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Question No. 4 (a) (i) Energy due to height

(ii) Initial position of the student in Diagram 9.1 is higherThe speed of the student on reaching the final position inDiagram 9.1 is higherThe higher the speed of the student, the higher the energyThe higher the position , the higher the energy gained

(iii) Principle of Conservation of Energy(b) (i) elastic potential energy to kinetic energy andgravitational potential energyelastic potential energy and kinetic energy togravitational potential energy

(ii) due to dampingwork done against air resistance // loss of energy due to the extension and compression of the molecules in the system

(c) strong // high strengthit wont break easilyhigh force constantstore higher elastic potential energystrong // high strength // any suitable materialit wont break easily

thin // aerodynamic // streamlineless air resistanceslightly above the targetthe path is parabolic / curved/downward // due to free fallmotion

Question No. 5

(a) (i) Priciple of conservation of momentum(ii) Air expels at high velocity

Produces momentum backwards / to the leftCauses thrust of momentum forward/ to the rightWith same magnitude of force but in opposite direction

(b) (i)

(ii) a = v ut

= 0 42

= 2 ms -2 (iii) s =

)2()4(21 2

= 16 m(c)Characteristics ReasonsShape Inverted aerofoil High pressure at top // greater

downward force//high stabilityRidges on tyres -none Less friction during motionEngine power -high Greater forward thrustMaterial light and stiff Does not break easily on impact

I choose SBecause shape inverted aerofoil , ridges on tyres none, enginepower high and material light and stiff

CHAPTER 3

Question No. 6

(a) (i) Sum of two or more forces to produce one resultant force.(ii)

1. Bottle in liquid Q floats lower

2. Weight and the buoyant force are equal and the same inboth cases.3. Density of liquid P is higher.4. As the density of liquid decreases, the lower the bottle

floats.5. When density of liquid decreases the volume of liquid

displaced increases to produce the same buoyant force.(b)

1. When force is applied to piston A2. Pressure is produced and transmitted uniformly throughout

the liquid towards piston B // Pascals Principle3. Pressure multiply by the surface area of piston B will

produce the output force that lift load M.4. Cross-sectional area of piston A is smaller than piston B to

produce large output force.

(c)

Suggestion ReasonsFix a long handle onpiston A

Small force can producebigger force /torque to presspiston A

Equip the hydraulic jack with valves

Liquid can flow in onedirection and does not back flow

Use released valves Liquid flows back to thestorage reservoir

Piston A is madesmaller // piston Bbigger.

To produce large outputforce.

Oil as liquid Does not evaporate easily // does not flow out easily// prevent from rust.

Question No. 7 (a) (i) Force per unit area

(ii) Pressure on piston Q = pressure on piston RCross sectional area of piston Q < Cross sectional area of piston RForce acted on piston Q < force acted on piston RThe greater the area, the greater the forcePascals principle

(b) aerofoil shapeThe air speed on the upper surface > the air speed on the lower

surface.The pressure on the lower surface > pressure on the uppersurface.Different in pressure produce lift force.

(c)

Design Reasons

Thick wall Water pressure increase withdepth

High density material Increase mass / can submergeeasilyStrong material Does not break easilyAerodynamic shape Reduce water resistanceBallast tank To increase or reduce weightSonar transmitter andreceiver To estimate distance and depth

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Question No. 8 (a) Upwards force cause by weight of fluid displaced by an object

when the object is immerse in the fluid.(b) Valve release air from ballast tank.

Sea water flooded ballast tank The weight of water displaced is smaller.

Buoyant force < Weight of the submarine(c)Choices ReasonsThe volume of ballasttank is higher

To produce bigger buoyant force / easy to rise up / can carrygreater weight inside

More number of airtanks cylinder carried

Can stay longer time under thewater / Can rise and submergemany times / more air supply forrespiration of crews

Can withstand highermaximum waterpressure

Safe when the submarinesubmerge very deep in the sea / The body will not break due tohigh water pressure

The suitable shape of submarine isaerodynamic

Less water resistance in themotion / can move faster

Submarine XBecause the volume of ballast tank is higher, more number of air tanks cylinder carried, can withstand higher maximum waterpressure and the suitable shape of submarine is aerodynamic.

(d) (i) V = 0.2 x 0.8= 0.16 m 3

(ii) B = V g = 0.16 x 1000 x 10= 1600N

(iii) mass = 1600 10 = 160 kg

Question No. 9 (a) Aerofoil

(b) (i) The shape of cross section of the wing causes the speedof airflow above the wings to be higher than the speed of airflow below.When the speed of moving air is higher the pressure islower.Hence air pressure below the wings is higher compare toabove the wings.

(ii) Bernoullis Principle(c)

The most suitable wing is Y.Because it has a shape of cross section which is upper side islonger than the bottom, large the area of the wing, the lowdensity of the wing materials and the high the difference inspeed of air.

(d) (i) P=F/A

F = 400 x 50F = 20000 N(ii) Resultant Force = 20 000 900(10)

= 11000 NDirection of force : upwards

Question No. 10 (a) (i) Buoyant force equal to weight of fluid displaced

(ii) Volume of air displaced equal to volume of aballoon

Density of air decreased as a altitude increaseWeight of displaced air become smallerAt certain height weight of displaced air equal toweight of the balloon

(iii)Large balloonTo produce bigger buoyant / up thrust // Increase thevolume of the air displacedLow density of gas filled in the balloonLighterSynthetic nylonLight-weight, strong and air-proof materialLow mass of radiosonde instrumentTo produce a bigger upward resultan forceR is choosen

Large ballon/low density of gas /synthetic nylon / low mass of radiosonde instrument

(b) (i) mass = density x volumem=0.216kg

(ii) V= 1.2m3m= 1.56kg= 15.6N

Question No. 11(a) (i) Buoyant force equal to weight of fluid displaced

(ii) Volume of air displaced equal to volume of aballoonDensity of air decreased as a altitude increaseWeight of displaced air become smallerAt certain height weight of displaced air equal toweight of the balloon

(iii)

Characteristics ExplainationLarge ballon To produce bigger buoyant /

upthrust // Increase thevolume of the air displaced

Use 2 burners // Manyburners

To produce bigger flame // heat up the gas in theballoon faster

Synthetic nylon Light-weight/ strong /air-proof material

High temperature of the air in the balloon Reduce the density /weightof the air in the balloonHot air balloon Q ischosen

Large balloon, use 2 burners / many burners, usesynthetic nylon and has hightemperature of the air in theballoon.

(b) (i) mass = density x volumem=0.216kg

(ii) V= 1.2m3m= 1.56kg= 15.6N

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CHAPTER 4

Question No. 12 (a) (i) The temperature in which a solid substance change to

liquid at atmospheric pressure(ii) The mass of substance in Diagram 9.1 < in Diagram 9.2

Time taken to reach the melting point in Diagram 9.1 9.2Size of image in Diagram 9.1 > 9.2Power of lens is same

The shorter the object distance ,the greater the size of image.(b) (i)

(ii) Range greater than two focal length ( u 2f)

(c) (i)Modification Reason

f u 2f Image magnified and realDistance = f 0 + f e Eyes are not strained// to get

normal adjustment microscope25 cm

f o = 10 cm and f e= 20 cm First image is magnifiedPosition of the firstimage < f e

Final image is magnified manytimes

(ii) use less power// longer focal length of eye piece lensMove the object nearer to the focal point (still in between(1F and 2F)

Question No. 21 (a) (i) Distance from the optical centre to a focal point.

(ii) Lens K is thicker than lens JLight ray is refracted more in lens K than lens J.Focal length of lens K is shorter than lens J.Therefore the thicker the lens, the greater the refractionof light and with that the shorter the focal length of a lenswill be

(b) (i) As a magnifying glass.

(ii) Enlarge/magnifiedInverted/upside downVirtual

(c) (i) Choose lens S as objective lensChoose lens Q as eyepieceLens S is placed in front of lens QThe two lenses are adjusted so that they are in normaladjustment where distance between the two lenses isequal to (fo + fe)

(ii)Modification Explanation

1. Use low powerconvex lens as theobjective lens.

- Magnification of telescope =

e

o

f f

,

Low power lens has a longerfocal length, fo , magnification

2. Use high powerconvex lens as the eyelens

High power lens has a shorterfocal length, fe , magnification

3.Use bigger diameterof objective lens

More light permitted to enter thetelescope and a clearer image isseen

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Question No. 22 (a) (i) Refraction

(ii) Ratio of sin i / sin r // The ratio of speed of light invacuum relative to that speed through a medium

(b) Refractive index of the glass is higher.The density of glass is higherThe angle of refraction of light ray in glass is shorterThe higher the density of medium, the smaller the angle of refraction of light.The higher the density of medium, the higher the refractiveindex.

(c)

(d) (i)

(ii)

Modifications ReasonsObjective lens withlargerdiameter.

More light passesthrough the lens

Eyepiece lens withhigherpower // Thickereyepiece lens

Act as a strongmagnifying glass

WAVES

Question No. 23 (a) (i) Interference of waves

(ii) wavelength depend on depth of waterincreasing/decreasing of depth will cause the wave torefractrefraction will increase/decrease the length of wavelengththe nodal/antinodal line will be affected

(iii) wavelength produce by the two spherical dipper indiagram 10.1 is the same as in diagram 10.2the distance between the two consecutive antinodesis greater in diagram 10.1 than 10.2the distance between the two consecutive nodes isgreater in diagram 10.1 than 10.2increase in the distance between the two coherent

source will decrease the distance between the twoconsecutive nodes or antinodes(b) (i) sonar is a reflection from an ultrasonic waves. (ultrasonic

echoes)ultrasonic wave has a higher frequency waves , it hasmore energy so it can move furtherit do not produce noise

(iii) attach ultrasonic transmitter to a shipuse a microphone receiver to detect ultrasonic pulsesdirect the ultrasonic pulses from the transmitter to theseabeduse microphone receiver to pick up the reflected pulsesfrom the seabedmeasure the time taken by the pulses to travel to theseabed and return

calculate the depth of the water using the formula

2txv

d = Question No. 24 (a) (i) Refraction

(ii) wave move from deeper to shallow areathe wavelength decreasethe speed decreasethe direction of wave bends towards normal

(b)

(c) (i) f = v

= 6

0.8= 7.5 Hz

(ii) V = f = 0.5 x 7.5= 3.75 cm

Question No. 25 (a) (i) To ensure constant velocity of water wave / ripple.

(ii) Interference(iii) Zero or minimum(iv) zero // minimum // smaller than the original amplitude

(b) (i) The distance in Diagram 10.1 is higher(ii) The distance in Diagram 10.1 is smaller(iii) Same(iv) The higher the a , the smaller the x // a inversely

Proportional(c) (i) Increase the number of loudspeakers // use more

Loudspeakers

Characteristics ReasonBay Wave is calmerConcrete retainingwall

Stronger/ / lasting

High wall Prevent highwave

Smaller opening/ aperture / slit of retaining wall

Diffractionobvious // lowwave energy

P At bay, concreteretaining wall ,high wall andsmaller slit

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To produce louder sound // Sound transmitted covers abigger

(ii) Placed at all corners // different locationsCover a bigger area // the spectators around the area canhear clearly

(iii) High amplificationto produce louder sound // sound can travel further

Large/bigLess diffraction // spreading // sound travels straightLarge/bigVibrates more air // produces louder sound

Question No. 26 (a) When two crests or two troughs meet each other to form

maximum amplitude(b) (i) Distance between two coherent sources in Diagram 10.2

is bigger than diagram 10.1(ii) The wavelength are the same(iii) The distance between two adjacent antinodal lines in

diagram 10.2 is smaller than diagram 10.1(c) As the distance between two coherent sources increases, the

distance between two adjacent antinodal lines decreases.

(d) Refraction of light rayLight travels from less dense to a denser mediumtowards the normal / bending closer.In the observer eyes, the sun is still not setting as they can stillsee the image from refraction.

(e)

Question No. 27 (a) A wave in which the vibration of particles in the medium is

parallel to the direction of propagation of the wave

(b) (i) The amplitude in Diagram 10.2 is higher

(ii) The peak value, a 2 in Diagram 10.2 is higher(iii) The higher the amplitude of vibration of tuning forks, the

higher the peak value(iv) The higher the peak value, the louder the sound(v) The higher the amplitude, the louder the sound

(c) - Use ultrasound- Ultrasound is transmitted to the sea bed- a receiver will then detect the reflected the reflected pulses- the time taken by the pulse to travel to the seabed and returnto the receiver being recorded, t- the depth of the sea can be calculated using the formula,

2

vtd =

(d)

Suggestions ReasonLoudspeakers are

positioned at quite adistance away.

So that the distancebetween consecutiveconstructive / destructive interference issmaller.

The two main

loudspeakers are notpositioned oppositeto each other

To prevent multiple

reflections

Fix soft boards/ wooden/ materialswhich are soundabsorbers

Reflection effects can bereduced

Use thick carpet/ Wooden floor/ Rubber floor

To prevent echo

Assemble a highpower speakersystem

To produce a highamplitude of sound wave

Assemble thespeaker at a high place

Wide coverage // thewave is not blocked

CHAPTER 7

Question No. 28 (a) Distance between focal pint and the optical centre of a lens

(b) 1. The convex lens is aimed/focused to a distant object(infinity)2. The screen is adjusted until a sharp image is formed on thescreen3. The distance between the screen and the lens is measured4. Focal length = distance between the screen and the lens

(c)

(d) (i)1 = 1 + 1

f u v1 = 1 1v 5 400v = 5.063 cm

(ii)h2 = v 2h1 v1

h2 = 5.063100 400

h2 = 1.27 cm(iii) Real , inverted and diminished.

Aspect ExplanationLonger focallength

To produce real, inverted andsmaller image

Highmagnification

Produce bigger image

Distance = fo + fe Produce image at normaladjustment // image at infinity

Bigger diameter More light can enter objectivelens // more brighter

J is chosen Longer focal length, highermagnification, Distance between

two lenses = fo + fe and biggerdiameter

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Question No. 29 (a) High melting point

Can withstand high temperature / heat // does not melt easily Not easily oxidizedCan last longer

(b) (i) Maximum current flowing through the fuse is 0.5 A(ii) 1 Low melting point

2 Can melt easily3 Low specific heat capacity4 Can heat up easily5 Small6 High resistance // more heat released // easy to break 7 High8 High resistance // more heat released9 Z10 Low melting point, low specific heat capacity, small

diameter and high resistivity

(c) (i)240500

= 2.08 A(ii) Not suitable

(iii) 500 x (10x60)300 000 J

Question No. 30 (a) A fuse is a very thin wire, which either melts or vaporizes when

too much current flows through it(b) - A parallel circuit can run several devices using the full

voltage of the supply.- If one device fails, the others will continue running normally- If the device shorts, the other devices will receive no voltage,preventing overload damage.- A failure of one component does not lead to the failure of theother components.- More components may be added in parallel without the needfor more voltage .- Each electrical appliance in the circuit has it own switch.

(c) (i) The electrical appliance use 240 V of voltage to generates500 W of power.

(ii) Current = Power/VoltageCurrent = 500/240 = 2.08 A

(iii) Efficiency = Output Power x 100 %Input PowerOutput Power = 85 x 500

100Output power = 425 W

(d)Characteristics Explanation

Thin fuse wire Less space needed/ to carry a limitedelectrical current/ less mass hence lowheat capacity/ shorter time to heat up tomelting point and blow.

Ceramic cartridge Can withstand higher temperaturebecause sparks created by high voltage,240V can be huge/

Fuse rating is 13 A Maximum rating must be higher thannormal current.

Low melting point For fast blow/ Melting faster whenexcessive current flow/ Easy to cut thecurrent flow.

R is chosen because Because it has thin fuse wire, ceramiccartridge, fuse rating is 13 A and lowmelting point.

CHAPTER 8

Question No. 31 (a) An electromagnet is a coil wrapped around a soft iron core

which will be magnetized when a current flows through the coiland demagnetized when the current is switched off

(b) Amount of iron filings attracted in Diagram 10.1 is smaller /

less// amount of iron filings attracted in Diagram 10.2 is higher / moreCurrent flow in Diagram 10.1 is smaller / less

// current flow in Diagram 10.2 is higher / moreMagnetic field strength in Diagram 10.1 is weaker

// Magnetic field strength in Diagram 10.2 is strongerWhen the current is higher, the magnetic field strength isstronger, or vice versaWhen the magnetic field strength is stronger, the amount of iron filings attracted is more, or vice versa

(c) When current flow through the solenoid, a magnetic field isproducedThe (soft) iron core will be magnetizedThe scrap metal attracted to the iron coreWhen the current is switched off, the soft iron core will be

demagnetised and the scrap metal falls down(d) Transformer step down

To reduce the voltage from 240 V to 12 VThe ratio of number of turns in primary coil to the number of turns in secondary coil is 240 : 12 // 20 : 1To reduce the voltage from 240 V to 12 VType of core is soft iron core // laminated iron coreSoft iron core is easy to magnetized and demagnetized // Laminated iron core to reduce eddy currentDiode is connected to the outputTo convert alternating current output to direct currentUse copper wireGood electrical conductor // has low resistance // produce low

heat

Question No. 32 (a) Electrical energy light energy(b) (i) 1. When an a.c. voltage is supplied to the primary coil,

the soft - iron core is magnetized2. The magnet produced varies in magnitude and

direction3. This causes a changing magnetic flux to pass through

the secondary coil4. Induced e.m.f across the secondary coil is produced.

(c)Aspect Explanation

Soft iron core Easy to magnetized anddemagnetised

Laminated Less eddy current/ reduce energy

lostThick wire Reduce the resistance/morecurrent

Copper wire Low resistance/ reduce the lost of heat

Q Soft iron core, Laminated , thick wire, Copper wire

(d) (i) 12 V(ii) Np = 240 x 200 = 4000

12(iii) Efficiency = Po x 100

Pi= 240 x 0.2 x 100 = 100%

48

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Question No. 33 (a) Magnet produce by current(b) (i) Current, I directly proportional to the number of paper

clips attached, n The strength of electromagnet increases(ii) The magnitude of the current / number of batteries (dry

cells)The use of soft iron core

(c) High strength magnetsLarger force acting on the coilHigh number of turns of coilLarger force acting on the coil / h igher electromagneticfield strengthLow density materialCoil has smaller mass / lighterUsing 4 diodes and a capacitor4 diodes to produce full-wave rectification and capacitoras a smootherY and high number of turns of coil, low density, highstrength of magnets and 4 diodes and a capacitor.

(d) (i) Aluminium

(ii)

30

240

= 8

(iii)

30240

= 1920 W

Question No. 34 (a) (i) Step-down

(ii) An alternating current flows through primary coilThe soft -iron core is magnetized.The magnet produced varies in magnitude and direction.This causes a changing magnetic flux to pass through the

secondary coil.An induced e.m.f across the secondary coil is produced.

(b) I = 24/12= 2A (with unit)

Efficiency = Output power x 100 %Input power

= 24 x 10040

= 60W(c)

Characteristics ReasonsThick

Soft iron

Laminated

Low// near//shorter

Reduce the resistance of the coil

Reduce the hysteresis los s// can be magnetizedand demagnetized easily// Little energy used tobe magnetized and demagnetized .

Make the iron core as insulation.Eddy current are not able to flow through thelayers of insulation// reduce eddy current

Reduce leakage of magnetic flux

Choose P because copper wire thick, soft iron core, laminated coreand distance is low.

Question No. 35 (a) (i) Current produced when the magnetic flux is cut by

conductor.(b) (i) No. of turns of the solenoid in Diagram 10.2 is greater.

(ii) the deflection of the galvanometer pointer in Diagram10.2 is bigger / larger.

(iii) the rate of cutting of the magnetic flux in Diagram 10.2

is greater.(iv) Magnitude of induced current in Diagram10.2 is greater.(c) (i) No of turns increases, rate of cutting of magnetic flux

increases.(ii) Rate of cutting increases, magnitude of induced current

increases.(d) Faradays Law(e) (i) The greater the transmission voltage, the smaller the

current in the power lines.Power loss during transmission due to resistance will besmaller. / P=I 2R

(ii)

1. laminated soft ironcore

2. to reduce the effect of eddy current

3. thick copper wire 4. reduce heat loss due toresistance

5. the ratio of turns inprimary coil to thenumber of turns insecondary coil is < 1

6. to reduce output voltage to18.5 V

7. Wind the secondarycoil on top of theprimary coil

8. to reduce the leakage of magnetic flux

9. 2/4 diodes 10. to produce full-waverectification

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Question No. 36

Question No. 37 (a) (i) Electromagnet is a device in which magnetism is

produced by an electric current // temporary magnetmade by winding a coil round a soft iron core andmagnetic field produce when current flow.

(ii) Current flow in diagram 10.2 is more than 10.1 // viceversa.The amount of iron filing attracted by iron core inDiagram 10.2 is more than 10.1 // vice versa.The magnetic strength in diagram 10.2 is more than 10.1

// vice versa.Amount of iron filing attracted increase when currentincreaseThe magnetic field strength increase when currentincrease

(b) When too high current flow, magnetic field strength becomevery strong / wire expandelectromagnet pull the soft iron armature / pulled to the rightby spring P.release the catch, contact separate and current does not flow

When reset button is pressed, spring Q pulls the soft ironarmature back to its original position(c) (i) When the coil rotates the coil cut across the magnetic

field linesInduced current flow in the coil.The current maximum when the coil cut the magneticfield at right angle // current decreased (become zero)when the coil move in parallel with magnetic field linesThe direction of current flow determine by usingFlemings right hand rule

After 90o

the direction of current in the external circuitreversed/ diagram(ii)

Modification ExplanationChange slip rings withcommutator

To reverse contact withbrushes so that the current flowin same direction in externalcircuit

Use stronger magnet To increase the magnetic fieldstrength

Use more number of turn forthe coil/ Increase the speed of rotation

Increase the rate of change of magnetic field/increase theinduced current

TOPIC 9

Question No. 38 (a) Semiconductors are a group of materials that can be conduct

electricity better than insulators but not as good as metalconductors

(b) n- type semiconductorp-type semiconductorFor n-type, majority charge-carries are electrons / minoritycharge-carries are positive holes but for p-type, majoritycharge-carries are positive holes / majority charge-carries areelectronsFor n-type, the doping substance is pentavalance but for p-type,the doping substance is trivalance

(c)

Substance Q is chosenBecause Q has low resistivity at 00C, high melting point,valency of the doping substance is 3 and the size of the dopingsubstance is almost the same as the size of the substance

(d) (i) Thermionic emission

(ii) Velocity of the electron,2

21

mveV =

31

19

101.94000106.12

= x

x x xv

= 3.75 x 10 7 m s -1

Characteristics ExplanationThe resistivity at 0 0C islow

A current can flow through itmore easily// Increasing the conductivity of thesubstance

High melting point Not melting at high temperatureThe valency of thedoping substance are 3and 5

Valency of 3 is used to make p-type semiconductor and valencyof 5 is used to make n-typesemiconductor

Size of the atom of thedoping substance isalmost the same as thesize of the substance

Can maintain the crystal-linestructure of the substance// Give good effect in the dopingprocess

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Question No. 39 (a) Doping is a process of adding a certain amount of specific

impurities to semiconductors to increase their conductivity(b) Diagram 10.1, the p end of diode is connected to negative

terminal of dry cell // Diagram 10.2 the p end of diode isconnected to positive terminal of dry cellBulb in Diagram 10.1 does not lights upNo current flow in Diagram 10.1 // Current flow in Diagram10.2The bulb will lights up when the p end of diode is connected tothe positive terminal of dry cell // vice versaCurrent only flow in the circuit when p end of diode isconnected to positive terminal of dry cell or in forward bias

(c) (i) Draw correct symbol and correct direction(ii) Draw smooth waveform

Capasitors store charge when current flowCapasitor discharge when current does not flow

(d) 1 LDR is replace by termistor / diagram

2 resistance decrease when temperature increase3 Termistor is place at R 1 and R 1 replace LDR / diagram4 So that V across R 1 increase when the room is hot5 Relay switch replace LED6 To switch on secondary circuit /to switch on the fan7 Fans are arranged parallel8 All fans received 240 V power supply / other fans still

functioning even though one fan did not function9 Resistor connected to the base of transistor10 limit the current to the transistor

Question No. 40 (a) (i) Circuit consists transistor, resistor and cell

(ii) 10.1 : Microammeter no reading // 0 A,milliammeter no reading // 0 A

(iii) 10.2 : Microammeter has a reading // 1 x 10 -6 A ,milliammeter has a reading // 1 x 10 -3 A

(iv) Change in microammeter reading small, change inmilliammeter reading is biggerIb Increase, I c increase // I c depends on I b A small change in I b caused a big change in I c

(b) At night resistance LDR increasesVBE increasesIb increases and switch on transistorIc increases and lights up bulb

(c)

Modification Explaination

Replace LDR with termistor To detect heat when temperatureis high

Replace bulb with siren / bell To produce sound

Connect relay switch to outputtransistor

To switch on the siren

Interchange the position of resistor R and termistor

To increase base voltage / / voltage across R // base current

Use 240 V power supply Siren is function at high voltage

Question No. 41 (a) Semiconductor is a material with electrical conductivity better

than an insulator but weaker than a conductor.

(b) (i)

(ii) connect a capacitor in parallel to the output(iii)

(iv)

(c)

(d) (i) current a.c(ii) 4V(iii) T = 4 x 0.1 =0.4s

Frequency, f = 1/T= 1/0.4= 2.5 Hz

(iv)

Question No. 42 (a) Unstable isotopes which decay and emit radioactive particles /

ray(b) Radioisotope is injected into the pipe

G-M tube as detector is used to find the leakageReading on detector increases when near a leakage

(c)1. type of ray isgamma

2. high penetration powercan penetrate through thepipes

3. short half life 4. Decay faster anddetection can be donefaster

5. in liquid state 6. Easy to flow/ dissolve

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