.physicsashok.in 127 9. Yes. In .-decay, the element moves back two places on the periodic table. In .-decay, the element moves forward one place on the periodic table. 10. The lawof radioactive decay is statistical in nature. Hence, individial experimental resultswillshowslight variations.The averages over a large number of experimental results conformexactlywith the theory. 11. No. Here c2 appears only as a constant and does not suggest motion. 12. This is a neutronwithenergyof the order of(3/2) kT,whereTis the absolute temperature of the surroundings and k is Boltzmann�s constant. This follows fromcomparisonwith the law of equipartition of energy as applied to gasmolecules. �Thermalisation� of a neutron brings down its energy froma high value of about (3/2) kT. 13. For fission, theneutronmust be absorbed bythe fissionable nucleus.This is possible onlywith slowneutrons. 14. Fusion occurs onlyat temperatures of the order of 106K. Thismakes it extremely difficult to control fusion processes. 15. Thomsonmodel,Rutherfordmodel. 16. E . Z2 / n2 .When E is constant, Z2 . n2 . Z.n . 3/1 = n/1 . n = 3 17. The emission of a-particles is caused by quantummechanical tunnelling through the repulsive Coulomb barrier. They bounce to and fro in the potentialwell bounded by the barrier before tunnelling through it. Hence the probability of escape is not the same for all the a-particles because all are not born inside the nucleus at the same time. 18. Spectral lines fromhydrogen arise fromthe relation hv = E0(1/n2 � 1/m2) withm> n. For the Lyman series n = 1, m= 2, ...., . . . (hv)max = E0 and (hv)min = (3/4) E0 or 34 E0/h < v < E0/h For the Balmer series, n = 2, m= 3, ....., . . . (hv)max = E0/4 and (hv)min = 5E0 /36 or 5E0/36h < v < E0/4h. Clearly, the same value of v cannot satisfy both the series. Hence, a spectral line cannot belong to both series. 19. Because theBalmer series lies inthe visible region and the Lyman series in the ultraviolet region. 20. Bohr�s principle of quantizationof angularmomentumis seen to be an essential conditionwhile considering de Broglie�smatterwave principle, that is, innature allmoving bodies have an associatedwave.We have p = h/.. In the stationary orbits thewaves associatedwiththe particlemust forma stationarywave. If r is the radius ofthe stationaryorbit, its circumference (equalto 2.r)must be inmultiples ofwavelength..Therefore, 2.r = n. or 2.r = n . h/p or pr = nh/2..Moment ofmomentumis angular momentum. Hence pr is the angularmomentum(L) of the electron. Thus finallywe find that I.= nh/2.. 21. .-particles have greater ionizing power than .-particles and so they lose their energymuch earlier than
.-particles because of collisions. This iswhy their track lengths are shorter than those of .-particles. 22. Theydiffer in respect of their origin. .-particles originate in the nucleuswhen a neutron is converted into a proton, whereas electrons in cathode rays are orbital electrons. 23. Yes. Since .-particles are heliumatoms, their emissionlowers the atomic number by2 andmass number by 4. Elements are arranged in the periodic table according to their atomic number. So emittion of .-particles lowers the position of the element by two places in the periodic table. 24. An atomconsists of a smallcentralizedmass containing positivelycharged particles.Otherwise the atomas awhole in empty. 25. 92 and 234.92U238 � 2He4 � 2 × �1e0 . 92U234
MODERN PHYSICS www.physicsashok.in 128 26. Neutrons, because theyare electricallyneutral and so they do not interactwithmatter electrically. 27. The only possible decay process is .-decay. The decayproduct would be 2He3. 28. In nuclear fission, two nuclideswith Z of the order of 40 to 50 are created. These are highly radioactive, with half-lives or thousands of years. These are called �radioactivewaste�. They have to be disposed of in sealed containerswhich can contain their radioactive emissions. 29. False. .-particles are deflectedmore due to their larger specific charge. 30. True.This follows fromthe exponentialnature ofthe decay. In N = N0e�.t, for N = 0, t = . 31. Naturaluraniumcontains less than1%ofU235mixedwithU238. The latter is not fissionable,while the former is fissionable. The proportion ofU235must be increased artificiallyfor the uraniumto be used in fission. This is called enrichment of uranium. 32. Fission yields greater energyper atom. Fusion yields greater energyper unit mass. 33. The neutrons emitted in fissionmust be slowed down inorder to cause further fission in other nuclei.This is called thermalization, and is performed bythemoderator. 34. When exactly one neutron, of the several produced by the fission of one nucleus, is permitted to cause further fission. This happens in a controlled chain reaction, e.g., ina nuclear reactor. 35. Themass number does not change. Themass is reduced. 36. Thomson�smodel. 37. For large quantumnumbers, quantummechanical results reduce to classical results. 38. Because large n corresponds to a verylarge value at which classical and quantumresults are identical by Bohr�s correspondence principle. 39. As the photonmoves out against strong gravitational attraction its energydecreases and so its frequencyis expected to decrease.This is gravitational shift.
MODERN PHYSICS www.physicsashok.in 129 ASSERTION-REASON (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statment-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True 1. Statement-1 : Electron capture occursmore often than positron emission in heavy elements. Statement-2 : Heavy elements exhibit radioactivity. 2. Statement-1 : In a hydrogen atomenergy of emitted photon corresponding to transition fromn = 2 to n = 1 ismuch greater as compared to transition fromn = . to n = 2. Statement-2 :Wavelengthof photon is directlyproportional to the energyof emitted photon 3. Statement-1 : Ionisation energyof atomic hydrogen is greater than atomic deuterium. Statement-2 : Ionisation energy is directly proportional to reducedmass 4. Statement-1 : For pair production, energy of . ray is greater than 1.02MeV. Statement-2 : In pair production, energy is converted into mass. 5. Statement-1 : The ratio of rate of production (R) of neutrons to the rate of leakage of neutron froma spherical body of 92U235 is directlyproportional to radius (r) Statement-2 : Rate of production of neutron is directly proportional to volume but rate of leakage of neutrons is directlyproportional to area. 6. Statement-1 : The nuclear energycan be obtained bythe nuclear fissionof heavier nuclei aswell as nuclear fusion oflighter nuclei. Statement-2 : The binding energy per nucleonwith increase in atomic number first increases and then decreases. 7. Statement-1 : Asmallmetal ball is suspended in a uniformelectric fieldwith an insulated thread. If high energyX-ray beamfalls on the ball, the ballwill be deflected in the direction of electric field. Statement-2 :Wavelength of L. X-raymust be greater then the wavelength ofK. X-ray for the same material. 8. Statement-1 : The difference in the frequencies ofseries limit ofLymanseries andBalmer series is equal to the frequency of the first line of the Lyman series. Statement-2 :Difference inenergyof two atomic levels is proportionalto the energyofemitted or absorbed photon. 9. Statement-1 :Work function of aluminiumis 4.2 eV. If two photons of eachof energy 2.5 eVstrike on an electron ofaluminium, the electron is not emitted. Statement-2 : In photoelectric effect, electron is emitted onlyif energyofeach of incident photonis greater than thework function. 10. Statement-1 : Ifthe acceleratingpotentialinanX-raytube is increased, thewavelengthsofthe characteristic X-rays do not change. Statement-2 :When an electron beamstrikes the target in anX-ray tube, part of the kinetic energy is converted intoX-rayenergy. [JEE, 07]
MODERN PHYSICS www.physicsashok.in 130 Match the Column 1. Column I Column II (A) Work function of copper is 4 eV. If two (P) �13.6 Z2/n2eV photons each of energy 2.5 eV strike an electron of copper emission of electrons (B) Cathode rays get deflected by (Q) 13.6 Z2 2 2 1 2 1 1 n n . . . . . . . eV (C) Ionisation energy of H like atom is (R) both electric and magnetic field (S) 1H1 (D) Greater wavelength in transition from n = 2 (T) 1H3 to n = 1 is for (U) Not possible (V) Possible 2. Column I Column II (A) Radius of orbit depend on principal quantum (P) Increase number as (B) Due to orbital motion of electron, (Q) decrease Magnetic field arises at the centre of Nucleus is proportional to principal quantum no. as (R) is proportional to 1/n2 (C) If electron is going from lower energy level to higher energy level then velocity of electron will (S) is proportional to n2 (D) If electron is going from lower energy level to higher energy level, then total energy of (T) is proportional to 1/n5 electron will 3. Column I Column II (A) Rate of disintegration, i.e. dN/dt is (P) Greater than Half life proportional to (B) Mean life of radioactive substance is (Q) Less than Half life (C) Intensity I of . ray of initial intensity I0 after (R) Number of atoms of parent radioactive transversing the thickness of x of the absorber is substance still undecayed at time t (related as) (D) The radioactive decay rate is not affected by (S) I = I0/x (T) I = I0e�.x (U) Temperature, pressure, volume 4. Column I Column II (A) Binding energy per nucleon for middle order (P) Optical Model or elements is (B) Nuclear force depends on (Q) Shell model (C) For nuclear fission Z2 A is (R) 8.8 MeV (D) Magic numbers are 2, 8, 20, 28, 50, 82, 126 (S) 2.5 eV explained by (T) Charges of Nucleons (U) Spin of Nucleons (V) Greater than 15 (W) Less than 15 5. Column I Column II (A) Radius of orbit is related with atomic number (Z) (P) is proportional to Z
(B) Current associated due to orbital motion (Q) is inversely proportion to Z electron with atomic number (Z) (C) Magnetic field at the centre due to orbital (R) is proportional to Z2 motion of electron related with Z. (D) Velocity of an electron related with atomic (S) is proportional to Z3 number (Z)
MODERN PHYSICS www.physicsashok.in 131 6. Match the followingColumns [JEE, 06] Column I Column II (A) Nuclear fusion (P) Converts somematter into energy (B) Nuclear fission (Q) Generallyoccurs for nucleiwith lowatomic number (C) .-decay (R) Generallyoccurs for nucleiwith higher atomic number (D) Exothermic nuclear reaction (S) Essentially proceeds byweek nuclear forces 7. In the following, column I lists some physical quantities&the column II gives approx, energy values associated with some of them. Choose the appropriate value of energy fromcolumn II for each of the physical quantities in column I andwrite the corresponding letterA, B, Cetc. against the number (i), (ii), (iii), etc. of the physical quantityin the answer book. In your answer, the sequence of column I should be maintained. Column I Column II [JEE, 97] (A) Energyofthermalneutrons (P) 0.025 eV (B) EnergyofX-rays (Q) 0.5 eV (C) Binding energyper nucleon (R) 3 eV (D) Photoelectric threshold ofmetal (S) 20 eV (T) 10 keV (U) 8MeV 8. Some laws/processes are given inColumnI.Match thesewiththe physicalphenomena giveninColumn II. Column I Column II (A) Transition between two atomic energylevels (P) CharacteristicX-rays [JEE, 07] (B) electron emissionfromamaterial (Q) Photoelectric effect (C) Mosley�s law (R) Hydrogen spectrum (D) Change of photonenergyinto kinetic (S) .-decay energyof electrons. 9. Column- II gives certainsystemsundergoing aprocess.Column-I suggests changes insomeofthe parameters related to the system.Match the statements inColumn I to the appropriate process(es) fromColumn-II. Column I Column II [JEE, 09] (A) The energy ofthe systemis increased (P) System: Acapacitor, initiallyuncharged (B) Mechanical energyis provided to the system, Process : It is connected to a battery which is converted into energy of random (Q) System: Agas inan adiabatic container fitted motion ofits parts with an adiabatic piston (C) Internal energyof the systemis converted into Process :The gas is compressed by pushing itsmechanical energy. the piston (D) Mass of the systemis decreased (R) System: Agas in a girid container Process :The gas gets cooled due to colder atmosphere surrounding it (S) System: Aheavynucleus, initiallyat rest Process :The nucleus fissions into two fragments of nearly equalmasses and some neutrons are emitted. (T) System:Aresistivewire loop Process :The loop is placed in a time varrying magnetic fieldperpendicular to its plane
MODERN PHYSICS www.physicsashok.in 132 Level � 1 1. An energy of 24.6 eV is required to remove one of the electrons from the neutral helium atom. The energy (in eV) required to remove both the electron from a neutral helium atom is (a) 38.2 (b) 49.2 (c) 51.8 (d) 79.0 2. The K X . ray . emission line of tungsten occurs at . = 0.021 nm. The energy difference between K and L levels in this atom is about (a) 0.51 MeV (b) 1.2 MeV (c) 59 keV (c) 136 eV 3. The electron in a hydrogen atom makes a transition n1 .n2 where n1 and n2 are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is 8 times that in the final state. The possible values of n1 and n2 are (a) n1 = 4, n2 = 1 (b) n1 = 8, n2 = 2 (c) n1 = 8, n2 = 1 (d) n1 = 6, n2 = 3 4. X-rays are produced in an X-ray tube operating at a given accelerating voltage. The wavelength of the continuous X-rays has values from (a) 0 to . (b) min . to . where min . > 0 (c) 0 to max . where max . < . (d) min . to max . where 0 < min . < max . < . 5. A particle of mass M at rest decays into two particles of masses m1 and m2, having non zero velocities. The ratio of the de Broglie wavelengths of the particles, 1 2 . . , is (a) 21 mm (b) 12 mm (c) 1.0 (d) 12 mm 6. The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true ? (a) Its kinetic energy increases and its potential and total energies decrease. (b) Its kinetic energy decreases, potential energy increases and its total energy remains the same. (c) Its kinetic and total energies decrease and its potential energy increases. (d) Its kinetic, potential and total energies decrease. 7. Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength . (given in terms of the Rydberg constant R for the hydrogen atom) equal to (a) 9/5R (b) 36/5R (c) 18/5R (d) 4/R 8. Electrons with energy 80 keV are incident on the tungsten target of an X-ray
tube. K shell electrons of tungsten have � 72.5 keV energy. X-rays emitted by the tube contain (a) a continuous X-ray spectrum (Bresmsstrahlung) with a minimum wavelength of about 0.155A. (b) a continuous X-ray spectrum (Bremsstrahlung) with all wave-lengths (c) the characteristic X-ray spectrum of tungsten (d) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of about 0.155A.and the characteristics X-rays spectrum of tungsten. 9. For a photoelectric cell, the graph in Figure. showing the variation of the cut-off voltage V0 with frequency (v) of incident light is O v V0 (a) O v V0 (b) O v V0 (c) O v V0 (d)
MODERN PHYSICS www.physicsashok.in 133 10. Monochromatic light of frequency v1 irradiates a photocell and the stopping potential is found to be V1. What is the new stopping potential of the cell if it is irradiated by monochromatic light of frequency v2? (a) 1 .v2 v1. eV . h . (b) . . 1 v2 v1 eV . h . (c) . . 1 v1 v2 eV . h . (d) . . 1 v1 v2 eV . h . 11. When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for photoelectric current is V0/2. When the same surface is illuminated by monochromatic high of frequency v/2, the stopping potential is V0. The threshold frequency for photoelectric emission is (a) 32v (b) 23v (c) 53v (d) 35v 12. The energy of a photon of frequency v is E = hv and the momentum of a photon of wavelength . is p = h / . . From this statement one may conclude that the wave velocity of light is equal to (a) 3 x 108 ms�1 (b) .. (c) .p (d) 2 p. .. . . .. . . 13. When a centimeter thick surface is illuminated with light of wavelength . , the stopping potential is V. When the same surface is illuminated by light of wavelength 2 . , the stopping potential is V/3. The threshold wavelength for the surface is (a) 3 4 . (b) 4 . (c) 6 . (d) 3 8 . 14. A star of mass M0, radius R0 contracts to radius R. Energy radiated by the star assuming uniform density in each case while temperature remains unchanged is (a) .0 c. (b) . . .. . . .
.
. ..
.
. ..
.
. . . 2 0 0 Rc 1 R (c) . .. . . .. . . . . 0 0 Rc 1 R (d) . . .. . . .. . .. . . .. . . . . 3 0 0 Rc 1 R 15. A sensor is exposed for time t to a lamp of power P placed at a distance . . The sensor has an opening that is 4d in diameter. Assuming all energy of the lamp is given off as light, the number of photons entering the sensor if the wavelength of light is . is (l >> d) (a) 2 2 hc N P d t . . . (b) 22 hc N 4P d t . . . (c) 2 2 4hc N P d t . . . (d) 2 2 16hc N P d t . .
. 16. If elements of quantum number greater than n were not allowed, the number of possible elements in nature would have been (a) n .n 1. 21 . (b) . . 2 2 n n 1 . . . . . .. (c) n.n 1. .2n 1. 61 . . (d) n.n 1. .2n 1. 31 . . 17. An electron is lying initially in the n = 4 excited state. the electron de-excites itself to go to n = 1 state directly emitting a photon of frequency v41 . If the same electron first de-excites to n = 3 state by emitting a photon of frequency v43 and then goes from n = 3 to n = 1 state by emitting a photon of frequency v31 , then (a) 41 43 31 . . . . . (b) 41 43 31 . . . . . (c) .41 . .43 . 2.31 (d) Data Insufficient 18. A photon of energy 10.2 eV corresponds to light of wavelength 0 . . Due to an electron transition from n = 2 to n = 1 in a hydrogen atom, light of wavelength . is emitted. If we take into account the recoil of the atom when the photon is emitted, (a) . = 0 . (b) 0 . . . (c) 0 . . . (d) the data is not sufficient to reach a conclusion.
MODERN PHYSICS www.physicsashok.in 134 19. When an electron moving at a high speed strikes a metal surface, which of the following are possible? (a) The entire energy of the electron may be converted into an X-ray photon. (b) Any fraction of the energy of the electron may be converted into an X-ray photon. (c) The entire energy of the electron may get converted to heat. (d) The electron may undergo elastic collision with the metal surface. 20. A star converts all of its 2He4 nuclei completely into oxygen nuclei. The energy released per oxygen nuclei is (given mass of nucleus = 4.0026 amu, mass of oxygen nucleus = 15.994 amu) (a) 7.26 MeV (b) 7 MeV (c) 15.252 MeV (d) 23.9 MeV 21. The graph showing the energy spectrum of . particles is : (a) E n(E)Y X (b) E n(E)Y X (c) E n(E)Y X (d) E n(E) 22. Binding energy per nucleon of 1H2 and 2He4 are 1.1 eV and 7.0 MeV respectively. Energy released in the process 1H2 + 1H2 = 2He4 is : (a) 20.8 MeV (b) 16.6 MeV (c) 25.2 MeV (d) 23.6 MeV 23. Two electrons are moving with the same speed V. One electron enters a region of uniform electric field while the other enters a region of uniform magnetic field . After sometime if the de Broglie wavelength of the two are .1 and .2 then : (a) .1 = .2 (b) .1 > .2 (c) .1 > .2 (d) .1 > .2 or .1 < .2 24. In a characteristic X-ray spectra of some atom superimposed on a continuous X-ray spectra : P Q .min . Relation intensity (a) P represents K. line (b) Q represents K. line (c) Q and P represents K. and K. lines respectively (d) Position of K. and K. depend on the particular atom 25. Difference between nth and (n + 1)th Bohr�s radius of �H� atom is equal to its (n �1)th Bohr�s radius. The value of n is (a) 1 (b) 2 (c) 3 (d) 4 26. A hydrogen atom is in an excited state of principal quantum number n. It emits a photon of wavelength . while returning to the ground state. The value of n is :
(a) .R(.R .1) (b) ( R 1) R . . . (c) R R 1 . . . (d) (R 1) . . 27. The binding energies of nuclei X and Y are E1 and E2 respectively. Two atoms of X fuse to give one atom of Y and an energy Q is released. Then : (a) Q = 2E1 � E2 (b) Q = E2 � E1 (c) Q < 2E1 � E2 (d) Q > E2 � 2E1 28. Two radioactive materials x1 and x2 have decay constant 10 . and . respectively. Initially they have the same number of nuclei. The ratio of the number of nuclei x1 to that of x2 will be 1/e after time : (a) 1/10. (b) 1/11. (c) 11/10. (d) 1/9.
MODERN PHYSICS www.physicsashok.in 135 29. At t = 0 activity of radioactive substance is 1600 Bq and t = 8 sec activity remains 100 Bq. The activity at t = 2 sec is : (a) 200 Bq (b) 400 Bq (c) 600 Bq (d) 800 Bq 30. There is a stream of neutrons with a kinetic energy of 0.0327 eV. If the half life of neutrons is 700 s. The fraction of neutrons will decay before they travel a distance 10 m. mn = 1.675 × 10�27 kg : (a) 3.96 × 10�5 (b) 3.96 × 10�6 (c) 2.96 × 10�4 (d) None 31. The count rate for 10 gram radioactive material was measured time (hrs) at different times and this had been shown in figure scale given. The half life of material and the total count in the first half value period respectively are : (a) 4 hrs. and 9000 approximately (b) 3 hrs and 14100 approximately (c) 3 hrs and 235 approximately 75 50 25 3 6 9 100 12.5 (d) 10 hrs and 157 approximately time in hr. 32. Assuming that about 200 MeV energy is released per fission of 92U235 nuclei. What would be mass of 92U235 consumed per day in the fission of reactor of power 1 MW approximately ? (a) 10 kg (b) 100 kg (c) 1 gram (d) 10�2 gm 33. The energy, the magnitude of linear momentum and orbital radius of an electron in a hydrogen atom corresponding to the quantum number n are E, P and r, according to the Bohr�s theory of hydrogen atom: (a) EPr is proportional to 1/n (b) P/E is proportional to n0 (c) Er is not constant for all orbits (d) Pr is proportional to n. 34. An electron is excited from a lower energy state to a higher energy state in a hydrogen atom. Which of the following quantities decrease in the excitation. (a) Potential enrgy (b) Angular speed (c) Kinetic energy (d) Angular momentum 35. The correct statement is l are : (a) density of nucleus is independent of mass number (A) (b) Radius of nucleus increases with mass number (A) (c) Mass of nucleus is directly proportional to mass number (A). (d) Density of nucleus is directly proportional to mass number. 36. A hydrogen like atom of atomic number Z is an excited state of quantum number 2n. It can emit a maximum energy photon of 204. eV. It makes a transition to quantum state n, a photon of energy 40.8 eV is emitted, then (a) Z = 2 (b) Z = 4 (c) n = 1 (d) n = 2 37. An electron with kinetic energy varying from 5 eV to 50 eV is incident on a hydrogen atom in its ground state. The collision : (a) may be elastic (b) may be partially elastic (c) must be completely inelastic (d) from zero to 13.6 eV be elastic and more than 27.2 eV be inelastic. 38. The wavelength of first Balmer line for 1H1 , 1H2 and 2He4 and .1, .2 and .3
respectively. The correct statements are (a) .2 ...1 (b) .3 ...2 (c) .1 ...2 (d) .1 ...2 > .3 39. Regarding X ray spectrum which of the following statement (a) The characteristic X ray spectrum is emitted due to excitation of inner electrons of atom (b) Wavelength of characteristic spectrum depend on the potential difference across the tube. (c) Wavelength of continuous spectrum is dependent on the potential difference across tube (d) None of these
MODERN PHYSICS www.physicsashok.in 136 40. If the wavelength of light in an experiment on photo electric effect is doubled : (a) The photoelectric emission will not take place. (b) The photoemission may or may not take place. (c) The stopping potential will increase (d) The stopping potential will decrease under the condition that energy of photon of doubled.Wavelength is more than work function of metal. 41. the binding energy per nucleon is : (a) Maximum for middle order element (b) Minimum for lighter elements (c) Binding energy per nucleon suddenly increases for some mass number called magic numbers. (d) Binding energy per nucleon is minimum for middle order elements 42. When Z is doubled in an atom, which of the following statement are consistent with Bohr�s theory : (a) Energy of a state is double. (b) Radius of an orbit is double. (c) Velocity of electrons in an orbit is doubled (d) Radius of orbit is halved. 43. Photons of wavelength 6620 Å are incident normally on a perfectly reflecting screen. Calculate the number of photons per second falling on the screen as total power of photons such that the exerted force is 1N : (a) 5 × 1026 (b) 5 × 1025 (c) 1.5 × 108 (d) None of these 44. The energy of . particles emitted by 210Po is 5.3 MeV. What mass 210Po is needed to power a thermoelectric cell of 13 watt output, What would be power output after 1 year : (The half life of 210Po is 138 days) (a) 8.85 × 10�2 gram (b) 0.159 watt (c) 0.179 watt (d) 8.85 × 10�4 gram 45. The atomic masses of 7N15, 8O15 and 8O16 are respectively 15.0001 a.m.u., 15.0030 a.m.u. and 15.9949 a.m.u. Then : (a) Binding energy per nucleon in 8O16 is 7.97 MeV (b) Energy is needed to remove one proton 8O16 is 12.13 MeV (c) Energy needed to remove one proton from 8O16 is 15.61 MeV. (d) All the above 46. A sample contains 10�2 kg each of two substances A and B with half lives 4 sec and 8 sec respectively. Their weights are in the ratio of 1 : 2. The amounts of a and B after an interval of 16 sec. (a) 6.25 × 10�4 kg (b) 12.5 × 10�4 kg (c) 2.5 × 10�3 kg (d) 1.25 × 10�5 kg 47. The wavelength and frequency of photons in transition 1,2 and 3 for H like atom are .1, .2, .3 and .1, .2, .3. then : 123 CBA .1 .2 .3 (a) .3 = .1 + .2 (b) .3 = .1 + .2 (c) 1 2 3 1 2 . . . .
. . . (d) 1 2 3 1 2 . . . . . . . 48. Which of the following pair constitute very similar radiations ? (a) Hard U.V. ray and soft X ray. (b) Soft U.V. ray and hard X ray (c) Very hard X ray and low frequency Y ray (d) Soft X ray and Y ray 49. The correct option are : (a) In uranium ore, the ratio of U235 to U238 is 1 : 40 (b) Critical mass of uranium is 10 kg (c) 92U235 : 92U238 = 1 : 4 (d) All the above
MODERN PHYSICS www.physicsashok.in 137 50. A star initially has 1040 deutrons. It produces energy via, the processes 1H2 + 1H2 ... 1H3 + p & 1H2 + 1H3 ... 2He4 + n. If the average power radiated bythe star is 1016W, the deuteron supplyof the star is exhausted in a time of the order of : [JEE, 93] (A) 106 sec (B) 108 sec (C) 1012 sec (D) 1016 sec 51(i). Fast neutrons can easily be slowed down by : (A) the use of lead shielding (B) passing themthroughwater (C) elastic collisionswith heavynuclei (D) applying a strong electric field (ii). Consider .�particles, .�particles & .-rays, each having an energy of 0.5 MeV. Increasing order of penetrating powers, the radiations are : [JEE, 94] (A) ., ., . (B) ., ., . (C) ., ., . (D) ., ., . 52. Which ofthe following statement(s) is (are) correct ? [JEE, 94] (A) The rest mass of a stable nucleus is less than the sumof the rest masses of its separated nucleons. (B) The rest mass of a stable nucleus is greater than the sumof the rest masses of its separated nucleons. (C) Innuclear fusion, energy is released byfusion two nuclei ofmediummass (approximately100 amu). (D) In nuclear fission, energyis released by fragmentation of a very heavy nucleus. 53. The binding energy per nucleon of 16O is 7.97 MeV & that of 17O is 7.75 MeV. The energy in MeV required to remove a neutron from17Ois : [JEE, 95] (A) 3.52 (B) 3.64 (C) 4.23 (D) 7.86 54. Themaximumkinetic energyofphotoelectrons emitted froma surfacewhenphotons of energy6 eVfallon it is 4 eV. the stopping potential inVolts is : [JEE, 97] (A) 2 (B) 4 (C) 6 (D) 10 55. Select the correct alternative(s). [JEE, 98] (i) Let mp be themass of a proton,mn themass of a neutron,M1 themass of a 20 10 Ne nucleus&M2 themass of a 40 20Ca nucleus. Then : (A) M2 = 2M1 (B) M2 > 2M1 (C) M2 < 2M1 (D) M1 < 10(mn + mp) (ii) The half-life of 131I is 8 days. Given a sample of 131I at time t = 0, we can assert that : (A) no nucleuswill decay before t = 4 days (B) no nucleuswill decay before t = 8 days (C) all nucleiwill decay before t = 16 days (D) a given nucleusmaydecayat anytime after t = 0 56(a). Binding energyper nucleon vs.mass number curve for nuclei is shownin the figure.W,X,Yand Zare four nuclei indicated on the curve. The process that would release energyis [JEE, 99] 30 60 90 120 Z Y X W 8.5 8.0 7.5 5.0 Mass Number of Nuclei in MeV
Binding Energy/nucleon (A)Y. 2Z (B)W. X + Z (C) W. 2Y (D) X .Y+ Z
MODERN PHYSICS www.physicsashok.in 138 (b) Order ofmagnitude of density ofUraniumnucleus is, [mp = 1.67 × 10�27 kg] (A) 1020 kg/m3 (B) 1017 kg/m3 (C) 1014 kg/m3 (D) 1011 kg/m3 (c) 22Ne nucleus, after absorbing energy, decays into two .-particles and an unknown nucleus.The unknown nucleus is (A) nitrogen (B) carbon (C) boron (D) oxygen (d) Which of the following is a correct statement ? (A) Beta rays are same as cathode rays (B) Gamma rays are high energyneutrons. (C)Alpha particles are singlyionized heliumatoms (D) Protons and neutrons have exactlythe samemass (E)None (e) The half-life period of a radioactive elementXis same as themean-life time ofanother radioactive element Y. Initiallyboth of themhave the same number ofatoms.Then (A) X&Yhave the same decay rate initially (B) X &Ydecay at the same rate always. (C)Ywilldecay at a faster rate thanX (D) X willdecay at a faster rate thanY 57. Aparticle ofmassMat rest decays into two particles ofmassesm1 andm2, having non-zero velocities. The radio of the de-Brogliewavelengths of the particles, .1/.2, is [JEE, 99] (A)m1/m2 (B)m2/m1 (C) 1.0 (D) 2 1 m / m 58.(a) Imagine an atommade up of a proton and a hypotherical particle of double themass of the electron but having the same charge as the electron.Apply theBohr atommodel and consider allpossible transitions of this hypotheticalparticle to the first excited level. The longest wavelength photon that willbe emitted has wavelength ..(given in terms ofthe Rydberg constant R for the hydrogen atom) equal to [JEE, 2000 Scr.] (A) 9/(5R) (B) 36/(5R) (C) 18/(5R) (D) 4/R (b). The electronin a hydrogen atommakes a transition froman excited state to the ground state.Whichof the following statements is true ? [JEE, 2000 Scr.] (A) Its kinetic energy increases and its potential and total energies decrease. (B) Its kinetic energydecreases, potential energyincreases and its total energyremains the same. (C) Its kinetic and total energies decrease and its potential energy increases. (D) Its kinetic, potential and total energies decrease. 59. The potentialdifference applied to anX-ray tube is 5 kV and the current through it is 3.2mA. Then the number of electrons striking the target per second is [JEE, 02 Scr.] (A) 2 × 1016 (B) 5 × 1016 (C) 1 × 1017 (D) 4 × 1015 60. A Hydrogen atom and Li++ ion are both in the second excited state. If lH and lLi are their respective electronic angularmomenta, and EH and ELi their respective energies, then [JEE, 02 Scr.] (A) lH > lLi and |EH| > |ELi| (B) lH = lLi and |EH| < |ELi| (C) lH = lLi and |EH| > |ELi| (D) lH < lLi and |EH| < |ELi| 61. Givena sample ofRadium-226 having half-life of 4 days. Find the probability, a nucleus disintegrateswithin 2 half lives. [JEE, 06] (A) 1 (B) 1/2 (C) 3/4 (D) 1/4
MODERN PHYSICS www.physicsashok.in 139 62. The graph between 1/. and stopping potential (V) of three metals havingwork functions .1, .2 and .3 is anexperiment of photoelectric effect is plotted as shown in the figure. Which of the following statement(s) is/are correct? [Here . is thewavelength of the incident ray]. V 0.001 0.002 0.004 1/ nm�1 metal 1 metal 2 metal 3 (A) Ratio ofwork functions .1 : .2 : .3 = 1 : 2 : 4 (B) Ratio ofwork functions .1 : .2 : .3 = 4 : 2 : 1 (C) tan ..is directly proportional to hc/e, where h is Planck�s constant and c is the speed of light (D) The violet colour light can eject photoelectrons frommetals 2 and 3. [JEE, 06] 63. In the options given below, let E denote the rest mass energy of a nucleus and n a neutron. The correct option is : [JEE, 07] (A) .236 . .137 . .97 . 92 53 39 E U . E I . E Y . 2E(n) (B) .236 . .137 . .97 . 92 53 39 E U . E I . E Y . 2E(n) (C) . 236 . .140 . .94 . 92 56 36 E U . E Ba . E Kr . 2E(n) (D) . 236 . .140 . .94 . 92 56 36 E U . E Ba . E Kr . 2E(n) 64. The largestwavelengthinthe ultraviolet regionofthe hydrogenspectrumis 122 nm.The smallestwavelength in the infrared region of the hydrogen spectrum(to the nearest integer) is [JEE, 07] (A) 802 nm (B) 823 nm (C) 1882 nm (D) 1648 nm 65. Electronswith de-Brogliewavelength . fallon the target in anX-ray tube. The cut-offwavelength of the emittedX-rays is [JEE, 07] (A) 2 0 2mc h. . . (B) 0 2h mc . . (C) 2 2 2 0 2 2m c h . . . (D) .0 = . 66. Which one of the following statements isWRONGin the context ofX-rays generated fromaX-raytube? (A)Wavelength of characteristicX-rays decreaseswhen the atomic number ofthe target increases (B) Cut-offwavelengthof the continuousX-rays depends on the atomic number of the target. [JEE, 08] (C) Intensityof the characteristicX-rays depends on the electricalpower given to theX-rays tube. (D) Cut-offwavelengthof the continuousX-rays depends on the energyof the electrons intheX-rays tube.
67. Assume that the nuclear binding energyper nucleon(B/A) versusmass number (A) is as shown inthe figure.Use this plot to choose the correct choice(s) given below: Figure 100 200 A 02468 B/A (A) Fusionof two nucleiwithmass numbers lying in the range of 1 <A< 50 will release energy. (B) Fusionof two nucleiwithmass numbers lying inthe range of 51 <A< 100 will release energy. (C) Fission of a nucleus lying inthemass range of100 <A< 200will release energywhen broken into two equalfragments. (D) Fission ofa nucleus lying inthemass range of 200 <A< 260will release energywhenbroken into two equalfragments. [JEE, 08] 68. The quantumnumber n of the state finally populated inHe+ ions is [JEE, 08] (A) 2 (B) 3 (C) 4 (D) 5
MODERN PHYSICS www.physicsashok.in 140 69. Thewavelength oflight emitted in the visible region byHe+ ions after collisionswithHatoms is [JEE, 08] (A) 6.5 × 10�7 m (B) 5.6 × 10�7 m (C) 4.8 × 10�7 m (D) 4.0 × 10�7 m 70. The ratio of the kinetic energy of the n = 2 electron for theHatomto that ofHe+ ion is [JEE, 08] (A) 14 (B) 12 (C) 1 (D) 2 71. Aradioactive sampel S1 having an activity of 5µCi has twice the number of nuclei as another sample S2 which has an activity of 10 µCi. The half lives of S1 and S2 can be : [JEE, 08] (A) 20 years and 5 years, respectively (B) 20 years and 10 years, respectively (C) 10 years each (D) 5 years each 72. Photoelectric effect experiments are performed using three different metal plates p, q and r havingwork functions .p = 2.0 eV, .q = 2.5 eVand .r = 3.0 eV, respectively.Alight beamcontaining wavelengths of 550 nm, 450 nmand 350 nmwith equal intensities illuminates each of the plates. The correct I�Vgraph for the experiment is [JEE, 09] (A) I pqrV (B) I p q r V (C) I pqr V (D) I r q p V Passage PASSAGE : 1 Figure 1 below depicts three hypothetical atoms. Energy levels are represented as horizontal segments. The distance between the segments is representative of the energy difference between the various levels . All possible transitions between energy levels are indicated by arrows Atom #1 Atom # 2 Atom # 3 Scientists can observe the spectral lines of atoms that are dominant in far-away galaxies. Due to the speed at which these galaxies are travelling, these lines are shifted, but their pattern remains the same. This allows researchers to use the spectral pattern to determine which atoms they are seeing
. Table 1 below shows spectroscopic measurements made by researchers trying to determine the atomic makeup of a particular faraway galaxy. Light energy is not measured directly, but rather is determined from measuring the frequency of light. Which is proportional to the energy.
MODERN PHYSICS www.physicsashok.in 141 Table 1 Frequencies Measured 868440 880570 879910 856390 1. For each of three hypothetical atoms (Atom 1, Atom 2 and Atom 3), Figure 1 depicts the (A) number of electrons and the amount of energy the atom contains (B) distance an electron travels from one part of the atom to another (C) energy released by the atom as an electron as it moves from one energy state to another (D) frequency with which the atom�s electrons move from one energy state to another 2. In which of the three hypothetical atoms depicted in Figure 1 does the energy of the light released by the atom vary the least (A) Atom 1 (B) Atom 2 (C) Atom 3 (D) It is impossible to tell 3. Scientist observing an actual atom similar to hypothetical Atom 1 in the figure might observe � (A) three spectral lines close together and two other spectral lines close together (B) light blinking at six different frequencies (C) a much brighter light emanating from one electron than from any other. (D) four distinct spectral lines emanating from six different electrons. 4. Based on the spectroscopic measurements shown in Table 1, which of the atoms in Figure 1 (Atom 1, Atom2, or Atom 3) is most similar to the one the scientists were observing, and why ? (A) Atom 2, because it contains four different energy levels (B) Atom 3, because it contains four different energy levels (C) Atom 1, because the frequencies listed in Table 1 indicate a high level of atomic activity. (D) Atom 3, because there is a comparatively small difference between exactly two of the four frequencies listed in Table 1 5. The laws of atomic physics prohibit electron movements between certain energy states. In atomic physics. these prohibitions are called �forbidden transitions.� Based on Figure 1, which of the following is most accurate ? (A) Atom 2 has the same number of forbidden transitions than Atom 3 (B) Atom 2 has more forbidden transitions than atom 3 (C) Atom 3 has the same number of forbidden transitions as Atom 1 (D) Atom 1 has fewer forbidden transitions than Atom 2 PASSAGE : 2 In quantum mechanics, some quantities are discrete and cannot be continuous. One of these quantities is the energy. Energy can only take certain values � E1, E2, E3, E4......., which are called energy levels. The energy cannot take any values between E1 and E2, or E2 and E3 or E3 and E4 etc. Certain transitions from one energy level to another result in the emission of a photon of radiation, whereas others can only take place if a photon is absorbed. The energy levels in a newly discovered gas are expressed as:
2 1 n 2 E E z n . . in which �E1z2 is the ground state energy. Take z = 1 for simplicity, but do not assume that the gas is hydrogen. An experiment is designed to measure the energy as a functions of the level. The results obtained are as follows : n En(eV) 2 �144 3 �64 4 �36 6. The ionization energy of the gas must be (A) 244 eV (B) 576 eV (C) 144 eV (D) +13.6 eV 7. The ground state energy is � (A) �144 eV (B) + 144 eV (C) �244 eV (D) none of the above
PASSAGE : 3 The power per unit area reaching the Earth�s surface from the sun, averaged over 24 hours, is 0.2 kW/m2. This solar energy can be converted directly into electrical energy via the photoelectric effect. For example, in the photoelectrical cathode (emitter) incidentlight Anode (collector) photocell A Ammeter cell shown in figure 1, a cathode emits electrons when PASSAGE : 3 The power per unit area reaching the Earth�s surface from the sun, averaged over 24 hours, is 0.2 kW/m2. This solar energy can be converted directly into electrical energy via the photoelectric effect. For example, in the photoelectrical cathode (emitter) incidentlight Anode (collector) photocell A Ammeter cell shown in figure 1, a cathode emits electrons when kinetic energy of Emax will not be able to reach the anode therefore the current will stop altogether. The value of this stopping voltage is dependent only on Emax . 11. The most efficient modern photovoltaic cells can covert the Sun�s energy into electrical energy with an efficiency of 35%. Approximately what area would have to be covered by such cells in order to supply a household with 20 kW-hourse of electrical energy per day � (A) 0.5 m2 (B) 12 m2 (C) 285 m2 (D) 6850 m2 12. Light intensity is defined as the energy flowing per unit area per unit time for an area perpendicular to the direction of energy flow. In an experiment, the frequency of light incident on the cathode of a MODERN PHYSICS 8. Which of the following shapes is most likely to represent the graph of En versus 1/n2 ? (A) (B) (C) (D) 9. A transition from the n = 2 state to the n = 3 state results : (A) in emission of a photon of energy 144 eV (B) in emission of a photon of energy 80 eV (C) in emission of an ultraviolet photon 10. A transition from the n = 4 state to the n = 3 state results : (A) in emission of a photon of energy 28 eV (C) in emission of an infrared photon (D) only accomplished if a photon is absorbed (B) in emission of a photon of energy 13.6 eV (D) only accomplished if a photon is absorbed. electrons travel to the anode, and a small electric current flows. An electron w
ithin the cathode requires a minimum energy to break free from the cathode surface. This minimum energy is known as the work function, W, and is a constant intrinsic to the material of which the cathode is composed. An individual photonincident on the cathode collides with an electron and is absorbed, transferring all of its energy to the electron. The energy of each incident photon is given by Ep = hf, where f is the frequency of incident light and h is Planck�s constant. If Ep is less than W, then no electrons will be ejected from the cathode at all. The maximum kinetic energy Emax� of an electron liberated from the cathode is given by illuminated by light of a high enough frequency. The ejected
:E =E �W
max p
A voltage source can be connected across the photoelectric cell to oppose the current flow. At a critical applied voltage, called the stopping voltage, even an electron ejected from the cathode with a
photoelectric cell is held constant, but the intensity is varied. As the intensity of the incident light is increased, the stopping voltage. it increases.
(A) increases, because more electrons are ejected from the cathode as the number of photons striking (B) remains the same, because the energy supplied to one electron depends only on the energy of an individual photon. (C) increases, because the electrons in the cathode absorb more energy per unit time (D) remains the same, because each incident photon shares its energy between several electrons in the cathode. 13. The behavior of light is sometimes explained in terms of particles and sometimes in terms of waves. Which of the following CANNOT be explained by a theory that refers to light in terms of waves alone � (A) Current flow in a photoelectric cell can be stopped by reducing the intensity of the incident light while maintaining the same frequency. (B) An electron requires energy to escape from the surface of a photosensitive cathode. (C) Current flow in a photoelectric cell can be stopped by reducing the frequency of the incident light while maintaining the same intensity. (D) The angle through which light is refracted when it moves from one medium to another is a function of frequency, rather than intensity. www.physicsashok.in 142
MODERN PHYSICS www.physicsashok.in 143 14. Under which of the following conditions will the stopping voltage across a photoelectric cell be greatest- (A) The wavelength of the incident light is short, and the work function of the cathode material is low. (B) The wavelength of the incident light is short, and the work function of the cathode material is high. (C) The wavelength of incident light is long, and the work function of the cathode is low. (D) The wavelength of the incident light is long, and the work function of the cathode material is high. PASSAGE : 4 An x- ray tube consists of two metal electrodes, a heated filament cathode, and an anode containing the metal target sealed under high vaccum in a glass envelope. The heated filament in the cathode emits electrons which are accelerated by a high DC voltage and collide with the positive anode target. Two different types of x-ray spectra may be seen. The continuous or bremsstrahlung� spectrum that is always present is produced when the electron Kp Ka I .min . penetrates through the outer electron cloud and is abruptly accelerated by the large positive charge on the nucleus of a heavy atom. The production of x-rays increases with increasing atomic number but is typically no more than 1% efficient, the remaining energy appearing as heat in the target metal. The sharp line spectra that can be seen at higher voltages occur when the incident electrons eject an inner shell electron, such as n = 1 shell electron. The spectral line is produced when an electron, say from n = 2, fills the vacancy in the n = 1 shell, emitting an x-ray photon whose energy corresponds to the energy difference between the n = 2 and n = 1 shells. �The intensity of x-rays is proportional to the number of photons created. The photon energy E = hf = hc/. where h is Planck�s constant and c is the speed of light. Figure 1 is a sketch of intensity versus wavelength for a molybdenum target with an accelerating voltage of 35,000 V. 15. Figure shows that the continuous x-ray spectrum has a minimum (cut-off) wavelength. No shorter wavelengths are emitted from the tube. This minimum wavelength corresponds to : (A) the smallest number of emitted photons (B) the highest energy photons emitted (C) the type of cathode used (D) the type of anode material used. 16. The sharp K. peak in figure 1 corresponds to an electron transition from state n = 2 to n = 1, where K. corresponds to a transition from state n = 3 to n = 1. The K. line peak is higher than the K. because� (A) it is due to a higher atomic number target metal (B) K. is the more energetic transition (C) the K. transition is more probable than the K. transition, so its intensity is higher (D) the K. transition occurs in the valence shells instead of the inner shells 17. The current to the heated filament in the cathode is increased while the accelerating voltage is kept constant. This increased current increases the number of electrons striking the target increasing the overall intensity. What effect does this have on the minimum wavelength value ?
(A) The minimum value will move to shorter wavelength values. (B) It will move to longer wavelength values (C) There will no longer be a cutoff wavelength. (D) The minimum wavelength will remain the same. 18. If the accelerating voltage, V0, increased while keeping the filament current constant, the overall intensity will also increase. What effect will this have on the wavelength position where the two peaks are observed? (A) They will occur at the same wavelengths (B) The peaks occur at shorter wavelengths due to the higher wavelengths due to the higher energies available. (C) They may diappear because all energies may exceed those of the n = 3 to n = 1 transition (D) The K. occur at longer wavelengths but the K. occur at shorter wavelengths. 19. Because the efficiency of x-ray production increases with increasing atomic number, Z, it would seem that lead (Z = 82) would be a better target material than tungsten (Z = 74). However, tungsten (also used in ordinary light bulb filaments) is the most common target meal, whereas lead is never used. this is because- (A) lead has too many electrons (B) lead becomes radioactive (C) lead would melt, whereas tungsten has a very high melting point (D) lead will not get hot enough to produce and x-rays.
MODERN PHYSICS www.physicsashok.in 144 PASSAGE : 5 Student in a medical physics class are given the assignment of planning a nuclear medicine facility. They not only design the rooms and choose the major equipment they also will have to solve elementary problems dealing with treatment, doses, radiation protection and safety, and the general principles of physics of typical isotopes that might be used in diagnostic nuclear medicine. They are required to be familiar with concepts of half life, half-thickness for shielding and the decay schemes of representative isotopes. 20. The most common isotope used in diagnostic work is Technicium. It is furnished from a generator or �cow� in which the negative beta decay of Molybdenum -99 produces the desirablemetastable isotope of Technicium according to the following decay scheme ? 42Mo99 . ZTcA + �1e0 What are the atomic number , Z and mass number A, of the Tc isotope ? (A) 41, 99 (B) 42, 99 (C) 43, 98 (D) 43, 99 21. If the Mo99 has a physical half-life of 67 hours, about what fraction is left after 5.5 days ? (A) 0.10 (B) 0.25 (C) 0.40 (D) 0.45 22. This isotope of Technicium has a physical half life of 6 hours. When it is tagged onto a polyphosphate carrier used for a certain procedure, there is a biological half-life of 12 hours (for the biological excretion of the carrier). What is the �effective half life� in this case ? (A) 4.0 hours (B) 7.5 hours (C) 10 hours (D) 14 hours 23. The �cow� was milked at 8.00 A.M. At 2.00 PM the activity of the milked sample is measured by a technician and found to be 200 millicuries. What was the activity of the Technicium at 8.00 A.M. (A) 100 mCi (B) 150 mCi (C) 300 mCi (D) 400 mCi 24. Thallium -201 is used for myocardial prefusion studies of the heart. It decays by electron capture when the nucleus captures one of the atom�s own orbital electrons (converting a proton in the nucleus into an uncharged neutron), with the emission of gamma rays used for the imaging. What are the atomic number Z, the mass number, A, of the Mercury isotope produced in the decay of the Thallium -201 ? 81Tl201 + 1e0 . ZHgA + Is (A) 80, 201 (B) 80, 202 (C) 81, 201 (D) 81, 203 25. The Thallium -201 half-life is 74 hours. If the sample has an activity of 80 millicuries initially, what will be the activity after 9.25 days ? (A) 2.5 mCi (B) 5 mCi (C) 10 mCi (D) 20 mCi 26. One advantage of the Thallium isotope is the �low whole body absorbed dose per millicurie�, which for Tl-201 is 70 millirads/millicurie. If the recommended amount to be injected for a heart scan is 10 microcurie per kg of body mass, what would be the whole body dose in millirads for a 70 kg person ? (A) 34 mrad (B) 49 mrad (C) 72 mrad (D) 134 mrad 27. Another feature that makes Technicium a desirable isotope for diagnostic nuclear medicine use is that it is a �pure gamma emitter.� What is the meaning of the terminology �pure gamma emitt
er�? (A) The gamma radiation stays in the patient�s tissue while the electrons are detected. (B) Particles emitted cannot escape tissue while the gama radiation escapes (C) The isotope decay emits only penetrating gamma radiation that can escape from tissue and be detected. (D) The isotope decay emits energetic electrons that act like gamma rays. 28. A patient is injected with Technicium and is estimated to have received a whole body dose of 400 millirads. If the �Quality Factor� is 1 for gamma radiation and 3 for low energy neutrons, what was the doese received by the patient in rem units ? (A) 4 mrems (B) 400 mrems (C) 1200 mrems (D) 4000 mrems 29. The half-thickness of lead for the absorption of the gamma radiation from a particular isotope is 0.4 cm of lead. How many half thicknesses are necessary to reduce the radiation penetration to less than 1% and how thick would the lead be ? (A) 2 half - thickness, 1.2 cm (B) 4 half-thicknesses, 3.2 cm (C) 7 half-thicknesses, 2.8 cm (D) 11 half-thicknesses , 4.4 cm PASSAGE : 6 Medical researchers and technicians can track the characteristic radiation patterns emitted by certain inherently unstable isotopes as they spontaneously decay into other elements. The half life of a radioactive isotope is the amount of time necessary for one-half of the initial amount of its nuclei to decay. the decay curves of isotopes 39Y90 and 39Y91 are graphed below as functions of the ratio of N, the number nuclei remaining after a given period to N0, the initial number of nuclei.
MODERN PHYSICS www.physicsashok.in 145 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1 2 3 4 5 6 Time (days) N/N0 39Y90 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 30 60 90 120 150 180 Time (days) N/N0 39Y91 30. The half-life of 39Y90 is approximately : (A) 2.7 days (B) 5.4 days (C) 27 days (D) 48 days 31. What will the approximate ratio of 39Y90 to 39Y91 be after 2.7 days if the initial samples of the two isotopes contain equal numbers of nuclei ? (A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 10 : 1 32. When inhaled by humans, 39Y90 accumulates in the gastrointestinal tract, whereas 39Y91 accumulates in the bones. If the total amount of each isotope inhaled goes to the specified area, which of the following situations will exist three days after a patient inhales these substances, assuming none of the isotopes leave the assuming none of the isotopes leave the specified areas due to physiological factors ? (A) The amount of 39Y91 in the gastrointestinal tract will be approximately equal to the total amount inhaled. (B) The amount of 39Y90 in the bones will be approximately one-half of the total amount inhaled (C) The amount of 39Y90 in the gastrointestinal tract will be approximately one-half of the total amount inhaled (D) None of the 39Y91 inhaled will be left in the bones. 33. Approximately how many 39Y91 nuclei will exist after three half -lives have passed, if there are 1,000 nuclei to begin with ? (A) 50 (B) 125 (C) 250 (D) 500 34. Which of the following conclusions is / are supported by the information given in the passage ?
I. 39Y90 is less stable than 39Y91 II. Only one-quarter of the original amount of 39Y90 will remain after 116 days. III. 39Y90 and 39Y91 are both radioactive (A) I only (B) III only (C) I and II only (D) I and III only PASSAGE : 7 Scientists are working hard to develop nuclear fusion reactor. Nucleiof heavy hydrogen, 21 H , known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D�D reaction is 21 H + 21 H . 32He + n+ energy. In the core of fusion reactor, a gas of heavyhydrogen is fullyionized into deuteron nuclei and electrons. This collection of 21 H nuclei and electrons is known as plasma. The nuclei move randomlyin the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no materialwall can be used to confine the plasma. Special techniques are used which confine the plasma for a time t0 before the particles fly away fromthe core. Ifn is the density(number/volume) ofdeuterons, the product nt0 is calledLawsonnumber. In one of the criteria, a reactor is termed successful ifLawson number is greater than 5 × 1014s/cm3. It may be halpful to use the following : Boltzmann constant k = 8.6 × 10�5 eV/K; 2 0 e 4.. = 1.44 × 10�9 eVm. [JEE, 09]
MODERN PHYSICS MODERN PHYSICS 35. In the core of nuclear fusion reactor, the gas becomes plasma because of (A) strongnuclearforceactingbetweenthedeuterons (B) Coulomb force acting between the deuterons (C) Coulomb force acting between deuteron-electron pairs (D) the high temperature maintained inside the reactor core. 36. Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each withkinetic energy1.5 kT, whenthe separationbetweenthemis large enoughto neglect Coulomb potential energy.Also neglect any interaction from other particles in the core. The minimum temperature T required for them to reach a separation of 4 × 10�15 m is in the range. (A) 1.0 × 109 K < T < 2.0 × 109 K (B) 2.0 × 109 K < T < 3.0 × 109 K (C) 3.0 × 109 K < T < 4.0 × 109 K (D) 4.0 × 109 K < T < 5.0 × 109 K 37. Results of calculations for four different designs of a fusion reactor using D�D reaction are given below. Which of these is most promising based on Lawson criterian ? (A) deuteron density = 2.0 × 1012 cm�3, confinement time = 5.0 × 10�3 s (B) deuteron density = 8.0 × 1014 cm�3, confinement time = 9.0 × 10�1 s (C) deuteron density = 4.0 × 1023 cm�3, confinement time = 1.0 × 10�11 s (D) deuteron density = 1.0 × 1024 cm�3, confinement time = 4.0 × 10�12 s www.physicsashok.in 146
MODERN PHYSICS www.physicsashok.in 147 Level � 2 1. When a surface is irridated with light of . = 4950A., a photo current appears which vanishes if a retarding potential greater then 0.6V is applied across the photo tube. When a different source of light is used, it is found that the critical retarding potential is changed to 1.1 V. what is the work function of the surface and the wavelength of the second source ? If the photo-electrons (after emission from the source) are subjected to a magnetic field of 10 tesla what changes will be observed in the above two retarding potentials? 2. A particle of charge equal to that of an electron and mass 208 times the mass of the electron (called a mu-meson) moves in a circular orbit around a nucleus of charge +3e. (Take the mass of the nucleus to be infinite.) Assuming the Bohr model of the atom to be applicable to this system, (i) derive an expression for the radius of the nth Bohr-orbit, (ii) find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom, and (iii) find the wavelength of the radiation emitted when the .�meson jumps from the third orbit to the first orbit. (Rydberg�s constant = 1.097 x 107 m�1 ) 3. Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 975A.. How many different lines are possible in the resulting spectrum ? Calculate the longest wavelength amongst them. You may assume the ionization energy for hydrogen atom as 13.6 eV. 4. An X-ray tube works at a potential difference of 100,000 V. Only 0.1% of the energy of cathode rays is converted into X-ray radiation and heat is generated in the target at the rate of 120 calorie per second. What current does the tube pass and what is the energy and velocity of an electron when it reaches the target? 5. A laser emits a light pulse of duration . . 0.13 ms and energy E = 10 J. Find the mean pressure exerted by such a light pulse when it is focussed into a spot of diameter d = 10 .m on a surface perpendicular to the beam and possessing a reflection coefficient . = 0.5 6. A short light pulse of energy E = 7.5 J falls in the form of a narrow and almost parallel beam on a mirror plate whose reflection coefficient is . = 0.60. The angle of incidence is 30°. In terms of the corpuscular theory find the momentum transferred to the plate. 7. The binding energy of an electron in the ground state of He atom is equal to E0 = 24.6 eV. Find the energy required to remove both electrons from the atom. 8. A stationary He+ ion emitted a photon corresponding to the first line of the Lyman series. That photon liberated a photoelectron from a stationary hydrogen atom in the ground state. Find the velocity of the photoelectron.
9. Taking into account the motion of the nucleus of a hydrogen atom, find the expressions for the electron�s binding energy in the ground state and for the Rydberg constant. How much (in per cent) do the binding energy and the Rydberg constant, obtained without taking into account the motion of the nucleus, differ from the more accurate corresponding values of these quantities? 10. Calculate the separation between the particles of a system in the ground state, the corresponding binding energy, and the wavelength of the first line of the Lyman series, if such a system is (a) a mesonic hydrogen atom whose nucleus is a proton ( in a mesonic atom an electron is replaced by a meson whose charge is the same and mass is 207 that of an electron); (b) a positronium consisting of an electron and a positron revolving around their common centre of masses.
MODERN PHYSICS www.physicsashok.in 148 11. In accordance with classical electrodynamics an electron moving with an acceleration a loses its energy due to radiation as : 2 32 a 3c2e dt d . . . Estimate the time during which an electron moving in a hydrogen atom along a circular orbit of radius r = 50 pm would have fallen onto the nucleus. Assume a to be directed permanently towards the nucleus. 12. The Earth revolves round the Sun due to gravitational attraction. Suppose that the Sun and the Earth are point particles with their existing masses and that Bohr�s quantization rule for angular momentum is valid in the case of gravitation. (a) Calculate the minimum radius the Earth can have for its orbit. (b) What is the value of the principal quantum number n for the present radius? Mass of the Earth = 6.0 x 1024 kg, mass of the Sun = 2.0 x 1030 kg, Earth-Sun distance = 1.5 x 1011 m. 13. A parallel electron beam is obtained from the application of accelerating voltage V0 = 2 x 104 V. These electrons are sent travelling in the direction normal to an infinitely long straight copper wire of radius r0 = 10�8 m as shown in figure. The wire carries uniform positive charge with charge density . = 4.4 x 10�11 c/m, the distance of the electrons closest b L Electron Beam Screen approach to the wire if uncharged is represented by bmax = 10�4 m. The electrons after passing the charged copper wire land on the screen located at distance L = 0.3 cm ( L >.b) from the wire. At the beginning of the experiment the electron beam is confined within the normal distance to the wire of b ( collision parameter) (see Figure). (a) Determine electric field E due to the charged copper wire and sketch a graph of electric field E as a function of distance measured from the axis of the wire to inside as well as outside the wire. (b) Calculate the angle of deflection of electron beam which has the collision parameter b and the electron does not collide with the wire. If f . is a small angle between the direction of the original velocity of the electron and the direction of final velocity of the electron arriving at the screen. Calculate f . . (c) Show that two plane waves representing deflected beams in the upper and lower parts give interference pattern on the screen. Calculate the number of bright bands in the interfence pattern. 14. To stop the flow of photoelectrons produced by electromagnetic radiation incident on a certain metal, a
negative potential of 300 volts is required. If the photoelectric threshold of the metal is 1500A., what is the frequency of the incident radiation? Is this radiation visible? 15. A toy truck has dimensions as shown in Figure and its width normal to the plane of this paper is d. Sun rays are incident on it as shown in figure. If intensity of sun rays is . and all surfaces of truck are perfectly black, calculate tension in thread used to keep truck stationary. Neglect fraction. . Thread b a h 16. Amonochromatic beam of light ( . = 4900 A°) incident normally upon a surface produces a pressure of 5 x 10�7 N/m2 on it. Assuming that 25% of the light incident is reflected and the rest absorbed, find the number of photons falling per second on a unit area of thin surface. 17. A beam of light consists of four wavelength 4000A., 4800A., 6000A.and 7000A., each of intensity 1.5 x 10�3 wm�2. The beam falls normally on an area 10�4 m2 of a clean metallic surface of work function 1.9 eV.Assuming no loss of light energy calculate the number of photo electrons liberated per second. 18. A single electron orbits around a stationay nucleus of charge +Ze, where Z is a constant and e is the magnitude of the electric charge. It requires 47.2 eV to excite the electronfromthe second Bohrorbit tothe third Bohr orbit. Find :
MODERN PHYSICS www.physicsashok.in 149 (i) the value of Z (ii) the energy required to excite the electron from the third to the fourth Bohr orbit. (iii) The wavelength of the electromagnetic radiation required tomove the electron from. (iv) Thekinetic energy, the potentialenergy and the angular momentumof the electron in the first Bohr ortib. (v) the radius of the first Bohr orbit. 19. Radiations from hydrogenic C gas corresponding to 2nd line of Lyman series falls on a hydrogenic atom where rotating particle�s mass and charge are unknown. Nucleus contains one proton only. This atom is excited to 4th excited energy level. (a) Find a relation for valid values of mass and charge of the rotating particles. (b) Find the maximum K.E.of the electron ejected due to aforesaid radiation falling on a hydrogenic potassium having some of the atoms in the ground energy level & some in the 4th energy level. 20. Stopping potentials of 24, 100, 110 and 115 kV are measured for photoelectrons emitted from a certain element when it is radiated with monochromatic x-ray. If this element is used as a target in an x-ray tube, what will be the wavelength of K line? 21. Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photo-electrons emitted form sodius is 0.73 eV. The work fraction for sodium is 1.82 eV. Find : (i) the energy of the photons causing the photoelectric emission. (ii) the quantum numbers of the two levels involved in the emission of these photons. (iii) the charge in the angular momentum of the eletron in the hydrogen atom in the above transition. (iv) the recoil speed of the emitting atom assuming it to be at rest before the transition. (Ionization potential of hydrogen is 13.6 eV). 22. A gas of hydrogen like atoms can absorb radiations of 68 eV. Consequently, the atoms emit radiations of only three different wavelengths. All the wavelengths are equal or smaller than that of the absorbed photon. (a) Determine the initial state of the gas atoms. (b) Identify the gas atoms. (c) Find the minimum wavelength of the emitted radiation. (d) Find the ionization energy and the respective wavelength for the gas atoms. 23. According to the Thomson model, a helium atom consists of a cloud of positive charge, within which two electrons sit at equlibrium positions. Assume that the positive cloud has a charge +2e uniformly distributed over the volume of a sphere of radius 0.50 A.. (a) Find the equilibrium position of the two electrons. Assume that the electrons are symmetrically placed with respect to the centre. (b) What is the frequency of small radial oscillations of the electrons about th
eir equilibrium positions. Assume that the electrons move symmetrically with identical amplitudes. 24. The Rydberg constants for hydrogen and singly ionized helium are R1 and R2 respectively. Find the ratio of the mass of the proton to that of the electron in terms of R1, R2 and R. . 25. A sample contains 10�2 kg each of two substances A and B with half lives 4 second and 8 second respectively. Their weights are in the ratio of 1.2. Find the amounts of A and B after an interval of 16 second. 26. A sample of uranium is a mixture of three isotopes 92U234, 92U235 and 92U238 present in the ratio of 0.006%, 0.71% and 99.284% respectively. The half-lives of these isotopes are 2.5 x 105 years, 7.1 x 108 years and 4.5 x 109 years respectively. Calculate the contribution to activity (in %) of each isotope in this sample.
MODERN PHYSICS www.physicsashok.in 150 27. Polonium ( 84Po210 ) emits 2He4 particles and is converted into lead (82Pb206). This reaction is used for producing electric power in a space mission. Po210 has half life of 138.6 days. Assuming an efficiency of 10% for the thermoelectric machine, howmuch Po210 is required to produce 1.2 x 107 J of electric energy per day at the end of 693 days. Also find the initial activity of the material. (Given : masses of nuclei Po210 = 209.98264 a.m.u., Pb206 = 205.97440 a.m.u., 2He4 = 4.00260 a.m.u., 1 a.m.u. = 931 MeV and Avogadro number = 6x1023/mol. 28. 10�3 kg of radioactive isotope (atomic mass 226) emits 3.72 x 1010 . -particles in a second. Calculate the half-life of the isotope. If 4.2 x 102 joule is released in one hour in this process, what is the average energy of the . -particle? 29. In an ore containing uranium, the ratio of U238 to Pb206 nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of U238. Take the half-life of U238 to be 4.5 x 109 year. 30. A7 Li target is bombarded with a proton beam current of 10�4 A for 1 hour to produce 7Be of activity 1.8 x 108 disintegrations per second. Assuming that one 7Be radioactive nuclei is produced by bombarding 1000 protons, determine its half-life. Level � 3 1. Asmall quantityof solution containing 24Na radionuclide (half life 15 hours) or activity1.0microcurie is injected into the blood of a person.Asample of the blood of value 1 cm3 taken after 5 hours shows an activity of 296 disintegrations per minute. Determine the total value of blood in the bodyof the person. Assume that the radioactive solutionmixes uniformly in the blood of the person. (1 Curie = 3.7 × 1010 disintegrations per second) [JEE, 94] 2. At a given instant there are 25%undecayed radio-active nuclei in a sample.After 10 sec the number of undecayed nuclei remains to 12.5%. Calculate : [JEE, 96] (i)mean-life ofthe nuclei and (ii)The time inwhich the number ofundecayed nuclearwill further reduce to 6.25%ofthe reduced number. 3. Consider the following reaction ; 2H1 + 2H1 = 4He2 + Q. [JEE, 96] Mass of the deuteriumatom= 2.0141 u;Mass of the heliumatom= 4.0024 u This is a nuclear _______ reaction inwhich the energyQis released is _______MeV. 4. The element Curium 248 96 Cmhas a mean life of 1013 seconds. Its primary decaymodes are spontaneous fission and . decay, the formerwith a probabilityof 8%and the laterwitha probabilityof92%. Eachfission releases 200MeVof energy.Themasses involved in . decayare as follows : 248 96 Cm= 248.072220 u, 244 94 Pu = 244.064100 u & 42He = 4.002603 u. Calculate the power output froma sample of 1020 Cmatoms. (Iu = 931MeV/C2) [JEE, 97]
5. Nuclei of a radioactive elementAare being produced at a constant rate .. the element has a decay constant ..At time t = 0, there areN0 nucleiof the element. [JEE, 98] (a) Calculate the number Nof nuclei ofAat time t. (b) If . = 2N0., calculate the number of nucleiofAafter one halflife ofA&also the limiting value ofNas t ....
MODERN PHYSICS www.physicsashok.in 151 6. Photoelectrons are emittedwhen 400 nmradiationis incident on a surface ofwork function 1.9 eV. These photoelectrons pass througha regioncontaining .-particles.Amaximumenergyelectron combineswith an .-particle to formaHe+ ion, emittinga single photoninthis process.He+ ions thus formed are intheir fourth excited state. Find the energies in eVofthe photons, lying in the 2 to 4eVrange, that are likelyto be emitted during and after the combination. [Take, h = 4.14 × 10�15 eV�s] [JEE, 99] 7(a). Ahydrogen-like atomof atomic number Z is in an excited state of quantumnumber 2 n. It can emit a maximumenergyphoton of 204 eV. If itmakes a transition to quantumstate n, a photon ofenergy40.8 eV is emitted. Find n, Z and the ground state energy(in eV) for this atom.Also, calculate theminimumenergy (in eV) that can be emitted bythis atomduring de-excitation. Ground state energy of hydrogen atomis � 13.6 eV. [JEE, 2000] (b). When a beamof 10.6 eV photon of intensity 2W/m2 falls on a platinumsurface of area 1 × 104 m2 and work function5.6 eV, 0.53%ofthe incident photons eject photoelectrons. Find the number ofphotoelectrons emitted per sec and theirminimumandmaximumenergies in eV. [JEE, 2000] 8. Ahydrogenlike atom(described bytheBohrmodel) is observed to emit sixwavelengths, originating from all possible transitionbetween a group of levels. These levels have energies between � 0.85 eVand �0.544 eV(including both these values) [JEE, 02] (a) Find the atomic number of the atom. (b) Calculate the smallest wavelength emitted in these transitions. 9. Two metallic platesAand B each of area 5 × 10�4m2, are placed at a separation of 1 cm. Plate B carries a positive charge of 33.7 × 10�12 C.Amonochromatic beamof light,with photons of energy 5 eVeach, starts falling onplateAat t = 0 so that 1016 photons fall on it per squaremeter per second. assume that one photoelectron is emitted for every106 incident photons.Also assume that all the emitted photoelectrons are collected by plateB and thework dunction of plateAremains constant at the value 2 eV. Determine (a) the number of photoelectrons emitted up to t = 10 sec. (b) themagnitude of the electric field between the platesAand B at t = 10 s and (c) the kinetic energy of the most energetic photoelectron emitted at t = 10 s when it reaches plate B. (Neglect the time taken by photoelectron to reach plate B) [JEE, 02] 10. Frequencyof a photon emitted due to transition ofelectronofa certain elemrnt fromLtoKshellis found to be 4.2 × 1018 Hz. UsingMoseley�s law, find the atomic number of the element, given that the Rydberg�s constant R = 1.1 × 107 m�1. [JEE, 03] 11. In a photoelectric experiment set up, photons of energy 5 eVfalls on the cathode havingwork function 3 eV. (a) if the saturation current is iA = 4µA for intensity10�5W/m2, then plot a graph between anode potential and current. (b)Also drawa graph for intensity of incident ratiation of 2 × 10�5W/m2 ? [JEE, 03]
12. Aradioactive samle emits n .-particles in 2 sec. In next 2 sec it emits 0.75 n .-particles, what is themean life of the sample ? [JEE, 03]
MODERN PHYSICS www.physicsashok.in 152 13. The potential energy of a particle ofmassmis given by 0 E 0 x 1 V(x) 0 x 1 . . . . . . . . . . .1 and .2 are are the de-Brogliewavelengths of the particle, when 0 . x . 1 and x > 1 respectively. If the total energy of particle is 2E0, find .1/.2. [JEE, 05] 14. Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of heliumnucleus is (14)1/3. Find [JEE, 05] (a) atomic number of the nucleus (b) the frequency ofK. line of the X-ray produced. (R = 1.1 × 107 m�1 and c = 3 × 108 m/s) 15. Inhydrogen-like atom(z=11), nth line ofLymanseries haswavelength. equalto thede-Broglie�swavelength of electron in the level fromwhich it originated.What is the value of n ? [Take :Bohr radius (r0) = 0.53Å and Rydberg constant (R) = 1.1 × 107m�1] [JEE, 06] Answer Key Assertion-Reason Q. 1 2 3 4 5 6 7 8 9 10 Ans. B C D B A A B A A B Match the Column 1. (A) . (U), (B) . (R), (C) . (P), (D) . (S) 2. (A) . (S), (B) . (T), (C) . (Q), (D) . (P) 3. (A) . (R), (B) . (P), (C) . (T), (D) . (U) 4. (A) . (R), (B) . (U), (C) . (V), (D) . (Q) 5. (A) . (Q), (B) . (R), (C) . (S), (D) . (P) 6. (A) . (PQ), (B) . (PR), (C) . (PS), (D) . (PQR) 7. (A) . (P), (B) . (T), (C) . (U), (D) . (R) 8. (A) . (PR), (B) . (QS), (C) . (P), (D) . (Q) 9. (A) . (PQT), (B) . (Q), (C) . (S), (D) . (S)
MODERN PHYSICS www.physicsashok.in 153 Level � 1 Q. 1 2 3 4 5 6 7 8 9 10 Ans. D C D B C A C D D A Q. 11 12 13 14 15 16 17 18 19 20 Ans. B B B D A D A C ABC C Q. 21 22 23 24 25 26 27 28 29 30 Ans. C D D C D C B D D B Q. 31 32 33 34 35 36 37 38 39 40 Ans. B C AD BC ABC BD ABD AB AC BD Q. 41 42 43 44 45 46 47 48 49 50 Ans. ABC CD AC AB AB AC AD AC AB C Q. 52 53 54 Ans. AD C B Q. 57 59 60 61 62 63 64 65 Ans. C A B C AC A B A Q. 66 67 68 69 70 71 72 Ans. B BD C C A A A 56 (a-C), (b-B), (c-B), (d-E), (e-C) 58 (a-C), (b-A) 51 (i-B), (ii-A) (i-CD), (ii-D) 55 Passage Q. 1 2 3 4 5 6 7 8 9 10 Ans. C B A D B B D A D A Q. 11 12 13 14 15 16 17 18 19 20 Ans. B B C A B C D A C D Q. 21 22 23 24 25 26 27 28 29 30 Ans. B A D A C B C B C A Q. 31 32 33 34 35 36 37 Ans. B C B D D A B Level � 2 1. 1.9 eV 0 . . , . = 4125A.2. (i) 8.426 x 10�14 n2 (ii) 25 (iii) 5.478 x 10�11 m 3. n = 4; . = 18800A.4. 50.6 .A 5. (p) . 4.1. .. . .d2c. . 50 atm 6. P . .E/c. 1. .2 . 2. cos 2. . 35 nN.s 7. 4 R 79 eV. 0 . . . . . . 8. .. 2 .R/m.3.1.106 m/s, where m is the mass of the electron. 9. e4 / 2 3 , b . . . . where . is the reduced mass of the system. If the motion of the nucleus is not taken into account, these values (in the case of a hydrogen atom) are greater by m/M . 0.055%, where m and M are the masses of an electron and a proton.
MODERN PHYSICS www.physicsashok.in 154 10. (a) 0.285 pm, 2.53 keV, 0.65 nm; (b) 106 pm, 6.8 eV, 0.243 . m. 11. ps 13 e 4 / r c m t 2 3 2 2e. . 12. (a) 2.3 x 10�138 m (b) 2.5 x 1074 13. (a) 2 r E q L0 .. . . for r > r0 = 0; (b) 0 0 f 4 V q ... . . ; (c) 500 14. v = 7.45 x 1016 cycles/s, no 15. . . . . . .. . . b cos acos hsin sin cd 16. 2.93 x 1020 17. 11.72 x 1011 18. (i) x = 5 ; (ii) 16.53 eV (iii) 36.36 A ; (iv) 1.056 x 10�34 kg m2 s�1 (v) 1.06 x 10�11 m 19. (a) 1.6 x 10�17 2 2 0 . h (b) The radiation connot eject electron. 20. 0.163 A 21. 2.55 eV, electron jumps from 4th to 2nd orbits, 2.11 x 10�34 Js 0.8144 m/s 22. (a) n = 2 (b) z = 6 (c) 28.5 A. (d) 25.32 A. 23. (a) 0.25 A (b) 1.76 x 1016 Hz 24. .. . .. . . . 1 R2 1 RR 10.75 25. NA = 6.25 x 10�4 kg, NB = 2.5 x 10�3 26. 51.41%, 2.13%, 46.45% 27. A0 = 4.57 x 1021 per day, 320 gm 28. T½ = 1573 year, Energy per . particle = 19.5 MeV 29. 1.868 x 109 year 30. T½ = 8.63 x 106 s. Level � 3 1. 6 litre 2. (i) t1/2 = 10 sec., tmeans = 14.43 s; (ii) 40 seconds 3. Fusion, 24 4. . 33.298 µW 5. (a) t t 0 N . 1 [.(1. e.. ) . .N e.. ] . ; (b) 0 0 3N , 2N 2 6. during combination = 3.365 eV; after combination = 3.88 eV(5 . 3) &2.63 eV(4 . 3) 7.(a) n = 2, z = 4; G.S.E.� 217.6 eV;Min. energy = 10.58eV; (b) 6.25×1019per sec, 0, 5 eV 8. (a) 3; (b) 4052.3 nm 9. (a) 5 × 107, (b) 2000 N/C, (c) 23 eV
10. z = 42 11. 8µA I 4µA �2V VP I=2×10�5W/m2 I = 10�5W/m2 12. 1.75n = N0(1�e�4.), 6.95 sec, 2 ln(4 / 3) 13. 2 14. v = 1.546 × 1018 Hz 15. n = 24 �X�X�X�X�
NUCLEAR PHYSICS
DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125
NUCLEAR PHYSICS www.physicsashok.in 1 RADIO ACTIVITY INTRODUCTION OF ATOMIC NUCLEUS (a) The atomic nucleus consists oftwo types of elementaryparticles, viz. protons and neutrons.These particles are collectivelycalled nucleons. (b) The electrons surround this nucleus to formthe atom. (c) This structure ofatomwas revealed by the experiments ofRutherford inwhich a beamof .-particleswere made to strike a thin gold foil. (d) Nucleus can be regarded as a small spherical volume situated at the centre of atom. (e) Most of themass of the atomis concentrated in the nucleus. (f) Nucleus has a positive charge. PROPERTIES OF AN ATOMIC NUCLEUS Composition: (a) All nucleiare composed of two types of particles protons and neutrons. (b) The onlyexceptionis the ordinary hydrogen nucleuswhich is just a single proton. (c) A proton has a mass of 1.6726 × 10�27 kg and charge +e (= 1.6 × 10�19 C). (d) Aneutron has amass of 1.6750 × 10�27 kg and is electrically neutral. The atomic number Z: This is equal to the number of protons in the nucleus. The neutron numberN: This is equal to the number of neutrons in the nucleus. Themass numberA: This is equal to the number of nucleons (protons + neutrons) in the nucleus. Thus,A= Z + N Symbolically a nucleusXshall be represented as A ZX . Types of Nuclei : Nuclei are of three types: (i) Isotopes: Nuclei having the same Z value but different NandAvalues are called isotopes of the element. eg. 1H1, 1H2 and 1H3 are isotopes of hydrogen. (ii) Isobars: Nuclides (nucleiof different elements) having the sameAvalue but different Z andNvalues are called isobars. e.g. 1H3 and 2He3 are isobars. (iii) Isotones: Nuclides having the same Nvalue but differentAand Z values are known as isotones. e.g. 1H3 and 2He4 are isotones. Radius of atomic nucleus: The size of the nucleus is of the order of fermi (fm). 1 fm = 10�15. Most nucleus are almost sphericalwith an average radius Rgiven by R = R0A1/3 whereAis the mass number and R0 is a constant. R0 ~. 1.2 fm = 1.2 × 10�15 m
NUCLEAR PHYSICS www.physicsashok.in 2 Mass : (a) Nuclearmasses have been accurately determinedwith the help of themass spectrometer. (b) It is expressed in a.m.u. i.e. atomicmass unit. (c) This unit is such that mass of the carbon isotope 6C12 is exactly 12 amu. (d) 1 a.m.u. = 1.66056 × 10�27 kg (e) Matter can be viewed as a condense formof energy. (f) The energy corresponding to themass of a particlewhen it is at rest is called rest mass energy. (g) Rest mass energyis E =mc2, fromEinstein relation. (h) Energy equivalent of 1 a.m.u. is equal to 931.5MeV. i.e. 1 a.m.u. = 931.5MeV/c2. Example 51: (a) Calculate the value of 1 a.m.u. fromAvogadro�s number. (b) Determine the energy equivalent of 1 a.m.u. Sol: (a) Onemole of C12 has a mass of 12 g and containsAvogadro number, NA of atoms. Bydefinition eachC12 atomhas a mass of 12 a.m.u. Thus, 12 g corresponds to (12NA) a.m.u.whichmeans, 1 a.m.u. = A 1g N = 3 23 10 6.022055 10 . . kg or 1 a.m.u. = 1.66056 × 10�27 kg (b) FromEinstein relation, restmass energy E = mc2 Hence, energy equivalent of 1 a.m.u., E = (1 a.m.u.) × c2 = (1.66056 × 10�27 kg) × (3 × 108 m/s)2 = 1.4924 × 10�10 J Since, 1 eV = 1.6 × 10�19 J, . E = 931.5MeV Hence, 1 a.m.u = 931.5MeV/c2 Density: Mass of a nucleus can be taken approximatelyAm, wheremis the mass of proton or a neutron andAis mass number . Mass =Am Also, assuming the nucleus to be a sphere of radius R, its volume is V = 43 .R3 = 30 4 R A 3 . , [. R = R0A1/3] The nuclear densityis thus given by, . = mass volume = 30 Am
4 R A 3 . = 30 3m 4.R It is thus independent of themass numberAand is therefore nearly the same for allnuclei. Putting R0 = 1.2 fm = 1.2 × 10�15mand m= 1.67 × 10�27 kg, we get, . = 2.3 × 1017 kg/m3. This is almost 1014 times the densityofwater. (a) Nuclear density . = 2.3 × 1017 kg/m3 (b) It is nearly the same for all nuclei.
NUCLEAR PHYSICS www.physicsashok.in 3 C48: Find themass density of the oxygen nucleus 8O16. Sol: Volume = 43 .R3 = 30 4 R A 3 . [. R = R0A1/3] = 43 .(1.2)3 × 16 × 10�45 m3, [. R0 = 1.2 × 10�15, A= 16] = 1.16 × 10�43 m3 Mass of oxygen atoms (A= 16) is approximately 16 a.m.u. Therefore densityis . = mass volume = 43 3 16 amu 1.16.10. m = 27 43 3 16 1.66 10 1.16 10 m. . . . . kg/m3 = 2.3 × 1017 kg/m3 NUCLEAR STABILITY (a) High densityof the nucleus suggests a very tight packing of protons and neutrons in it. (b) The Coulomb�s repulsive force between two protons in a nucleus is about 1036 times as large as the gravitational force between them. It is therefore surprising that a nucleus should be so stable. (c) Nuclei are stable because of the presence of another force, called the nuclear force. Nuclear force : It arises due to interaction betweenprotons, protonwith neutrons, and neutronwith neutrons.This force is essentially a very strong attractive force and overcomes the electrostatic repulsion between the proton inside the nucleus. Properties of Nuclear forces : (a) These are strong attractive forces. (b) These are about 100 times stronger than the Coulomb�s force. (c) These are short ranged forces (effective upto 10 fm). (d) They contain a small component of repulsive force which is effective up to a distance of the order of 0.5 fmor 0.5 × 10�15mor less. This repulsive component prevents the collapse of the nucleus. (e) These forces are charge independent. (f) Let d be the range of the effectiveness of nuclear forces, then 0.5 fm . d . 10 fm (g) Let Fpp, Fpn and Fnn denote themagnitude of the nuclear force by a proton on
a proton, by a proton on a neutron and bya neutron ona neutron respectively. Then for a separation of 1 fm, Fpp = Fpn = Fnn. N/Z ratio : (a) N/Zratio inside a nucleus is responsible for stability of a nucleus. N = Z line N v/s Z curve N (b) It must be greater than or equal to unity and less than 1.6 Z i.e. 1 . NZ . 1.6.
NUCLEAR PHYSICS www.physicsashok.in 4 (c) Reason forN/Zratio to be greater than unityis due to the fact that protons are positivelycharged and repel on another electrically.This repulsion becomes so great in nucleiwithmore than 10 protons or so that an excess of neutrons, which produce onlyattractive nuclear forces, is required for stability. ThusN/Z ratio increaseswith increase in Z. (d) When excess of neutrons or protons in a nuclide is there then the nuclide .-decayor .-decayto achieve the requiredN/Zratio for stability.This causes radioactive disintegrations of nuclides. Binding Energy (B.E.) : The binding energy is equal to thework thatmust be done to split the nucleus into particles constituting it. Hence, Energyof nucleus+B.E. = Energyofeach nucleon individually. Let Mass of nucleus =M, Mass of neutron =mn, and, Mass of proton = mp . Rest mass energyof nucleus =Mc2, . Rest mass energyof neutron =mnc2, . Rest mass energyof proton =mpc2. Thus, Mc2 + B.E. = Z mpc2 + (A � Z)mn c2 . B.E. = [M� {Zmp + (A� Z)mn}]c2 The quantity,M� {Zmp + (A� Z)mn} is called mass-defect (.m) . Mass � defect = .m=M� {Z mp + (A � Z)mn} . B.E. = .m.c2 If .mis in a.m.u., then B.E. = .m× 931.5MeV or B.E. = [M� {Z mp + (A � Z)mn}] × 931.5MeV NOTE: Negative sign of B.E. represents boundedness of nucleons inside the nucleus. C49: Find the binding energy of 56 26Fe .Atomicmass of 56Fe is 55.9349 u and that of 1His 1.00783 u.Mass of neutron = 1.00867 u. Sol: Z = The number of protons in 56 26Fe = 26 and the number of neutrons,A� Z = 56 � 26 = 30 Then binding energy of 56 26Fe = [M� {Zmp + (A� Z))mn}]c2 = � [26 × 1.00783 u + 30 × 1.00867 u � 55.9349 u]c2 = � (0.52878 u)c2 = � (0.52878 u) (931MeV/u) = � 492 MeV Negative sign indicates boundedness of nucleons. . B.E. = 492MeVinmagnitude. Binding energy per nucleon : Binding energyper nucleonis obtainedbydividing the binding energyof the nucleus bythe numberA of nucleons in the nucleus. i.e. B.E. per nucleon = B.E. A 0 50 100 150 200 250 246 8 10
56 26Fe 4He 6Li Binding energy per nucleon, MeV Mass number, A
NUCLEAR PHYSICS www.physicsashok.in 5 (a) The adjacent figure shows the dependence of the B.E. per nucleon, B.E./Aonthemass number Aofthe nucleus. (b) Nucleons in nucleiwithmass number from50 to 60 have the highest B.E. TheB.E./Afor these nuclei amounts to 8.7MeVand gradually decreaseswith increasingA. (c) B.E. per nucleon is highest for 56 26Fe . Example 52: Find the binding energy of 126C ?Also find the binding energy per nucleon. Mass of 6C12 atom = 12 a.m.u. Mass of proton = 1.00759 a.m.u. Mass of neutron = 1.00898 a.m.u. Sol: M = mass of C12 atom = 12 a.m.u. mp = mass of proton = 1.00759 a.m.u. mn = mass of neutron = 1.00898 a.m.u. Z = number of proton = 6 A� Z = number of neutrons = 12 � 6 = 6 . Mass-defect, .m = M� {Zmp + (A� Z)mn} = 12 � (6 × 1.00759 + 6 × 1 × 1.00898) in a.m.u. = (12 � 12.009) a.m.u = �0.099 a.m.u. . B.E. = .m× 931.5MeV = �0.099 × 931.5 MeV = �92.22MeV Hence, Binding energy= �92.22MeV (Negative sign indicates boundedness of the nucleon) Binding energy per nucleon = B.E. A = 92.22 12 MeV= 7.68MeV NUCLEAR COLLISIONS (a) Anuclear reaction in which a collision between particle a and nucleus X producesY and particle b is represented as a + X Y+ b (b) The reaction is sometimes expressed inthe shorthand notationX(a, b)Y. (c) Reaction are subjected to the restrictions imposed by (i) The conservationof charge, (ii) The conservationof energy, (iii) The conservationofmomentum, and (iv) The conservationof angularmomentum. Q-Value or Energy of a reaction : Let m2,m3 are nuclearmasses ofXandYrespectively. a m1 K1 X m2 K2 Before collision . Y m3 K3 X m4 K4 After collision Initial energy: Ei = m1c2 + m2c2 + K1 + K2 Final energy: Ef = m3c2 + m4c2 + K3 + K4
NUCLEAR PHYSICS www.physicsashok.in 6 Since, Ei = Ef, (fromenergyconservation) . [(m1 + m2) � (m3 + m4)]c2 = (K3 + K4) � (K1 + K2) The energy, [(m1 + m2) � (m3 +m4)]c2, that is released or absorbed in a nuclear reaction is called the QValue or disintegration energyof the reaction. Hence, Q = [(m1 + m2) � (m3 + m4)]c2 J or, Q = [(m1 + m2) � (m3 + m4)] 931.5MeV, when masses are in a.m.u. Mass defect: The quantity [(m1 +m2) � (m3 +m4)] is called themass defect of the reaction and is given by .m= (m1 + m2) � (m3 + m4) ina.m.u. Q = .m. 931.5 MeV Example 53:Aneutron breaks into a proton and electron. Calculate the energy produced in this reaction in MeV. Mass of an electron = 9 × 10�31 Kg, Mass of proton = 1.6725 × 10�27 kg, Mass of neutron = 1.6747 × 10�27 Kg. Speed of light = 3 × 108m/s. Sol: 0n1 1H1 + �1e0 Mass defect, .m = [mass of neutron � (mass of proton +mass of �1e0)] = 1.6747 × 10�27 kg � (1.6725 + 0.0009) × 10�27 kg = 0.0013 × 10�27 kg . Energyreleased, Q = .m.c2 = 1.3 × 10�30 × (3 × 108)2 kg �m2/s2 = 1.17 × 10�13 J = 13 19 1.17 10 1.6 10 . . . . eV = 0.73MeV Example 54. Find theQ-value of the reaction 1H2 + 3Li6 . 3Li7 + 1H1 The rest masses of 1H2, 3Li6.. 3Li7, and 1H1are, respectively, 2.01410 amu, 6.01513 amu, 7.01601 amu and 1.00783 amu. Sol. Suppose 1H2 + 3Li6 . 3Li7 + 1H1 + Q. Totalmass of left-hand side = 2.01410 + 6.01513 = 8.02933 amu Totalmass on right-hand side = 7.01601 + 1.00783 = 8.02384 amu . 8.02933 = 8.02384 + Q or Q = 0.00549 amu = 0.00549 × 931 MeV (. 1 amu = 931MeV) Q = 5.1 MeV INTRODUCTION OF RADIOACTIVITY (a) In 1896, Becquerel discovered accidentally that uraniumsalt crystals emit an invisible radiationwhich affected a photographic plate even though it was properly covered. (b) In 1898,Marie and PierreCurie and otherworkers showed thatmanyother substances also emitted similar radiations.
NUCLEAR PHYSICS www.physicsashok.in 7 (c) The propertyof spontaneous emission ofradiation fromthe substance is called radioactivityand such type of substance is called radioactive substance. (d) Radioactivityis due to the decay or disintegrationof unstable nuclei. (e) The radiations arebeing emitted fromthenucleihence it is anuclear phenomenon, not anatomicphenomenon. (f) Some examples of radioactive substances are:U, Ra, Th, Po and Np. (g) Electronic configuration ofatomdoes not have anyrelationshipwith radioactivity. (h) Radioactivity is explained on the basis of quantummechanics. (i) No single phenomenon has played so significant role in the development of nuclear physics as radioactivity. (j) It is not influenced byexternal parameters such as pressure, temperature, chemical reaction (combination) or phase ofmatter. C50:Uraniumsalt crystals emit (a) visible radiation (b) invisible radiation (c) anytype of electromagnetic radiation (d) soundwaves Sol: This is according to discovery ofradioactivityofBecquerel. C51: The radioactivityis a/an (a) optical phenomenon (b)Atomic phenomenon (c) nuclear phenomenon (d) photoelectric phenomenon Sol: Nuclear radiations are obtained fromthe nuclei hence it is a nuclear phenomenon. RADIOACTIVE DECAY (a) The decayof radioactive substancemeans disintegration of nucleiofthe substance byemissionof different radiations. (b) Despite the strength of the forces that hold nucleons (protons and neutrons) together to forman atomic nucleus,many nuclides are unstable and spontaneouslychange into other nuclides byradioactive decay. (c) The energyliberatedduring radioactive decaycomes fromwithinindividualnucleiwithout externalexcitation, unlike the case of atomic radiation. (d) It is statisticalprocess that obeys the laws of chance. (e) The decayof nucleus takes place to achieve the stable end products. Kinds of Decay There are five kinds of radioactive decays. When radioactivitywas discovered, onlythree kinds of radioactive decays alpha(.), beta(.) and gamma)(.) were known.Whichwere eventuallyidentifiedas 42He nucleus, electronand highenergyphotonrespectively. Later two more kinds of radioactive decays namely positron emission and electron capturewere added. Alpha decay (a) In this type of decay, the unstable nucleus emits an alpha particle. (b) It reduces the proton number Z of the nucleus by 2. (c) It reduces the neutron number Nof the nucleus by 2.
NUCLEAR PHYSICS www.physicsashok.in 8 (d) It reduces themass number (i.e. Z +N) of the nucleus by 4. (e) It changes the element itself hence the chemical symbol of the residualnucleus is different fromthat of the originalnucleus. (f) The alpha decay processmay be represented as A A 4 4 Z Z 2 2 X� . Y He . . . (g) The nucleus before the decay is called the parent nucleus and that obtained after the decay is called the daughter nucleus. Ex: Let us consider the example given below: 212 208 4 83 81 2 Bi�. Tl. He In the above example of .-decay, the parent nucleus is bismuth (Bi) and the daughter nucleus is thallium (Tl). (h) Alpha decay occurs in all nucleiwithmassA> 210. (i) In this decay, the nucleus decreases itsmass number tomove towards stability. (j) On emission of .-particle, the binding energyper nucleon increases and the residualnucleus tends towards stability. Q-Value for .-decay: If .-decay process is given by AZX A 4 Z 2 . Y . + 42He , then Q-value = . A . . A 4 . . 4 . Z Z 2 2 m X m . Y m He . .. . . .. c2 NOTE: The quantity m. . AZX represent atomic mass of the particle X. Example 55. Aradon nucleus Rn86222, ofmass 3.6 × 10�25 kg, undergoes .-decay. The .-particle hasmass 6.7 × 10�27 kg and energy 8.8 × 10�31 J. (a)What is the resulting nucleus? (b) Find the velocity of recoilof the nucleus. Sol. (a) The atomic number will be reduced by 2 and the mass number by 4. . A = 222 � 4 = 218 and Z = 86 � 2 = 84 The resulting nucleus is 84Po218. (b) mass of resulting nucleus = m1 = 3.6 × 10�25 � 6.7 × 10�27 = 3.533 × 10�25 kg Let v1 = its velocity of recoil Mass of .-particle = m2 = 6.7 ×10�27 kg Velocity of .-particle = v2 = 1/2 2 2E m . . .. .. with E = 8.8 × 10�13 J Now m1v1 =m2v2 . 1/ 2 27 13 1/ 2
2 1 25 1 (2m E) (2 6.7 10 8.8 10 ) v m 3.533 10 . . . . . . . . . . , v1 = 3.1 × 10+5 ms�1
NUCLEAR PHYSICS www.physicsashok.in 9 Beta Decay (a) Beta decay is a process inwhich a neutron is converted into a proton. n p + e� + . (b) It increases the atomic number (Z) of nucleus by 1. (c) It does not alter themass number (A). (d) If a nucleus is formedwithmore number of neutrons than needed for stability, a neutronwill convert itself into a proton tomove towards stability. (e) When a neutron is converted into a proton, an electron and a newparticle named antineutrino are created and emitted fromthe nucleus. (f) The antineutrino is denoted bythe symbol .. It is supposed to have zero restmass like photon. It is chargeless and has quantumnumber ±½. (g) The electron emitted fromthe nucleus is called a beta particle and is denoted bythe symbol .� or �1e0. (h) Astreamof beta particles coming frombulk of unstable nuclei is called beta ray. (i) It is also called betaminus decayas negatively charged beta particles are emitted. (j) The beta decay processmay be represented by AZA A Z 1Y . + e� + . (. n p + e� + . ) AZX A Z 1Y . + .� + . (k) An example of beta decay is 6C14 7N14 + e� + . (antineutrino) Q-value for .-decay: (a) .�decay: If .� -decay process is given by AZX A Z 1Y . + .� + . , then Initial rest mass energy, REi = . . AZe ..m X . Zm .. c2 Final restmass energy, REf= . A . Z 1 e e m Y Zm (Z 1) m . .. . . . .. c2 . Q = REi � REf = . A . . A . Z Z 1 m X m Y . .. . .. c2 Because ofthe largemass, the residual nucleus A Z 1Y . does not share appreciable kinetic energy. Thus, the energyQis shared by the antineutrino and the beta particle. Depending on the fraction taken away by the antineutrino, the kinetic energyof the beta particle can be anything between zero and amaximumvalueQ. Positron emission (.+ decay) (a) .+ decay is a process in which a proton is converted into a neutron with emission of positron (e+) and neutrino (v). p n + e+ + v (b) It reduces the atomic number (Z) of nucleus by 1. (c) It does not alter themass number (A). (d) An isolated proton does not beta decay to a neutron. On the other-hand, an isolated neutron decays to a proton.
NUCLEAR PHYSICS www.physicsashok.in 10 (e) If the unstable nucleus has excess protons than needed for stability, a proton converts itselfinto a neutron. (f) When a proton is converted into a neutron, a positron and a neutrino are created and emitted fromthe nucleus. (g) The neutrino is denoted by the symbol v. It is charge-less particle. (h) The positron (e+) has a positive electric charge equalinmagnitude to the charge on an electrons and has a mass equal to themass of an electron. (i) Positron is called the antiparticle of electron. (j) When an electron and a positron collide, both the particles are destroyed and energyismade available. (k) Neutrino and antineutrino are antiparticles of each other. (l) The .+ decay process is represented as AZX A Z 1Y . + e+ + v [. p . n + e+ + v] AZX A Z 1Y . + .+ + v If the unstable nucleus has excess protons than needed for stability, a proton converts itselfinto a neutron. In the process, a positron and a neutrino are created and emitted fromthe nucleus, p n + e+ + v .....(iv) The positron e+ has a positive electric charge equalinmagnitude to the chargeon an electron and has amass equal to the mass of an electron. Positron is called the antiparticle of electron.When an electron and a positron collide, both the particles are destroyed and energy is made available. Similarly, neutrino and antineutrino are antiparticles of each other.When a proton in a nucleus converts itself into a neutron, the decay process is represented as AZX A Z 1Y . + e+ + v or AZX A Z 1Y . + .+ + v .....(v) This process is called beta plus decay. The positron so emitted is called a beta plus particle. . + - decay or position - emission : If the .+-decay or position-emission is given by AZX A Z 1Y . + .+ + v, then R.Ei = . . AZe ..m X . Zm .. c2 R.Ef = . A . Z 1 e e m Y (Z 1)m m . .. . . . .. c2 . Q = R.Ei � R.Ef = . A . . A . Z Z 1 e m X m Y 2m . .. . . .. c2 Q = . A . . A . Z Z 1 e m X m Y 2m . .. . . .. c2 .....(vi) Can an isolated proton decayto a neutron emitting a positron and a neutrino as suggested byequation (iv)?
Themass ofa neutron is larger than themass ofa proton and hence theQ-value of such a processwould be negative. So, an isolated proton does not beta decay to a neutron.On the other hand, an isolated neutron decays to a proton as suggested by equation (i). (m) The positronis also called beta plus particle. (n) An example of .+ decay is 29Cu64 28Ni64 + e+ + v
NUCLEAR PHYSICS www.physicsashok.in 11 Asimilar process, known as electron capture, takes place in certain nuclides. In this process, the nucleus captures one of the atomic electrons (most like an electron fromthe K shell). Aproton in the nucleus combineswith this electron and coverts itselfinto a neutron.Aneutrino is created in the process and emitted fromthe nucleus. Electron capture (a) When the nucleus has toomanyprotons relative to the number ofneutrons, the nucleus captures one of the atomic electrons (most likely an electron fromtheK-shell). (b) Aproton inthe nucleus combineswith this electron and converts itselfinto a neutron. (c) Aneutrino is created in this process and emitted fromthe nucleus. p + e� n + v (d) In this process, atomic number (Z) of the nucleus decreases by 1. (e) This process does not alter the value ofmass number (A). (f) When an atomic electron is captured, a vacancy is created in the atomic shell and X-rays are emitted following the capture. (g) This process is also calledK-capture. (h) The processmay be represented as AZX + e� A Z 1Y . + v [. p + e� n + v] (i) An example of electron capture is 29Cu64 + e� 28Ni64 + neutrino Q-Value of K-capture process: IfK-capture process is given by AZX + e� A Z 1Y . + v, then Q = . A . . A . Z Z 1 m X m Y . .. . .. c2 Example 56: Calculate theQ-value in the following decays- (a) 19O 19F + e + . [.� -decay] (b) 25Al 25Mg + e+ + v, [.+-decay] The atomic masses of 19O = 19.003576 a.m.u. 19F = 18.998403 a.m.u, 25Al = 24.990432 a.m.u.,25Mg = 24.985939 a.m.u. Sol: (a) The Q-value of .� -decay is Q = [m (19O) � m(19F)]c2 = [19.003576 a.m.u. � 18.998403 a.m.u.] (931.5MeV/a.m.u.) = 4.819MeV (b) The Q-value of .+ -decay is Q = [(mass of 25Al nucleus) � (mass of 25Mg nucleus) � (mass of positron)]c2 = [(24.990432 a.m.u � 13me) � (24.985939 a.m.u. � 12 me) � me]c2 = [(0.004593 a.m.u.) � 2me]c2 = 0.004593 a.m.u. × (931.5MeV/a.m.u.) � 2 × 0.511 2 MeV c .c2 = 4.276MeV � 1.022 MeV = 3.254MeV
NUCLEAR PHYSICS www.physicsashok.in 12 Gamma Decay (a) Nucleus has also energylevels like atoms have. (b) This decayprocess is related to the transitions between two nuclear energy levels. (c) The protons and neutrons inside a nucleusmove in discrete quantumstateswith definite energies. (d) In the ground state, the nucleons occupy those quantumstates whichminimise the total energy of the nucleus. (e) The higher energystates are also available to the nucleons and if appropriate energyis supplied, the nucleus maybe excited to higher energies. (f) The energydifferences in the allowed energylevels of a nucleus are generally large (in the order ofMeV). (g) It is difficult to excite the nucleus to higher energylevels byusualmethods of supplying energylike heating etc. (h) When an alpha or a beta decay takes place, the daughter nucleus is generallyformed in one of its excited states. The daughter nucleus in an excited state eventuallycomes to its ground state byemitting one photon ormore than one photon of electromagnetic radiation. (i) The process of a nucleus coming down to a lower energylevelbyemitting a photon is called gamma decay. (j) In this decay, atomic number (Z) aswell asmass number (A) of the nucleus remain constant. (k) In this decay, the quantumstates of the nucleons vary. (l) The electromagnetic radiationemitted in nuclear transitions is called gamma ray. (m) Thewavelength ofthis radiation is given bythe common relation. . = hc E where E = energy of the photon. (n) An example of gamma decayis shown in figure below: . . . 57Co .+ 136 keV 14 keV 0 keV 2nd excited state 1st excited state Ground state 57Fe When 57Co is taken in bulk, we can observe a streamof .+ particles, 136 keV photons, 122 keV photons and 14 keVphotons coming fromthe 57Co source. NOTE: The ., . and . rays are collectively called nuclear radiation. Comparison among the kinds of decay (a) The velocityof .-particles is relatively low: v. = (c/30 � c/15),when c is the velocity of light Mass m. = 4 amu charge q. = +2e For ..-particles: 0 . v. < c m. =mass of an electron q. = �e
NUCLEAR PHYSICS www.physicsashok.in 13 For .-rays: Since .-rays are electromagneticwaves, hence theypropagatewith the speed of light. i.e. v. = c Rest mass,m. = 0(like photon) q. = 0 (like photon) (b) The penetrability of .-rays is 0 - 100 times higher than the penetrability of b-rays and 1000 - 10000 times higher than the penetrability of .-rays. The penetrability of .-rays also exceeds the penetrabilityof x-rays. Carboard Al Lead .. . . . - particles from radioactive materials are stopped by a piece of cardboard. -particles penetrate the cardboard but are stopped by a sheet of aluminium. Even a thick slab of lead may not stop all the -rays. (c) In a magnetic field, a beamof ., . and .-rays splits into three parts. In amagnetic field, .-rays are undeviated and .-particles aremost deviated .-particles .-particles .-rays Magnetic field (d) Table for various decay directed into paper Decay Transformation Example .-decay AZX A 4 4 Z 2 2 . Y He . . 238 92U 234 4 90 2 Th . He .-decay AZX A Z 1Y . + e� + . 146C 147N + e� + . Positron emission AZX A Z 1Y . + e+ + v 64 29Cu 64 28Ni + e+ + v Electroncapture AZX + e� A Z 1Y . + v 64 29Cu + e� 64 28Ni + v Gamma decay A * Z X AZX + . 87 * 38Sr 87 38Sr + . The *denotes anexcited nuclear estate, .-denotes a gamma-rayphoton and v and . denotes neutrino and antineutrino particles respectively. C52:An a-particle is bombarded on 14N.As a result, a 17O nucleus is formed and a particle is emitted. This particle is a/an
(a) neutron (b) proton (c) electron (d) positron
NUCLEAR PHYSICS www.physicsashok.in 14 Sol: 14 4 7 N . 2He 17 1 8 1 O . p 11p is equivalent to 11H . C53: In a radioactive decay, neither the atomic number nor themass number changes.Which of the following particles is emitted in the decay ? (a) Proton (b) Neutron (c) .- particle (d) photon Sol: Photon is equivalent to .-radiation. Inthis decay, onlyquantumstates of the nucleons vary. C54: .-rays emitted bya radioactivematerial are (a) electromagneticwaves (b) electrons orbiting around the nucleus (c) charged particles emitted by the nucleus (d) neutral particles Sol: Aneutronin the nucleus decays emitting an electron. C55: Give an equation representing the decay ofa free neutron. Sol: 10 n 1 0 1 1 H e v. . . . C56: Howmanyelectrons, protons and neutrons are there in a nucleus of atomic number 11 andmass number 24 ? Sol: Number of electrons or protons, Z= 11 and number of neutrons, N=mass number �Atomic number = 24 � 11 = 12 C57: Fill up the blanks (i) hv e� + .............. (.-photon) (electron) (ii) 90Th234 �1B0 + 91Pa234 + ............. (Thorium) (.-particle) (Protactinium) (iii) 92U238 2He4 + ................ (Uranium) (.-particle) Sol: (i) e+(Positron) (ii) . (Antineutrino) (iii) 90Th234(Thorium) C58: Following the origin of gamma decay, calculate the value of wavelength of resulting photonwhen the nucleus of 27 13Al reaches the ground state (Eg = 0) from the state in which Eex. = 1.015MeV.The related figure is shown adjacent. . .� 1.015 MeV 27Mg 12 0 MeV 27Al Sol: .(in Å) = 13 12400 .E(eV) = 6 12400 (1.015 . 0).10 = 0.012217 Å LAW OF RADIOACTIVE DECAY When the radioactive substance is only disintegrating: Radioactive decay is a randomprocess. Each decay is an independent event, and on
e cannot tellwhen a particular nucleuswilldecay.When a given nucleus decays, it is transformed into another nuclidewhichmay
NUCLEAR PHYSICS www.physicsashok.in 15 or may not be radioactive. When there is a very large number of nuclei in a sample, the rate of decay i.e. dN dt . . . . . . . is proportional to the number of nuclei, N, that are present i.e. � dN dt . N . � dN dt = .N where . is called the decay constant. This equationmay be expressed in the form dN dt = �.dt and integrated to get 0 N N dN N . = �. t0 . dt , to yield, ln 0 N N = �.t, where N0 is the initial number of parent nuclei at t = 0. The number that survive at time t is therefore N = N0e�.t Definition of decay constant : The probability of decay per second for a particular process for a sample is called the decay constant for that process for that sample. Radioactivity lawof decay gives, . = dN Ndt . Half life : The time period for the number of parent nuclei to fall to 50%is called the half-life, T, andmay be related to .. ifwe put N =N0/2 at t = T, the exponential decay equation gives, 0.5N0 = N0e�.T . .T = ln|2| = 0.693 , T = 0.693 . (a)It takes one half-life to drop to 50%of anystarting value. (b) The half-life for the decayof the free neutron is 12.8min. (c) Other half lives range fromabout 10�20s to 1016 years. Mean life(Tm) : Mean life of a radioactive sample is defined as the average of the lives of all nuclei. . Tm = 0 0 N 0 tdN
N . = 1. = T 0.693 . Tm = 1. and T = 0.693 Tm NOTE : (i) Radioactive decay equation, N = N0e�.t, can also be written as N = N0 t /T 12 . . . . . . , where T is half-life of the sample. (ii) The decay constant is also given as, . = 2.303 t log 0 NN
NUCLEAR PHYSICS www.physicsashok.in 16 Activity of radioactive substance: Since the number of atoms is not directlymeasurable, wemeasure the decay rate or activity (A) A= � dN dt , but N = N0e�.t A = � dN dt = .(N0e�.t) = .N . A = .N =A0e�.t, whereA0 = .N0 is the initial activity. BothNandAdecrease exponentiallywith time. The activityis characterized by the same half-life. Units of activity: (a) The SI unit for the activityis the becquerel (Bq), but the curie (Ci) is often used in practice. 1 becquerel (Bq) = 1 disintegrations per second (dps) 1 curie (Ci) = 3.7 × 1010 dps 1 rutherford = 106 dps (b) Rate of decayof 1 gmsubstance is called specific activity. (c) Activity of 1 gmRa226 is 1Ci. Example 57: The half-life of 198Au is 2.7 days. Calculate (a) the decayconstant, (b) the average life and (c) the activity of 1.00mg of 198Au. Take atomicweight of 198Au to be 198 g/mol. Sol: (a) The half-life and the decay constant are related as T = 0.693 . . . = 0.693 T = 0.693 2.7 days = 0.693 2.7.24.3600s . . = 2.9 × 10�6 s�1. (b) Tm= T 0.693 = 2.7 0.693 days = 3.9 days. (c) Activity is given as,A= .N, whereNis the number of nuclei present in 1mg of 198Au. Atomic mass of 198Au = 198 g . N = 1 mg 198g ×Avogadro no . N = 10 3 198. × 6 × 1023 atoms . N = 3.03 × 1018 atoms Thus, A = .N = (2.9 × 10�9 s�1) × 3.03 × 1018 (atoms) = 8.8 × 1012 disintegrations per sec. = 12 10
8.8 10 3.7 10 .. Ci = 240 Ci
NUCLEAR PHYSICS www.physicsashok.in 17 Example 58. Two radioactive materialA1 andA2 have decayconstants of 10 .0 and .0. If initially they have same number ofnuclei, the ratio of number of their undecayed nucleiwill be (1/e) after a time (A) 0 1 . (B) 0 1 9. (C) 0 1 10. (D) 1 Sol. 1t 10 0t 1 0 0 N . N e.. . N e. .2t 0t 2 0 0 N . N e.. . N e.. 0 0 10 t 1 0 t 2 0 N N e N N e. . .. . 1 9 0t 2 N e N . . . . 9 0t 1 e e . . . e.1 . e.9.0t �1 = �9.0t 0 t 1 9 . . Hence (B) is correct. C59. Acertain element has a density of 10 g cm�3 and half-life of 140 days. Over a period of 140 days, the average number of .-emissions per day is found to be 12 × 1012, from a sample of initial mass 1 µg. Estimate the number of atoms in1 cm3 ofthe element. Sol. We assume that only one emission takes place per atom. In 140 days, no. of emissions = 140 × 12 × 1012 . initial no. of atoms present = 2 × 140 × 12 × 1012 (since 140 days is the half-life) . no. of atoms in 1 µg = 28 × 12 × 1013 . no. of atoms in 1 cm3, i.e., 10g = 3.36 × 105 × 107 = 3.36 × 1022 C60:Aradioactive sample has 3.2 × 1016 active nuclei at certain instant.Howmany of
these nucleiwill stillbe in the same active state after four half-lives ? Sol: In one half-life the number of active nuclei reduces to half the original number. Thus in four half-lives the number is reduced to 12 . . . . . . 12 . . . . . . 12 . . . . . . 12 . . . . . . i.e. 1 16 th of the original number. . The number ofremaining active nuclei is, = 1 16 × 3.2 × 1016 = 2 × 1015 nuclei
NUCLEAR PHYSICS www.physicsashok.in 18 C61: The activityof a radioactive sample falls from1200 s�1 to 1000 s�1 in 60minutes. Calculate its half life. Sol:We have, A=A0e�.t (1000 s�1) = (1200 s�1) e�.t . 5/6 = e�.t .t = ln(6/5) . . = ln(6 / 5) t = ln(6 / 5) 60min but, T = ln 2 . . ln 2 T = ln(6 / 5) 60min T = ln 2 ln(6 / 5) × 60 min = 0.693 0.182 × 60 min . half-life = 228min C62.At a given instant there are 25%undecayed radioactive nuclei in a sample.After 10 second themumber of undecayed nuclei reduces to 12.5%Calculate : (i)mean life of the nuclei (ii) the time inwhich the number of undecayed nucleiwill further reduce to 6.25%ofthe reduced number. Sol. (i) In 10 second, number of nuclei has been reduced to half (25%to 12.5%). Therefore, its half life is t1/2 = 10 s Relation betweenhalf life andmean life is 1/ 2 mean t t 10 s ln (2) 0.693 . . tmean = 14.43 s (ii) Frominitial100%to reductiontill6.25%, it takes four half lives. t1/ 2 t1/ 2 t1/ 2 t1/ 2 100%...50%...25%...12.5%...6.25% . t = 4 t1/2 = 4(10) s = 40 s Example 59.Asample initiallycontains 1020 radioactive atoms of half-life 130days. Calculate the activityof the sample after 260 days have elapsed.Also find the total energy released during this period if the energy released per disintegration is 8 × 10�13 J. In 260 days, i.e., two half-lives, the number of undisintegrated atomswill reduce to 1/4th. Sol. . no. of atoms present after 260 days = N = 14 × 1020 Also, disintegration constant = . = 0.6931/T where T = 130 days = 130 × 86400 s Now, activity = .N = 0.693 130 . 86400 ×
14 × 1020 = 1.54 × 1012 s Number of atoms present initially= 1020 Number of atoms present after 260 days = 14 × 1020 . number of disintegrated atoms = 3/4 × 1020 Energy per disintegration = 8 × 10�13 J . total energy released = 34 × 1020 × 8 × 10�13 = 6 × 107 J
NUCLEAR PHYSICS www.physicsashok.in 19 Example 60. Abodyofmassm0 is placed on a smooth horizontal surface. Themass of the body is decreasing exponentiallywith disintegration constant ..Assuming that themass is ejected backwardswith a relative velocity u. Initially the bodywas at rest. Find the velocityof it after time t. Sol. Mass of the bodyleft after time t is m = m0e�.t So dm dt . . . .. .. = m0.e�.t and thrust force on the body is Ft = ur dm dt . . . .. .. (in forward direction) or m dv dt . . .. .. = u(m0.e�.t) (ur = u) or (m0e�.t) dv dt = m0 u.e�.t or dv = u.dt or v t 0 0 . dv . u.. dt or v = u.t Example 61. Aradio nuclidewith half life T = 69.31 second emits .-particles of average kinetic energyE = 11.25 eV.At an instant concentration of .-particles at distance, r = 2mfromnuclide is n= 3 × 1013 perm3. (i) Calculate number ofnuclie in the nuclide at that instant. (ii) If a small circular plate is placed at distance r fromnuclide suchthat .-particles strike the plate normallyand come to rest, calculate pressure experienced bythe plate due to collision of .-particles. [Mass of .-particle = 9 × 10�31 kg)] Sol. Let activity (rate of decay) of the nuclide beAnuclie per second. It means A..-particles are emitted per second. If a spherical surface of radius rwith centre at position of nuclide be considered thenA .-particles cross this surface (v dt) r per second. Itmeans during an elemental time intervaldt a number (A. dt) of .-particles cross this surface. If velocity of .-particles be v then above calculated (A. dt) .-particles are in a space having shape of a spherical shell of radius r and radial thickness (v dt) as shown in figure. Volume of this space = 4.r2 (v dt) . Concentration of .-particles at distance r fromnuclide is 2 n A dt 4 r (v dt) . . or activityofthe nuclide, A = 4.r2 vn But activity, A= .NwhereNis number of nuclei Hence, N =
4.r2vn . but decay constant ..= log 2 T . 4 r2vnT N 0.6931 . . ...(1)
NUCLEAR PHYSICS www.physicsashok.in 20 Kinetic energy of .-particles, E 1 mv2 2 . , v 2E m . substituting this value in equation(1), 2 N 4 r nT 2E 9.6 1022 0.6931 m . . . . . (ii) At distance r fromthe nuclide (A/4.r2) .-particles cross unit area per second. Let area of small circular plate be S, then number of .-particles striking the plate per second 2 2 2 A S N S 0.6931 NS 4 r 4 r 4 r T . . . . . . . Momentumof eachparticle just before collisionismv andafter collisionparticles come to rest ormomentum becomes zero. . Momentumtransferred to plate due to collision is .p = mv � 0 = mv Due to transfer ofmomentum, the plate experiences a forcewhichis equal to rate oftransfer ofmomentum. . Force, F = .p × no. of particles striking per second or F = mv × 2 0.6931NS 4.r T Pressure, P = Force per unit area . P = FS = 2 0.6931N 4.r T mv but v = 2E m . P = 2 0.6931N 2mE 4.r T = 1.08 × 10�4 Nm�2 RADIOACTIVE DATING OR CARBON DATING Radiocarbon dating, also called carbon dating, is used to estimate the age oforganic samples. The technique is based on the .-activity of the radioactive isotope C14; 14 6 C 147N + .� + . ..� . beta particles, . . anti neutrino] (a) High energyparticles for outer space, called cosmic rays, induce nuclear reactions in the upper atmosphere and create carbon - 14. (b) The carbon dioxidemolecules of the earth�s atmosphere have a constant ratio (~.1.4 × 10�12) ofC14 to C12
isotopes. (c) All living organisms also showthe same ratio as theycontinuously exchange CO2withtheir surroundings. (d) However after its death, an organismcan no longer absorbCO2 and the ratio C14/C12 decreases due to the .-decay of C14. (e) Thus bymeasuring the .-activityper unit mass, it is possible to estimate the age of amaterial. (f) Half life of 14C is 5739 y.
NUCLEAR PHYSICS www.physicsashok.in 21 Example 62:When charcoalis prepared froma living tree it shows a disintegration rate of 15.3 disintegrations of 14C per gram per minute. Asample from an ancient piece of charcoal shows 14C activity to be 12.3 disintegrations per gramperminute. Howold is this sample ?Half life of 14C is 5730 y. Sol: 14C-activity of a living tree,A0 = 15.3 dis. per min. 14C-activity of the old sample, A= 12.3 dis. permin. Suppose, the sample is t year old, then (14C-activity of sample) = (14C-activity of living tree )e�.t i.e. A=A0e�.t .....(i) where . is decayconstant of 14C-activity Putting values ofAandA0 in equation (i),we get 12.3 = 15.3 e�.t ....t = ln 15.3 12.3 . ...t = ln(1.24) = 0.218 . t = 0.218 . = 0.218 0.693 T, [as, . = 0.693 T ] . t = 0.218 0.693 × 5730 y, [T = half-life of 14C -activity= 5730 y] . t = 1805 y Thus, the sample is 1805 y old. Example 63: The relative radiocarbon activityin a piece of charcoal fromthe remains of an ancient camp fire is 0.18 that of a contemporary specimen.Howlong ago did the fire occur ?Half-life of 14C-activity is 5730 y. Sol: Here, ratio between the 14C-activityof burnt charcoal and that of a living tree is given. 14 14 0 C activity of charcoal (say,A) C activity of a living tree (say,A ) . . = 0.18 . 0 A A = 0.18 Suppose, t year ago fire occurred and let . be the decayconstant of 14C-activity.We have, using radioactive decay law, A=A0e�.t ...e.t = 0 AA . .t = ln 0 AA . t = 1. ln 0 AA . t = T 0.693 ln 0 AA , [. =
0.693 T ] = 5730 0.693 ln 1 0.18 = 8268.3 ln|5.56| = 1.4 × 104 y Thus, fire occurred 1.4 × 104 years ago. Radioactivity law for different types of disintegration of the radioactive substance It is seenthat radioactive disintegration of a radioactive substance is not only ruled bythe radioactive law, N=N0e�.t (whichis applicable onlywhenthe radioactive substance is onlydisintegrating), but the radioactive lawchanges for various types ofdisintegrationofthe substance.Let us first enlist these types ofdisintegration as:
NUCLEAR PHYSICS www.physicsashok.in 22 (1) Radioactive substance only disintegrates: For this radioactive law is � dN dt = .N or N = N0e�.t (2) Disintegration with continuous production of the radioactive substance. It deals with the case when production and the decay of the radioactive substance are taking place simultaneously. Formation A decays (.) B (3) Successive disintegrations of the products: It dealswith the casewhen a substanceAdecays into a substance Band Bsuccessively decays into a third substance Cwith the same or different decay rates. B decays (. ) 2 decays C (. ) 1 A (4) Simultaneous disintegrations of parent nuclei: It dealswith the casewhen a parent nucleusmaydisintegrate in a number ofways into different products. A . B 1.2 C Parent nucleusAmay decay in B or Cwith decay constants .1 and .2 respectively. (5) Radioactive equilibrium: In a radioactive series, after a period of time, successive daughter nuclei decay at the same rate as it is formed. This situationis called radioactive equilibrium. (6) Disintegration of isotopes: Apreparationmayhave a number ofradioactive isotopes.Herewewilldealwiththe net rate ofdisintegration of the preparation. Let us discuss these types of disintegrations. Radioactive substance only disintegrates : Suppose disintegration ofAinto Bis taking placewithdecay constant ., then decayrate, � dN dt = .N . N = N0e�.t Disintegration with continuous production: Suppose a substanceAdecays into B with decay constant . and simultaneously the production ofAis taking place at a constant rate q. decays B (.) Formation (q) A Let, Nis the number of nuclei ofApresent at time t. q = constant rate of formation ofA. Disintegration ratewillbe given by � dN dt = �q + .N, where .N is the rate of decay ofA.
Rearranging,we get,
NUCLEAR PHYSICS www.physicsashok.in 23 dN .q . .N = �dt, Integration gives, 0 N N dN . .q . .N = � t0 .dt ,whereN0 is the no. of nuclei ofAinitially present. Finally, N = 1. [q + (.N0 � q)e�.t] Example 64. Aradionuclidewith disintegration constant . is produced in a reactor at a constant rate . nuclei per second.DuringeachdecayenergyE0 isreleased. 20%ofthis energyisutilised inincreasingthe temperature ofwater. find the increase in temperature ofmmass ofwater in time t. Specific heat ofwater is s.Assume that there is no loss of energythroughwater surface. Sol. Let Nbe the number of nuclei at any time t. Then Rate of formation = N Rate of decay = N . net rate of formationof nuclei at time t is dN dt = . � .N or N t 0 0 dN dt N . . . . . . or N = .. (1 � e�.t) Number of nuclei formed in time t = .t and number of nuclei left after time t .1 e..t . . . . . Therefore, number of nucleidisintegrated in time t t .1 e..t . . . . . . . . energy released till time . t . 0 t E t 1 e.. . . . . .. . . . . . . . But only20%of it is used in raising the temperature ofwater. So . t . 0 0.2 E t 1 e.. Q . . . .. . . . . . . . where Q = ms.. . ...= increase in temperature ofwater = Q ms . . t . 0 0.2 E t 1 e ms .. . . . .. . .. . .. .. . Example 65. Nuclei of a radioactive element Aare being produced at a constant ra
te .. The element has a decay constant ..At time t = 0 , there areN0 nucleiof the element. (a) Calculate the number Nof nuclei ofAat time t (b) If . =2N0., calculate the number of nucleiofAafter one half life ofA, and also the limiting value ofN as t ....
NUCLEAR PHYSICS www.physicsashok.in 24 Sol. (a) Let at time �t�, number of radioactive nuclei areN. Net rate of formation of nucleiofA. dN dt = . � .N or dN . . .N = dt or 0 N t N 0 dN dt N . . . . . Rate of formation = A Rate of decay = N t = t N = N Solving this equation,we get . . t 0 N . 1 .. . . . .N e.. . . . . ...(1) (b) (i) Substituting . = 2.N0 and t = t1/2 = ln (2) . in equation (1), we get 0 N 3 N 2 . (ii)Substituting . = 2.N0 and t ... in equation (1), we get N = .. = 2N0 or N = 2N0 Example 66. Aradionuclide with half life T is produced in a reactor at a constant rate q nuclei per second. During each decay, energyE0 is released. If production of radionuclide is started at t = 0, calculate. (i) rate of release of energyas function of time t and (ii) total energy released upto time t. Sol. To calculate rate of release of energy at time t and total energy released upto time t, rate of decayat that instant and totalnumber of decays upto that instantmust be known. Since, nucleiproduced are radioactive, therefore, their decay starts as soon as their production is started. Let at some instant number of nuclei in the radionuclide beN. Then rate ofits decay= .Nwhere . is decay constant which is equal to e log 2 T . . .. .. . Since, rate of production is q nuclei per second, therefore, at instant t, net rate of increase of nuclei e dN q N q Nlog 2 dt T . . . . . or e dN qT Nlog 2 dt T .
.
. e dN dt qT N log 2 T . . ...(1) Integrating above equationwith limits at t = 0, N= 0 and at t,N = ? N t 0 0 e dN dt qT N log 2 T . . . . . t loge 2 T e qT N 1 e log 2 . . . . . . . . .
NUCLEAR PHYSICS www.physicsashok.in 25 Hence, rate of decay, t loge 2 A N q 1 e T . . . . . . . . . . . Since, energyE0 releases during each decay, therefore, rate of release of energy at time t loge 2 T 0 0 t AE qE 1 e. . . . . . . . . . Total number of nuclei produced upto time t = q . t But the number of nuclei remaining undecayed at that instant isN. Therefore, total number of nucleiwhich decayed upto time t = (qt � N) Hence, total energy released upto this time = (qt �N)E0 t loge 2 0 T 0 e qTE qtE 1 e log 2 . . . . . . . . . . Successive disintegration : Suppose a radioactive substanceAdecays into B with decay constant .1 andB successivelydecays into another stable product Cwith a decay constant .2. decays C (stable product) .2 B .1 A decays Let N0 be the number of nuclei ofApresent at t = 0, N1, N2, N3 be the number of nuclei ofA, B and C respectively at anyinstant t. Decay rate ofAis given by dN dt . = .1N1 . N1 = N0e�.1t .....(i) Rate of change of no of nucleiofB is 2 dNdt = (Rate of decay ofA) � (Rate of decay of B) But, Rate of decay ofA= .1N1 and rate of decay of B = .2N2 hence, 2 dNdt = .1N1 � .2N2 .....(ii) Rate of change of number of nuclei ofC is 3 dNdt = (rate of decay ofB) = .2N2 . 3 dNdt = .2N2 .....(iii) Solving (i), (ii) and (iii),we get N1 = N0e�.1t N2 = 0 1 2 1 N .
. . . [e�.1t� e�.2t] N3 = N0 2t 1t 1 2 2 1 e e 1 .. .. . . .. . . . . . . . . .
NUCLEAR PHYSICS www.physicsashok.in 26 NOTE : In this case total number of nuclei remains constant, hence, N1 + N2 + N3 = N0 at any time. Example 67:Aradioactive nucleusAdecays to a nucleus B with a decay constant .1. B further decays to a stable nucleusCwith a decayconstant .2. Initiallythere are onlyAnuclei and their number isN0. Set up the rate equations for the populations ofA,BandC.The population ofBnucleus as function of time is given by N2(t) = 0 1 1t 2t 2 1 N (e e ) ( ) .. .. . . . . . . . . . . . . Calculate the population ofCas a function of time t. Sol: Let N1, N2 and N3 be the number ofA, B and C nuclei respectively present at a time t. Decay rate forAnucleiwill be � 1 dNdt = .1N1 . N1 = N0e�.1t Rate of change of the number of nucleiofB is 2 dNdt = .1N1 � .2N2 where .2N2 = decay rate of B and .1N1 = decay rate ofA Rate of change of the number of nuclei ofC is 3 dNdt = (Rate of decay of B) = .2N2 3 dNdt = .2N2 . N3 3 0 . dN = t 2 2 0 . . N dt , after integrating. . N3 = .2 t 2 0 . N dt but, N2 is given as, N2 = 0 1 1t 2t 2 1 N (e e ) ( ) .. .. . . . . . . . . . . . . N3 = 1 2 t 0 1 2 t t 2 1 0 N . . (e.. e.. )dt . . . . . = 0 1 2 2 1 N . . . . . 1 2
t t t 1 2 0 . e.. e.. . . . . . .. .. . = 0 1 2 2 1 N . . . . . 1t 2t 1 2 1 2 . e.. e.. . 1 1 .. . . . . .. . .. .. . .. . .. . N3 = 0 2 1 N . . . 2 t 1t 1 2 [. e.. . . e.. ] + N0 . N3 = N0 2t 1t 1 2 2 1 e e 1 .. .. . . .. . . . . . . . . . Simultaneous disintegration: Aradioactive nucleus can decay bytwo different processes. For example a nucleusAmay either a-decay to a nucleusB or .-decay to nucleus C. A . B 1.2 C Let .1 and .2 be the decay constants for these two decay processes.
NUCLEAR PHYSICS www.physicsashok.in 27 The probability that an active nucleus decays by the first process in a time interval dt is .1dt .As decay constant, . is defined as the probabilityofdecayper second for a particular process for a sample. Similarly, the probability that it decays by the second process is .2dt. Hence, the probability the it either decays by the first process or by the second process is .1dt + .2dt. If the effective decay constant is .eff, this probability is also equal to .effdt. Thus, .effdt = .1dt + .2dt . .eff = .1 + .2 For a number of different process for decay, .eff = .1 + .2 + ............ Example 68:Aradioactive nucleus can decay bytwo different processes. The half life for the first process is t1 and that for the second process is t2. Showthat the effective half-life t of the nucleus is given 1t = 1 1t + 2 1t . Sol: The decay constant for the first process is .1 = 1 ln 2 t For the second process, .2 = 2 ln 2 t Probability of decay by the first process = .1dt Probability of decay by the second process = .2dt Probability that it either decay by the first process or the second process = .1dt + .2dt This probability also equals to .effdt,where .eff is the effective decay constant. Thus, .effdt = .1dt + .2dt . .eff = .1 + .2 . ln 2 t = 1 ln 2 t + 2 ln 2 t . 1t = 1 1
t + 2 1t Proved. Example 69. Anumber N0 of atoms of a radio active element are placed inside a closed volume.The radiactive decayconstant for the nucleus of this element is .1. The daughter nucleus that formas a result of the decay process are assumed to be radioactive toowitha radioactive decayconstant .2.Determine the time variation of the number of such nucleus. Consider two limiting cases .1>> .2 and .1<< .2. Sol. In time intervaldt, number of increase ofdaughter nuclei are dN2 = .1N1dt � .2N2dt or dN2 = .1N0 e..1t dt � .2N2dt (N1 = N0 e..1t ) or 2 dNdt + .2N2 = .2N0 e..1t ...(1) Case-1 : When .1 > >..2 i.e. (t1/2)1 < < (t1/2)2 (t1/2 = half life) We can assume that N20 . N0 so that N2 = N0 e..2t (N20 = number of daughter atoms at time t = 0)
NUCLEAR PHYSICS www.physicsashok.in 28 Physicallythismeans that parent nucleipractivallyinstantlytransforminto daughter nuclei,which thendecay according to the lawof radioactive decaywith decay constant .2. Case-2 : When .1 < <..2 i.e. (t1/2)1 > > (t1/2)2 In this case number of parent nuclei can be assumed to remain constant over a sizable time interval and is equal to N0. This transforms equation (1) into 2 dNdt = � (.2N2 � .1N0) or N2 t 2 0 1 0 2 2 0 dN dt N N . . . . . . Which after integration gives N2 = 12 .. N0(1 � e..2t ) Example 70. Aradio nuclide consists of two isotopes.One of the isotopes decays by .-emission and the other by .-emissionwith half livesT1 = 405 second and T2 = 1620 second, respectively.At t = 0, probabilities of getting ..and ..particles fromthe radionuclide are equal. Calculate their respective probabilities at t = 1620 second. If at t = 0, total number of nuclie in the ratio-nuclide areN0, calculate time t when total number of nuclie remained undecayed becomes equal to 0 N2 . Given log10 2 = 0.30103, log10 13 = 1.11394 Sol. Since, at t = 0, probabilities of getting . and . particles fromthe radionuclide are equal, therefore, initial activities of two isotopes are equal. Let it beA0. Activity of first isotope at t = 1620 sec. t /T1 0 1 0 1 A A A 2 16 . . . . .. .. That of second isotope, t /T2 0 2 0 1 A A A 2 2 . . . . .. .. . Total activity of radionuclide at t = 1620 sec,A=A1 +A2 = 9 16 A0 . Probabilityofgetting .-particle, 1 1
P A 1 A 9 . . and that of getting .-particle, 2 2 P A 8 A 9 . . Let at t = 0, number of nuclei of two isotopes beN01 andN02, respectively. Initial activityof first isotope, 1 01 1 01 1 A N N log 2 T . . . That of second isotope, 2 02 2 02 2 A N N log 2 T . . .
NUCLEAR PHYSICS www.physicsashok.in 29 Since, A1 =A2, therefore 01 02 1 2 N N T T . or 01 1 02 2 N T 1 N T 4 . . Initially, total number of nuclei, N0 =N01 +N02 . 01 N 15 . N0 and 02 N 45 . N0 At time t, number of nucleiof first isotope that remain undecayed, t /T1 t / 405 1 01 0 N N 1 1 N 1 2 5 2 . . . . . . .. .. .. .. That of second isotope, t /T2 t /1620 2 02 0 N N 1 4 N 1 2 5 2 . . . . . . .. .. .. .. . Total number of nuclei remaining undecayed at time t, N = N1 + N2 = t / 405 0 N 1 5 2 . . .. .. + t /1620 0 4 N 1 5 2 . . .. .. = 4 t /1620 0 N 1 4 1 5 2 2 .. . . . . .. . . . . . ... . .. . . But it is equal to 0 N2 . . 4 t /1620 0 0 N 1 4 1 N 5 2 2 2 .. . . . . .. . . . . . . ... . .. . . or t /1620 1 8 2 13
. . . .. .. Taking log, t 1620 . log 2 = log 8 � log 13 or .log13 log8. 1620 t s log 2 . . . or .log13 3log 2. 1620 t s 1134 s log 2 . . . . Radioactive equilibrium : Decayof 238 92 U into a stable end product 206 82 Pb is aradioactive serieswhichcontains anumber ofintermediate members. The intermediatemembers of each decayseries havemuch shorter half-lives than their parent nuclide.As a result, ifwe start with a sample NA nuclei of a parent nuclideA, after a period of time an equilibriumsituationwill come about inwhich each successive daughter B, C, ........... decays at the same rate as it is formed. Thus the activities RA, RB, RC, ........... are all equal at equilibrium, and sinceR = .Nwe have.
NUCLEAR PHYSICS www.physicsashok.in 30 .ANA = .BNB = .CNC = ........... This situation is called radioactive equilibrium. The above equation can be used to establish the decay constant (or half-life) of anymember ofthe series if the decayconstant of anothermember and their relative proportions in a sample are known. Example 71: The atomic ratio between the uraniumisotope 238Uand 234Uin amineral sample is found to be 1.8 × 104. The half-life of 234U is 2.5 × 105 year. Find the half-life of 238U. Sol: The two isotopes are in radioactive equilibrium.Hence activities of the twowillbe equal. Thus, .1N1 = .2N2, where, .1, .2 are decay constant of 238U and 234Urespectively. N1 and N2 are number of atoms of 238U and 234U respectively, we have .2 = 5 ln 2 2.5.10 (year)�1, 12 NN = 1.8 × 104, .1 = ln 2 T , where T is half life of 238U. Now, .1N1 = .2N2 ... ln 2 T 12 NN . . . . . . = 5 ln 2 2.5.10 (year)�1 . T = 12 NN . . . . . . × 2.5 × 105 year = 1.8 × 104 × 2.5 × 105year . half life of 238U = 4.5 × 105 year Disintegration of isotopes : Suppose a sample is a mixture of three radioactive isotopesA, B and C. Let .1, .2 and .3 be the decay constant ofA, B and C respectively. N1, N2 and N3 be the number of nuclei of isotopesA, B and C respectively at any instant. IfA1,A2,A3 are decay rates ofA, B and C respectively then, the net decay rate is Anet = A1 + A2 + A3 but, A = .N .. .net N = .1N1 + .2N2 + .3N3 . .net = 1 1 2 2 3 3 N N N
N . . . . . Nis the total no. of nuclei in the sample at any instant . N = N1 + N2 + N3 hence net decay constant, .net = 1 1 2 2 3 3 1 2 3 N N N N N N . . . . . . . Example 72:Asample of uraniumis amixture of three isotopes 234 92 U, 235 92U and 238 92U present in atomic ratio of 0.006%, 0.71%and 99.284%respectively.The half life of these isotopes are 2.5 × 105 years, 7.1 × 108 years and 4.5 × 109 years respectively. Calculate the contribution of activity (in%) of each isotope in this sample.
NUCLEAR PHYSICS www.physicsashok.in 31 Sol: Let N1, N2 and N3 be the number of the three isotopes in the sample. we haveN1 : N2 : N3 = 0.006 : 0.71 : 99.284 .....(i) If .1, .2, .3 are decay constants of these isotopes and as activityA= .N, we have A1 :A2 :A3 = .1N1 : .2N2 : .3N3 A1 :A2 :A3 = 1 1 (ln 2)N T : 2 2 (ln 2)N T : 3 3 (ln 2)N T , [as . = ln 2 T ] . A1 :A2 :A3 = 1 1 NT : 2 2 NT = 3 3 NT .....(ii) where, T1, T2 and T3 are half lives of the isotopes. Comparing (i) and (ii) we get A1 :A2 :A3 = 1 0.006 T : 2 0.71 T : 3 99.284 T = 5 0.006 2.5.10 : 8 0.71 7.1.10 : 9 99.284 4.5.10 = 60 2.5 : 1 : 99.284 4.5 = 24 : 1 : 22.06 A1 :A2 :A3 = 24 : 1 : 22.06 = 24 100 24 1 22.06
.
. . : 1 100 24 1 22.06 . . . : 22.06 100 24 1 22.06 . . . = 2400 47.06 %: 100 47.06 %: 2206 47.06 = 51.00%: 2.12%: 46.88% Activityratio = 51%: 2.12%: 46.88% . Contribution of 234 92 U = 51%; Contribution of 235 92 U = 2.12%; and Contribution of 238 92 U = 46.88%in activityof the sample. Example 73. The isotopes of uraniumU238 andU235 occur in nature in the ratio 128 : 1.Assuming that at the time ofthe earth�s formationtheywere in equal ratio,make an estimate ofthe age of the earth.The half-lives of U238 and U235 are 4.5 × 109 years and 7.13 × 108 years, respectively. Sol. Let N0 be the initial number of atoms. Then fromN= N0e�.t, (ln 2/T1)t 1 0 N . N e. and (ln 2/ T2 )t 2 0 N . N e. . ln 2.1/T2 1/T1.t 1 2 N / N e . . ln 2.1/ 7.13 108 1/ 4.5 109 .t 128 e . . . . . 128 eln 2 1.18 10 9 t . . . . or 27 eln 2 1.18 10 9 t . . . . or t 7 1.18 . × 109 = 5.9 × 109 years
NUCLEAR PHYSICS www.physicsashok.in 32 Types of Nuclear Collision: Exoergic collision / reaction: IfQis positive, restmass energyis converted to kineticmass energy(K3,K4 etc), radiation energyor both, and the reactionis called exoergic. Note: (i) The kinetic energyKE. of the emitted .-particle is never quite equal to the disintegration energyQ because the nucleus recoils with a small amount of kinetic energywhen the .-particle emerges (since momentummust be conserved). . * KE A 4 Q A . . . ,Ais themass number of the parent nucleus. (ii) Themass numbers of nearlyallalpha-emitters exceed 210, and hence most ofthe disintegration energy appears as theKE of the .-particle. (iii) In the .�decay process, the energyQis shared bythe antineutrinos and the beta particle. The kinetic energy of the beta particle can be anything between zero andmaximumvalue ofQ. * K. + Ky = Q & p. = py p2 2m.. + 2y y p 2m = Q p2 2m.. y 1 mm. . . .. . .. . . = Q K. 1 4 A 4 . . . . . . . . = Q . K. = A 4 A. . . . . . .Q C63: Find the kinetic energy of the .-particle emitted in the decay 238Pu 234U+ .. The atomic masses of 238Pu = 238.04955 a.m.u.; of 234U = 234.04095 a.m.u; of 4He = 4.002603 a.m.u. Neglect any recoilof the residualnucleus. Sol: Using energyconservation,
m(238Pu)c2 = m(234U)c2 + m(4He)c2 + K or K = [m(238Pu) � m(234U) � m(4He)]c2 = (238.04955 a.m.u. � 234.04095 a.m.u. � 4.002603 a.m.u.]c2 = 0.0059970 a.m.u. × (931.5)MeV/a.m.u. = 5.59MeV C64: Neon-23 beta decay in the followingway: 23 10 Ne 23 11 Na + 01e . + . Find theminimumandmaximumkinetic energy that the beta particle 01e . canhave. The atomicmasses of 23Ne and 23Na are 22.9945 u and 22.9898 u, respectively. Sol: Reactant Product
NUCLEAR PHYSICS www.physicsashok.in 33 23 10 Ne 22.9945 � 10me 23 11 Na 22.9898 � 11me 01e . me Total 22.9945 � 10 me Total 22.9898 � 10 me Mass defect, .m= 22.9945 u � 22.9898 u = 0.0047 u . Q = .m c2 = (0.0047 u) × 931.5MeV/u) = 4.4MeV The .-particle and neutrino share this energy.Themaximumkinetic energyofa beta particle inthis decayis, therefore, 4.4MeVwhenthe antineutrino does not get anyshare. Energyof . particle can range from0 to 4.4MeV. Endoergic collisions (a) IfQis negative, the reaction is endoergic. (b) For endoergic reaction to take place aminimumenergy has to be supplied. (c) Threshold energyEth: Theminimumamount of energy that a bombarding particle must have in order to initiate an endoergic reaction, is called threshold energyEth. Usingmomentumconservationalso,we get Eth = �Q 12 m 1 m . . . . . . . , wherem1 =mass of the bombarding particle,m2 =mass of the target nucleus. m1c2 + m2c2 + K1 = (m3 + m4)c2 + K3 + K4 & p1 = p3 + p4; Q + K1 = K3 + K4 Q + 21 1 p 2m = 23 3 p 2m + 24 4 p 2m = 23 3 p 2m + 2 1 3 4 (p p ) 2m.
. 2Q + 21 1 p 2m = 23 3 4 p 1 1 m m . . . . . . . � 1 3 4 2p p m + 214 pm 23 3 4 p 1 1 m m . . . . . . . � 1 3 4 2p p m + 21 p 4 1 1 1 m m . . . . . . . � 2Q = 0 2124 4p m � 4 3 4 1 1 m m . . . . . . . 2 1 4 1 p 1 1 m m . . . . . . . . . . . . .
� 2Q . 0 21 1 p 2m . � Q 3 4 3 4 1 m m m m m . . . ; K1 . �Q 3 4 3 4 1 m m m m m . . . If m1 + m2 ~� � Q 12 1 mm . . . . . . . C65: Howmuch energymust a bombarding proton possess to cause the reaction 73 Li + 11H 74Be + 10 n atomicmasses of 7Li, 1H, 7Be and 10 n are 7.01600 u, 1.0783 u, 7.01693 u and 1.0866 u respectively. Sol: Since themass of an atominclude themasses of the atomic electrons, the appropriate number of electron
NUCLEAR PHYSICS www.physicsashok.in 34 massesmust be subtracted fromthe given values to getmasses of nuclei. . Q-value, = [m( 73 Li ) � 3me+m(11H ) � me]c2 � [m(74Be ) � 4me + m(10 n )]c2 = [m( 73 Li ) +m(11H ) �m(74Be ) �m(10n )]c2 = (8.02383 � 8.02559)u.c2 = � 0.00176 u × (931.5MeV/c2) = �1.65MeV Negative sign ofQindicates endoergic reaction. Energymust be supplied for this reaction to take place. The energyis supplied as kinetic energy of the bombarding proton. The incident protonmust havemore than this energy because the system must posses some kinetic energy even after the reaction, so that momentumis conserved.Withmomentumconservationtakeninto account, theminimumkinetic energyof the incident can be foundwiththe formula. Eth = � 1 mM . . . . . . .Q = � 1 17 . . . . . . . (�1.65 MeV) = 1.89 MeV. C66.Making use ofthe table of atomicmasses find the energies of the following reachings Li7(., n) B10 Sol. 3Li7 + 2He4 . 5B10 + 0n1 Q = (7.01601 + 4.00260) � (10.0124 + 1.00867) Q = � 0.00300 amu = � 2.79 MeV Example 74: Find the energyof the reactionN14(., p) O17, if the kinetic energy of the incoming .-particle is T. = 4.0MeV and protonoutgoingat anangle .=60º to themotiondirection of the alpha-particle has a kinetic energyTp = 2.09MeV. N . = 60º . TO Tp p X .. T. Y Sol: T..= 4.0 MeV, Tp = 2.09MeV Let TO is kinetic energyafter collisionof oxygen. Reaction is, 7N14 + 2He4 8O17 + 1p1 Let Q - value of reactionQ. Energy conservation gives, Q + T. = TO + Tp .....(i) Momentumconservationalong x-directiongives, 2m T . . = O O 2m T .cos. + p p 2m T .cos. . 2m T . . � p p 2m T .cos.. = O O 2m T .cos. .....(ii)
Momentumconservationalong y-direction gives, p p 2m T .sin..= O O 2m T .sin. .....(iii) Squaring the equation (ii) and (iii) on both sides and adding the result,we get, ( 2m T . . � p p 2m T .cos.)2+ 2mpTp sin2. = 2mOTO . 2m.T. � 2 p p (2m T )(2m T ) . . .cos. + 2mpTp(cos2. + sin2.) = 2mOTO . TO = O 1 m [m. T. + mpTp � 2 p p m m T T . . .cos.] .....(iv) Fromequation (i), Q = TO + Tp � T. .....(v) Putting the value TO fromequation (iv) in equation (v),we get,
NUCLEAR PHYSICS www.physicsashok.in 35 Q = Tp � T. + O 1 m [m. T. + mp Tp � 2 p p m m T T . . . cos.] = 2.09 MeV � 4.0 MeV + 1 16 [1 × 4 + 1 × 2.09 � 2 4.1.4.2.09 .co60º] MeV . Q = �1.14MeV NUCLEAR FISSION Continuous research or artificialtransmuttionand especiallythe studyof inducedradioactivity, culminated in the discovery of nuclear fissionwhich is accompanied by the release of enormous amounts of energy. In ordinarynuclear disintegrations, both natural and artificial, the nucleus isonlychipped off rather than broken and accordingly, the amount of energyreleased is comparatively less i.e. fromabout 10 to 23MeV. It was discovered in 1939 that the heavyunstable uraniumnucleuswhen bombarded byneutrons splits into two almost equalfragmentswhich flyapart with great speed and the amount of energyreleased per fission is about 200MeV. This division of a nucleus into two approximately equalparts as called nuclear fission. Discovery of fission The starting point in the discovery of nuclear fission can be traced to the attempts of Fermi in 1934, to produce transuranic elements by bombarding uraniumwith neutrons. However, the fission process itself was discovered in 1939 by German radio chemists Otto Hahn and his two associates Meitner and Strassmann.After bombarding uraniumwith neutrons, theyperformed a series of chemical separations to identify the products. To their great surprise, they found that the atoms produced bythe bombardment of uraniumbelonged to elementswhich lie near the centre ofthe periodic table.Obviously, a uraniumnucleus after capturing neutronhad become so unstable that instead ofdisintegrating byejecting oneor two particles, it had split up into two parts. The actualfissionprocess canbeunderstoodwiththe helpoffigurewhich shows a uraniumnucleus capturing a neutron. 0n1 92U236 92p 143n 92U236 .-rays Unstable Antimony Nuclide 51Sb133 51p 82n .-rays Unstable
Niobium Nuclide 41p 48n 41Nb99 (a) (b) (c) The newly-formed nucleus of figure (b) isunstable and starts breaking up into two parts. Inbreaking up, the uraniumnucleus, behaving like a liquid drop, splashes out smalldroplets, i.e. neutron and .-rays. So great is the release of energy that the two fission fragments fly apart in opposite directions with tremendous speeds. It amy, however, be noted that not all uraniumnuclei break into Sb and Nb as shown in figure.
NUCLEAR PHYSICS www.physicsashok.in 36 There are at least 30 different ways inwhich a fissile nuclide can divide itself. The experimental evidence seems to favour pairs of fissionfragments of unequalmasses (asymmetrical fission) accompanied byone to five or some time more neutrons. In general, fission fragments are unstable nuclei containing an excess number of neutrons.After a series of .-emissions in which neutrons are converted into protons in the nucleus, a stable nuclide results. Out ofall the neutrons ejected during the fission of uranium, about 99 per cent are ejected inan extremely short interval oftime and are called prompt neutrons. The remaining one per cent of neutrons are emitted a little later and are called delayed neutrons. The delayed neutrons originate fromunstable fragments that decay by neutron emission on theirway to becoming stable nuclei. Itmaybe noted that divisionofa fissile nucleus into three fragments ofcomparable sizes (ternaryfission) has been observed although it is a rare event, occurring about 5 times permillionbinaryfissions. Types of Fission Reactions Historicallyspeaking, uraniumwas the first element to undergo fission.However, soon after itwas found that other elementsof high atomicweight could also bemade to undergo fissionand that particles other than neutron could be equallyeffective inthis respect. Naturaluraniumcontains three principal isotopeswith the following relative abundance: U238 99.28% 4.51 × 109 Y U235 0.714% 7.1 × 108 Y U234 0.006% 2.48 × 105 Y It is found that slowneutrons cause fission ofU235 but not ofmore abundant isotopeU238whichrequires fast neutronswith energies exceeding 1MeV. Similarly,Th232 and Pa231 undergo fissionwhenbombardedwith fast neutrons. Fission can also be produced in uraniumand thoriumby high-energy .-particles, protons, protons, deuterons and .-rays etc.Two other nuclideswhich do not occur in nature but have proved to be fissionalbe by neutrons of all energies are 92U233 and 94Pu239. In 1947, successful fission of bismuth, lead, thallium,mercury, gold, platinumand tantalumwas achieved inUSAbymeans of ..-particles, deutrons and neutrons of 100 MeV and more.With bismuth (Z = 83) fission was detected with 50MeV deuterons whereas tantalum(Z= 73) required .-particles of 400MeVenergy. It isworthnothing that onlythree fissilematerialsU233,U235, Pu239 are important inthe large-scale application of nuclear energy. Finally, some heavynucleihave beenfound to undergo spontaneous fission. In this process, nucleus divides in the ground statewithout bombardment byparticles fromoutside. Mass distribution of Fission Products During uraniumfission, a large number of nuclides of intermediate charge andmass are found. Their study is a promising source of information about the mechanismof the fission process itself and also offers the
possibility of discovering hitherto unknown nuclides. Investigations of the fission products ofU235 have shown that the range of their mass numbers is from72 to 158.About 97%of theU235 nuclei undergoing fission give fragmentswhich fall into two groups as shown inthe fission yield curve of figure . (i) light groupwithmass numbers from85 to 104 and (ii) heavygroupwithmass numbers from130 to 149. Themost probable type of fissionwhich occurs in about 7%of the total cases, gives fission productswith mass numbers 95 and 139.
NUCLEAR PHYSICS www.physicsashok.in 37 60 80 100 120 140 160 180 10 100 1000 10000 95 139 Mass Number (A) Number of Fragments Asmentioned earlier, fission fragments have toomany neutrons in their nuclei for stability. Consequently, most ofthemdecay byelectron emission. Each fragment starts a short radio-active series involvingmany emission of .-particles. These series are called fission decay series and chain has threemembers on the average althoughlonger and shorter chains occur frequently.One suchfission decaychain is shownin figure which startswith one of the unstable fragments of the fission ofU235 nucleus. 51 82 51Sb133 5m Unstable Antimony 52 81 52Te133 60m Unstable Tellurium O . 53 80 53I133 5d Unstable Iodine O .� 54 79 54Xe133 5d Unstable Xenon O .� 55 78 55Cs133 Stable Cesium O . Energy Distribution of Fission Products Energy distribution among the fission products can be found bymeasuring their kinetic energywith the help of suitable ionization chambers. The results of such studyonU235 fissionhave shown that the energy distribution curve is not uniform; rather it is a doublepeaked
curvewithmaxima at 67MeVand 100MeV. It is seen that while the greatest probability is for a fragment of 100MeV, the areas under the two peaks which represent the total number of particles in the two groups are approximatelyequal. 0 100 200 300 40040 60 80 100 120 Energy (MeV) Number of frangments 67 100 Neutron Emission in Nuclear Fission One ofthe notable features ofnuclear fission is thatwhile it is initiated byneutrons it is also accompanied by the emissionof fast-moving neutrons. The number of neutrons released depends on themode of fission and on the energyof the neutronswhichinduce fission. The average values for the number of neutrons emitted per thermalneutron absorbed bythe three important fissilematerials are given below: U235 2.43 U233 2.50 Pu239 2.89 These neutrons are emitted bythe fission fragments and not bythe compound nucleus.
NUCLEAR PHYSICS www.physicsashok.in 38 The neutrons emitted as a result of fission process (i.e. fissionneutrons) can be divided into two groups: (i) Prompt Neutrons: Thesemake up about 99.36%of the total fission neutrons and are ejected by the product nucleiwithin 10�14 second of the fission process. Prompt .-rays are also emitted at the same time. (ii) DelayedNeutrons: These constitute about 0.64%of the totalneutrons fromthe fission ofU235.These are emittedwithgraduallydecreasing intensityfor severalminutes after actual fissionprocess.Although the number ofdelayed neutrons is small, theyhave a strong influence on the time-dependent behaviour of chainreacting systems based on fission and play an important role in the controlofnuclear-fission reactors. Fissile and Fissionable Nuclides Elements likeU235, U233 and Pu239 undergo fission by neutrons of energy fromalmost zero upwards. Such nuclei are referred to as fissile nuclides. On the other hand, U238 and Th232 nuclei which have a fission threshold at 1MeV are said to be fissionable nuclides. In general, fissile nuclides have either an even number of protons and an odd number of neutrons or odd numbers ofboth. Fissionalbe nuclides, onthe other hand, have either even number ofprotons and neutrons or an odd number of protons and an even number of neutrons. Fission Energy One ofthe striking features ofthe fissionprocess is themagnitude of the energyreleasedwhich is about 200 MeV per fission ofU235 nuclide. Before 1939, the largest known nuclear reaction energywas 22.2MeV associatedwith Li6 (d, .) He4 reaction. The amount ofenergyreleased per fissionofU235 nuclidemaybe calculated bythe following threemethods: (i) Binding-energymethod:Asmentioned above all stable fission products havemass numbers in the range 72 to 158where the average binding energyper nucleon is about 8.5MeV.However, in the neighbourhood of uranium, its value is 7.6MeV. Hence, average binding energy per nucleon is (8.5 � 7.6) = 0.9MeV greater inthe fission products thanin the compound nucleus ofU235. The excess energyis released as fission energy. Its value is 235 × 0.9 ~. 200MeVper fission ofU235 nuclide (which has 235 nucleons). (ii) MassDefectMethod: The energy released per fission can also be estimated bycomparing themass of the interacting particles and the final fission products. As mentioned, U235 splits in manyways and the nuclei obtained in the greatest yield in fission by slow neutrons havemass numbers of 95 and 139. The fissionproducts being initiallyradioactive, undergo many .-emissions to formultimatelystable nuclides. Ifmolybdenum-95 and lanthanum-139 are takenas pair of stable products fromfission ofU235, the fission reaction can bewritten as 92U235 + 0n1 62Mo95 + 57La139 + 20n1 Comparingmasses on both sides of the above equationwe get, mass ofU235 nuclide = 235.124 amu
mass of one neutron = 1.009 amu Total = 236.133 amu mass ofMo95 nuclide = 94.946 amu mass of La139 nuclide = 138.955 amu mass oftwo neutrons = 2.018 amu Total = 235.919 amu mass of defect = 236.133 � 235.919 = 0.214 amu Therefore, energy released per fission ofU235 nucleus = 0.214 × 931 ~. 200 MeV (iii) Kinetic energymeasurementmethod: The total amount ofenergyreleased per fissionis equal to the sum ofthe following energies: (a) the kinetic energy of fission fragments.As seenfromfigure the average value of this energy for U235 is 167MeV.
NUCLEAR PHYSICS www.physicsashok.in 39 (b) the kinetic energyof fissionneutrons. Since the average number of neutrons emitted per fission ofU235 is 2.43 or say 2.5 and the average kinetic energyof these neutrons is 2MeV, total kinetic energyof fission neutrons is 2.5 × 2 = 5MeV. (c) the kinetic energy of prompt .-rays. Its value is about 7MeV. (d) the total energy of the decay process in the fission decay chains. This includes the energy carried away byradiations like .-rays, .-rays and neutrons. Its value is nearly 21 MeV. The totalof all the above energies is = 167 + 5 + 7 + 21 = 200 MeV C67:AU235 nucleus is fissioned bya thermalneutron and two fission fragments and two neutrons are produced. Compute the fissionenergyreleased if the average binding energyper nucleon is 7.8MeVin fissionedU235 nucleus and 8.6MeVin the fission fragments. Sol: Greater binding energyof the fissionfragments indicates that there has been release ofenergyduring fission of low-binding energynucleusU235. Fission energy released is = (234 × 8.6 � 236 × 7.8) = 171.6 MeV Theory of Fission Process The first attempt to explain themechanismoffission processwasmade byBohr andWheelerwho accounted formanyof the properties offission on the basis of the liquid-dropmodelof the nucleus. 1 2 3 4 5 6 The shape of the drop depends on a balance betweenthe surface tension forces and Coulombic repulsive forces. The excitation energy given to the drop during the capture of the slow or thermalneutron sets up oscillationswithin the drop. These oscillations tend to distort the spherical shape so that the drop becomes ellipsoid in shape.The surface tension forces trytomake the dropreturn to its originalspherical shapewhile the excitation energytends to distort the shape stillfurther. Ifthe excitation energyandhence oscillations are sufficientlylarge, the drop pattains the �dumbbell� shape as shownin figure. TheCoulombic repulsive forces thenpush the two �bells� further apart untilthe dumbbell splits into two similar drops eachofwhich assumes a spherical shape. The sequence of steps leading to fission is shown in figure. However, if the excitation energy is not large enough, the ellipsoidwillreturn to the sphericalshape. In that case, the excitation energyis givenout in the formof .-rays and the process becomes a radioactive capture process rather than fission process. LECTURE � 5 Nuclear Reactors : (a) Anuclear rectoris a systemdesignedto controlthe chainreactionoffissionwithcontinuous energyproduction. (b) Ausefulfactor for describing the levelof operation of a reactor is the reproduction constant K. It is defined as the average of neutrons available fromeach fission that will cause another fission. For a controlled or self sustained chain reactionKmust bemaintained close to unit.
i.e. K. 1 for controlled chain reaction.
NUCLEAR PHYSICS www.physicsashok.in 40 (c) Fuel: This is the fissionablematerial.Commonly usedmaterials areU238 enriched inU235 and plutonium(Pu239). (d) Moderator:Fastmoving neutrons cannot triggerthe fission ofU235 and have a very high chance of being captured by U238 which is not fissionable. It is therefore necessary to usemoderators to slowdown the neutrons. Control rods Shield Moderator Fuel elements Nuclear Reactor (e) Coolant: Air, water or CO2 are used as a coolant to remove the heat released inside the reactor. (f) Control Rods: Cd(cadmium) which is a good absorber of neutrons is used to controlthe rate of fissionand also to shut down the reactor in case of emergency. (g) Types of reactors: (i) Thermal reactors: In these reactors fission is produced by slowneutrons ro thermalneutrons. (ii) Breeder reactor: Breeder reactors generally use fast neutrons in these reactorsU238 is converted into Pu239 by capture of neutrons. Pu239 is fissionable. Thus such reactors also produce fuel in addition to the energyreleased through fission. (h) Critical mass: For a fuel there is a criticalmass belowwhich the fissionablematerial is completely safe. But for amass above the criticalmassmore neutrons are produced than are lost so that the chain reaction builds up rapidlyand the systemexplodes. The atomic bombare therefore stocked as subcriticalmass such that the combinedmass is greater than the criticalmass resulting in a spontaneous explosion. Example 75: In a nuclear reactor, fission is produced in 1 g for U235 (235.0439u) by a slowneutron (1.0087 u). Assume that 35Kr92 (91.8973 u) and 56Ba141 (140.9139 u) are produced inall reaction and no energy is lost. (a) Write the complete reaction, (b) Calculate the energy released per fission, (c) Calculate the total energy produced in kilowatt hour.Given 1 u = 931.5MeV/c2. Sol: (a) The nuclear fission reaction is 92U235 + 0n1 56Ba141 + 36K92 + 3 0n1 (b) Mass defect, .m= [(mu + mn) � (mBa + mKr + 3mn) .m= 256.0526 u � 235.8373 u = 0.2153 u Energyreleased per fission, Q = 0.2153 u × 931.5MeV/u = 200.6MeV (c) Number of atoms in 1g = 6.02 1023 235 . = 2.56 × 1021 Energy released in fission of 1 g ofU235 = 200.6 × 2.56 × 1021 MeV = 5.14 × 1023 MeV
= (5.14 × 1023) × (1.6 × 10�13)J = 8.2 × 1010 J = 8.2 × 1010 W-s = 10 6 8.2 10 3.6 10 .. KWh = 2.28 × 104 KWh
NUCLEAR PHYSICS www.physicsashok.in 41 Example 76: In neutron-induced binary fission of 92U235 (235.044) two stable end-products usuallyfound are 42Mo98 (97.905) and 54Xe136 (135.917).Assuming that these isotopes have come fromthe original fission process, find (i) what elementary particles are released (ii)mass defect of the reaction (iii) the equivalent energy released. Sol: (i) The reaction can bewritten as 0n1 + 93U235 = 43Mo98 + 51Xe136 It is seen that the totalZ-value of the two stable fission products is (42 + 54) =96. It is 4 unitsmore than that of 92U235. For balance, the original unstable fission products must have got Z = 92. Obviously, the originalunstable productsmust have emitted 4 .-particles before becoming stable.Now,mass number on right-hand side is 2 units less than on the left-hand side. It means that towfission neutronsmust have been produced.Hence, the fission reaction canbe represented by the following equation: 0n1 + 92U235 = 42Mo98 + 54Xe136 + 4 �1e0 + 20n1 (ii) .m= LHS mass � RHS mass LHSmass = (1.009 + 235.055) = 236.053 amu RHSmass = (97.905 + 135.917 + 4 × 0.0055 + 2 × 1.009) = 235.842 amu . .m = 236.053 � 235.842 = 0.211 amu (iii) Energy released = 0.211 × 931 = 196MeV Example 77. About 180MeVenergy is releasedwhen one nucleus of 92U235 undergoes fission. estimate the energy released from1 kg ofU235, assuming that each nucleus undergoes fission. Sol. 1 kg of U235 = 1000 g = 1000 235 mole . number of atoms = 1000 235 × 6.02 × 1023 . total energy released = 6.02 235 × 1026 × 180MeV = 6.02 18 235. × 1027 × 106 × 1.6 × 10�19 J = 7.37 × 1013 J C68. Calculate the energy released in slowneutron capture by Pu239.Mass of Pu239 = 239.127 amu, Pu240 = 240.1291 amu, 0n1 = 1.008665 amu. Sol. Energy released =mass defect in energy units = (239.127 + 1.008665 � 2401) amu = 0.006565 × 931MeV = 6.1 MeV Example 78.Anuclear reactor generates P =20MWpower at efficiency. =60%bynuclear fission of a radionuclidewhose half life isT= 2.2 years. If each fission releases energyE = 200MeV, calculate time during which µ = 10mole of the radionuclidewill be consumed completely. (Avogadro number, N = 6 × 1023, loge 2 = 0.693, 1 year = 3.15 × 107 s)
NUCLEAR PHYSICS www.physicsashok.in 42 Sol. To operate the nuclear reactor, let the number of fissions required per second be n0. Thenenergy released per second byfission reactions = n0E Since, efficiencyofthe reactor is ., therefore, power output fromthe reactor = .n0E. But it is equal to P therefore, P = .n0E or 0 n PE . . Let at an instant number of nuclei of radionuclide be n then rate of decay= .nwhere . is decay constant which is equal to e log 2 T . Hence, net rate of decrease of nuclei 0 dn n n dt . . . . . . .. .. or e e dn n log 2 P n Elog 2 PT dt T E ET . . . . . .. . . . . . . . . . e dn dt PT n E log 2 Et . . . . . ...(1) At t = 0, number of nuclei are n = µNand time t is to be calculated when all the nuclei are consumed or when n = 0, t = ? Integrating equation (1)withthese limits, 0 t µN 0 e dn dt PT n E log 2 ET . . . . . . . . e 8 . . e e e T µN Elog 2 t log 1 10 log 1.0576 s log 2 PT . . . . . . . . . . Example 79. The element Curium 248 96 Cmhas a mean life of . = 1013 second. Its primary decaymodes are spontaneous fission and .-decay, the former with a probabilityof P1 = 8%and latter with a probability of P2 = 92%. each fission releasesE = 200MeVenergy. The masses involved in .-decay are as follows 248 96 Cm= 248.072220 u, 244 94 Pu = 244.06400 u and 42He = 4.002603 u.
Calculate the power output froma sample ofN = 1020 Cmatoms. (1 u = 931MeV/c2) Sol. Decayconstant, 1 10 13 sec 1 mean life ( ) . . . . . . Rate of decay froma sample of N atom,A= .N = 107 sec�1 Since, probabilities of fission and .-decay are P1 and P2 respectively, therefore, rate of decay due to fission, A1 = P1A or A1 = 8 × 105 sec�1 and rate of decay due to .-emission,A2 = P2A= 9.2 × 106 sec�1. Since, each fission releases energyE, therefore, rate of release of energydue to fission =A1 . E
NUCLEAR PHYSICS www.physicsashok.in 43 Equation of .-decay is 248 244 4 96 94 2 Cm... Pu . He Massmost during each .-decay, ..m= [248.072220 � (244.06400 + 4.002603)] u = 5.617 × 10�3 u . Energy released during each .-decay, E´ = 5.617 ×10�3 × 931 MeV E´ = 5.23MeV . Rate of release of energy due to .-decay =A2 . E´ . Total rate of release of energy =A1E +A2E´ But totalrate of release of energy is equal to power output. Therefore, power output, P = A1E + A2E´ = 3.33 × 10�5W Nuclear fusion It is the process of combining or fusing two lighter nuclei into a stable and heavier nuclide. Inthis case also, large amount of energyis released becausemass of the product nucleus is less than themasses oftwo nuclei which are fused. Many reactions between nuclei of lowmass numbers have been brought about by accelerating one or the other nucleus in a suitable manner. These are often fusion processes accompanied by release of energy. However, reactions involving artificially-accelerated particles cannot be regarded as ofmuch significance for the utilizationof nuclear energy.To have practicalvalue, fusion reactionsmust occur in suchamanner as tomake themself-sustaining, i.e.more energymust be released thanis consumed in initiating the reaction. It is thought that the energyliberated inthe Sun and other stars of themainsequence type is due to the nuclear fusion reactions occurring at the very high stellar temperature of 30million ºK. Suchprocesses are called thermonuclear reactions because they are temperature-dependent. Steller Thermonuclear Reactions: Following two sets of thermonuclear reactions have been proposed as sources of energy in the Sun and other stars of themain sequence: (i) proton-proton (p - p) chain and (ii) carbon-nitrogen (C-N) cycle. At lowtemperatures corresponding to those in the Sunwhenit was first formed, the proton-proton chain was predominant. In the present state of the Sun with its higher central temperature and larger He4 concentration, the C-Ncycle is supposed to be the main source of its energy. Proton-Proton Chain It is so called because the step involves the combination of two protons.When two protons fuse, they produce a deuteronnucleus, a positron and a neutrino thus: 1H1 + 1H1 = 1D2 + 1e0 + v × 2 The deutron thencombineswith another proton to yield helium-3. 1D2 + 1H1 = 2He2 + . × 2 The two helium-3 nuclei fuse to produce helium-4 thus 2He3 + 2He3 = 2He4 + 21H1 + 1e0 It should be noted that for the third reaction to occur, each of the first two reactionsmust occur twice. The net effect ofthe reactions is 41H1 = 2He4 + 21e0 + 2. + 2v
NUCLEAR PHYSICS www.physicsashok.in 44 Obviously, four hydrogen atoms are fused to produce one heliumatomwith a total energyrelease of about 26.7 MeV.When the kinetic energy of neutrions is substracted, the energy is 26.2 MeV. The emitted positrons are annihilated by free electronswith the production of .-rays. Carbon-Nitrogen Cycle It was proposed byH.A. Bethe to account for the energy production in the Sun and other stars ofmain sequence. In this cycle, carbon acts as a nuclear catalyst. The cycle startswhen a proton (hydrogen atom) first interactswith carbon-12 nucleuswiththe release of fusionenergythus 6C12 + 1H1 = 7N13 + . The product N13 is known to be radioactive, emitting a positronwith a half-life of 10minutes. Hence, it decays in a very short time according to the relation 7N13 = 6C13 + 1e0 + v The stable C13 nucleus reactswith another proton, therebyliberatingmore energy 6C13 + 1H1 = 7N14 + . The stable product N14 combineswith third proton thus 7N14 + 1H1 = 8O15 + . TheO15 nucleus is a positron emitterwith a half-life of 2.06minuteswhich decays by the process 8O15 = 7N15 + 1e0 + v Finally, the resultingN15 nucleus interactswiththe fourthproton thus: 7N15 + 1H1 = 6C12 + 2He4 By adding up the above six equations and cancelling out those nucleiwhich appear on both sides, it is seen that four hydrogen atoms are consumed and, in return, 2 positrons, 3 .-rays and one heliumnucleus are created. In otherwords, hydrogen is burned and heliumis created. The overall processmay bewritten as 41H1 = 2He4 + 21e0 + 24.7 MeV The annihilation of positrons supplies an additional energy of 2MeVso that total energyreleased is 26.7 MeV. The fusion energy releasedmayalso be found by the loss ofmass during the above reaction: 41H1 = 4 × 1.008144 = 4.032576 amu 2He4 = 4.003873 amu 21e0 = 2 × 0.000558 = 0.001115 amu . mass loss = 4.032576 � (4.003873 + 0.001115) or .m = 0.028857 amu . energy released = 931 × 0.028857 = 26.7MeV It is worthnoting that the above energy release is less than that in nuclear fission. However, its value is 26.7/4 = 6.7MeVper nucleon as compared to less than 1MeVper nucleon in fission process. Controlled Thermonuclear Reactions The fact that nuclear fusion reactions release large amounts of energy, as in stars, has attracted much attention and continuous search is beingmade for finding practicalmeans or controlled release of such energy. It has however, been found that reactions ofC-Ncycle and proton-proton chain occur too slowly to be of any practical use.Other thermonuclear reactionswhich occurmuchmore rapidlyand depend on aboundant hydrogen isotopes like deuteron (1D2 or 1H2) and tritium(1T3 or 1H3) a
nd hence seemmore practicalproposition, are as under: (i) 1H2 + 1H2 = 2He3 + 0n1 + 3.3 MeV
NUCLEAR PHYSICS NUCLEAR PHYSICS or 1D2+ 1D2 = 2He3+ 0n1+3.3MeV
(ii) 1H2+ 1H2 = 1H3+ 1H1+4MeV or 1D2+ 1D2 = 1T3+ 1H1+4MeV (iii) 1H2 + 1H3 = 2He4 + 0n1 + 17.6 MeV or 1D2+ 1T3 = 2He4+ 0n1+17.6MeV The most important aspect of the above nuclear fusion reactions is that deuterium is available easily and aboundantly. It occurs in nature with an aboundance of one part in six thousands of hydrogen and can be separated fromthe lighter isotope quite cheaply. Five litres of water contain about 1/8 gram of deuterium but its energy content if it could be used as a fuel in a thermo-nuclear reactor, would be equivalent to 130 litres of petrol! The more than 5 × 1019 kg of water present in the oceans could thus supply world�s power requirement for severalmillion years at negligible cost ifthe deuteriumcould be utilized to provide energyby fusion reactions. However, as discussed below, there are some difficult problems to be solved before man-made controlled fusion reactors can become a reality. Condition for Controlled Fusion In order to provide useful energy, the fusion process must be self-sustaining. Once the temperature of deuterium (or a mixture of D2 and T3) has been raised to the point at which fusion occurs at an appreciable rate, the energyreleased must be sufficient, at least, to maintain that temperature. The minimumtemperature is known as critical ignition temperature and may be defined as that temperature above which the rate of energy production by fusion exceeds the rate of energy loss. Its value is about 5 keV (i.e. 50 million ºK) for a D-D-reaction. At these temperatures, the atoms are entirely stripped of their electrons. The result is a completely ionized gas or plasma consisting of atomic nuclei (like deuterons, tritons and protons) and electrons in rapid random motion. It is practically impossible to contain such plasma in walls of ordinary materials. In the Sun, the fusion reactions are contained bya tremendous gravitational pressure. Such a high pressure is yet not available for controlled thermo-nuclear reactions on earth although the plasma can be contained in a magnetic field. Hence, main problem is to devise an apparatus in which plasma can be obtained by meansofa magnetic field at the kinetic temperaturesrequired for the fusionreactions to proceed. Another necessary condition for a self-sustaining thermonuclear system is known as Lawson criterion. It is based on the requirement that in the operation of a fusion reactor, the total usefulrecoverable energyshould be at least sufficient to maintain the temperature of te reacting nuclei. Lawson criterion can be expressed in terms of the product �nt� where nis the number of reacting nuclei per m3 and t is the time inseconds in which the thermonuclear reaction takes place. The minimum value of �nt� for D-T system is 7 × 1019 and D-D system is 2 × 1021. In controlled thermonuclear reactions, t is taken as the time during which the high-
temperature plasma can be confined. It will be seen from above that both the critical ignition temperature and Lawson criterion are much more favourablefor D-TsystemthanforD-Dsystem. Buttheformersystemhasthedraw-backthat itrequires tritium which has to be obtained by nuclear reactions (because it does not occur in nature). Tritium can, however, bemadebybombardinglithiumwithslowneutronsinareactorthus:
3Li6+ 0n1= 1H3+ 2He4+4.8MeV Thisreactioncanbemadetoservetwousefulpurposes.InthethermonuclearreactorusingD-T reaction, the escaping neutrons carry off much of the energy (about 14 MeV per neutron). This energy can be converted into heat byslowing down these fast neutrons in a blanket of berylliumsurrounding the reactor in whichplasmaisproduced. Theslowneutronsarecapturedbylithiumwhichproducestritium.Theblanket couldthusconsistofmoderator (i.e.beryllium),coolantandlithium.Theheatgeneratedbythemoderation and absorption of neutrons could thus be transferred by coolant to external heat exchangers and then to turbines whichcould run alternators.
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NUCLEAR PHYSICS www.physicsashok.in 46 Hydrogen bomb This bomb is 1000 timesmore powerfulthan the atomic bombwhich is based onnuclear fission.Hydrogen bomb is based on the fusion if the hydrogen atoms into heavier ones by the thermonuclear reactionswith release ofenormous energy.The essentialconditions for the operationof the hydrogen bombare extremely high temperatures and pressures required for the fusion to start.Once started, the fusionitselfmaintains the temperatures to keep the process going. For this purpose, the atombomb (fission bomb) is used as a primer which, byfirst exploding, provides the high temperature and pressure necessaryfor the successful working of the hydrogen bomb (fusion bomb). Afusion bomb is superior to a fission bomb because of the following reasons: (i) The energy release in a hydrogen bomb is open-ended i.e. it has no upper limit. It depends on howmuch fusiblematerial is present in the bomb. (ii) It hasno limitation of a critical size of the fusiblematerialunlike anatomic bomb. If the activematerial in an atomic bombexceeds the critical size, spontaneous explosion results.Hydrogen bomb cannot explode unless �ignited� i.e., heated to critical ignition temperature and anyamount of fusiblematerial is safe until ignited. Thus the amount of fusiblematerialin a hydrogen bombis not limited. Cobalt Bomb It consists of a hydrogen bomb which is encased in a sheath of metallic cobalt and is more lethal and destructive thana simple uncased hydrogenbomb.When the hydrogenbomb explodes, it gives off neutrons which act on the cobalt cover and render it intensivelyradioactive due to the formation ofCo60 that is 300 timesmore powerfulthan radium.During explosion, the radioactive cobalt is pulverised and converted into a gigantic radioactive cloudwhich canspread over thousands ofkilometres killing everything living in that area. Fission and Fusion: One thing common between the two nuclear processes si that they release very large amounts of energy. But there aremanydifferences inthemechanisms of the two processes. (i) Fission involves breaking up of a heavy nucleus into lighter nuclei. Fusion, on the other hand, involves combining oftwo lighter nuclei into on heavynucleus. (ii) The links of the fission process are neutronswhile the links of a fusion process are protons. (iii) Fission proceeds best with thermalneutronswhere thermalmeans roomtemperature. Fusion proceeds best withthermalparticleswhere thermalmeans temperatures ofmillions of ºK. Example 80: the masses of 1H1 and 2He4 atoms are 1.00813 amu and 4.00386 amu respectively. Howmuch hydrogenmust be converted to heliumif solar constant is 1.35 kW/m2 anthe earth is 1.5 × 1011mfromthe Sun. Sol: This thermo-nuclear reactionmay bewritten, inits essentials as 41H1 = 2He4 + 1e0 Neglecting the two positrons, it is seen that 4 hydrogen atoms fuse to produce o
ne atomof helium. Mass of 4 hydrogen atoms = 4 × 1.00813 = 4.03252 amu Mass ofone heliumatom = 4.00386 amu Decrease inmass, .m = 4.03252 � 4.00386 = 0.002866 amu energy produced = 0.02866 × 931 = 26.68 MeV This is the energy released when four hydrogen atoms fuse. Hence, energy produced by one hydrogen atom. = 26.68 4 = 6.67 MeV = 6.67 × 1.6 × 10�13 = 10.67 × 10�13J = 6.02 × 1026 × 10.67 × 10�13 = 6.42 × 1014 J
NUCLEAR PHYSICS www.physicsashok.in 47 The solar constant represents the amount of energy received per second by 1m2 area held perpendicular to the Sun�s rays at a distance equal to the mean distance of the earth fromthe Sun. Sun 1.5 × 1011 Given value of solar constant = 1.35 kW/m2 = 1.35 × 1000 W/m3 = 1350 J/s/m2 Total energyemitted bythe Sunis equal to the energyreceived bythe inner surface of the imaginary sphere drawnwith Sun as centre and radius = 1.5 × 1011m. Surface area of the sphere = 4.R2 = 4. × (1.5 × 1011)2 = 28.28 × 1022 m2 . energy received by this surface area per second is = 1350 × 28.28 × 1022 = 38.18 × 1025 J/s It also represents the energy emitted by the Sun per second. Mass of hydrogen consumed is = 25 14 38.18 10 6.42 10 . . = 5.59 × 1011 kg/s = 5.95 × 108 tones/second C69: Calculate the energyliberatedwhen aHeliumnucleus is formed bythe fusion of two deuteriumnuclei. The mass 1H2 = 2.01478 a.m.u. and mass of 2He4 = 4.00388 amu. Sol: The reactionmaybewritten as 1H2 + 1H2 = 2He4 + Q . 2.01478 + 2.01478 = 4.00388 + Q , . Q = 0.02568 amu = 0.02568 × 931 = 23.9 MeV C70. In the fusion reaction 1H2 + 1H2 . 2He3 + 0n1, deuteron, heliumand the neutron havemasses 2.015 amu, 3.017 amu and 1.009 amu, respectively.Estimate the total energyreleased if 1 kg of deuteriumundergoes complete fusion. Sol. Mass difference = 2 × 2.015 � (3.017 + 1.009) = 0.004 amu . energy released = 0.004 × 931MeV = 3.724MeV Energy released per deuteron = 12 × 3.724MeV No. of deuterons in 1 kg = 6.02 1026 2. . energy released per kg = 1.862 × 3.01 × 1026 MeV = 5.6 × 1026 × 106 × 1.6 × 10�19 . 9 × 10 C71. In some stars, three 2He4 nuclides fuse together to form6C12 ofmass 12.0000 amu. Howmuch energy is released per fusion of 6C12 ? Rest mass of 2He4 = 4.002603 amu. Find also the rate of consumption of heliumto maintain the radiative power of the star at 4 × 1021MW.mass of 2He4 atom= 6.9 × 10�27 kg. Sol. Energy released per fusion =mass defect in amu = (3 × 4.00263 � 12.0000) amu = 0.007809 × 931MeV = 7.27 MeV . energy released by n atoms7.27 1.6 10 19 106 n J
3 . . . . . . = 4 × 1021 × 106 (given) , n = 1.103163 × 1040 . mass of heliumatoms burnt per second = 1.03163 × 1040 × 6.9 × 10�27 kg = 7.1 × 1013 kg
