1978-AL-Phy MCQ Sol (Lantco)

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HONG KONG ADVANCED LEVEL EXAMINATION

(B)

a~

)

(D)

/

a . ~ '

1978

. { Sg - T = SaT - 3g = 3a:. Sg - 3g = Sa + 3a

2 . = Ba

a=!

PHYS ICS

Suggested Solutions

5g

2. The force exerted by cord on pulley = 2 T. . Sg T = Sa

: . Sg - T = sc!)

T = 4

· Force = 2(154g)

= 15 g2

1.

3. The change in velocity = Vy - Vx

. . Vy = Vx = V~ V = Vy 2 + Vx 2 - 2VyVx cos 30°

2V 2- 2V2 cos 30 °

2V2(1 - cos 30°)

~ = 2V sin 15°



(D)

\,

4. The centripetal force= m w2

The component of weight= m g cos e· The total tension=m w2 + m g cos e

reT

I

I ?w

I

I (E)mg

5. At the north pole, gp = go

At the equator, due to the rotation of the sphere

ge - go =rw

2

gp > ge

· The weight decreases from north pole to equator.

(C)

6. In the insulated container, the work done incompressing the gas is used to increase the internal

energy of the gas. So the t e ~ e r a t u r € would increase.(E)

obeys the ideal gas equation. Because the gasPV = nRT

The slope for Cc)

2 P:

P V = ~ R TM

V_mR

T - ~ P

2 m of gas of pressure

f. V = 2m RT2 M

The slope for

.2



(C)

(C)

(A)

V 2 m RT = M E.

2V _ 4 m oRI-MP

· The slope of the line is 4 times that of (c)

From the graph, line (A) is correct.

8. By ideal gas equation, PV = n R TInitially, PlY = mlRT . . . . . (1)

Finally, the pressure of the gas = Pv - P2:. (P2 - Pv)V = m2R T . . . . . (2)(2) P2 - Pv _ ID2

TIT PI - ml= (P2 Pv)ml

ffi2 PI

9. Dew is formed as the temperature of the air falls inthe night, so that the saturated vapour in the airwill cQndense on the grass. But cloud will prevent

the radiation of heat energy from the air and thetemperature does not fall to the dew point.

10. The question is out of syllabusP = A 0 T4

• P•• u = AT4

The unit of 0 = w M-2 K-4 (D)

11. Usually, the width of acar is about 2 ffi.

25 x 10- 3 =

tan 3 x

tan 2.5 x 10- 3 = !x

x = 400 IDx

2m

9.-3

5xl0 rad

(B)

3



12. By lens fonnula,1 + 1 _ 1Y y + 20 - 24Y + 20 + Y _ 1y(y + 20) - 2448y + 480 = Y + 20y

Y - 28y - 480 = 0

y = 40 cm or -12cm (rejected)(E)

(B)

13. l ~ e n i = A, the light ray is symmetry in the prism,so the angle of deviation is the minimum.

Angle of deviation = B = minimum

(E)

(D)

15_ In microscope, the lenses used are converging lens.For high regular magnification, the focal length ofthe objective is shorter than the eyepiece.

14. When angle i increases from '0° to 8, the angle r isgreater than the critical angle, so total internalreflection occurs. Then when the angle i increases

from to I ' deviation cl decreases to a minimum

within 8 and I ·

16. For S.H.M. of the springIk 2 7T I ill

w = 1 ill T = W = 27T1 k

: . T a vm(C)

4



17. The veloci!y of sound in string,

v = / = 84 illS - 1

For stretched string,

f- 1. IT- TI ID

... {140 = 1 (84 ) 0.6 IT\ J'_ 1

210 - I2 (84) I ~ {

AI = 0.6 ID _.

A2 = 0.4 m

· The length of the string is o. 6 m

(A)

18.

(D)

When the wavelength shifts to the blue end, thewavelength decreases. As wavelength i, frequency +,so the star is approaching from the earth.By Doppler's Effect,

f' = (V c)fc _ ( c )c"f' - v-::c J:

1 3 x 10 1(480 - 0.2) x 10-9 = Cv - 3 x 10 8)(480 x 10-9)

480 3 x 10 8

479.8 = V - 3 x 10 8

3 x 108

(479.8) 480V - (3 x 108

)480V = 125000 m s-1

= 125 km s-1

19. Because when the mica is placed at S I , the light raythrough SI travels longer than that of 52, the pathdifference for light rays are largero So in order

to produce interference pattern, the light ray from52 has to travel longer, the central fringe moves toA.

(A)

5



20. Let. end-correction be cA4 - c = 19 (1)

{ 31t

'4 - c = 59 (2)

(2) - (1) ¥- := 40

It = 80 an

21. Let the thickness of the foil be d

2d = nit• • . • • • . •

(1){ 2 (2d) := n' (3t) . . (2)

( 2 ) . 2 = ~ ( ~ )l IT· n 5

n' 10 n- -3-

(E)

(E)

19cm

24v'

3.0..

_...----ot I 11••\ 1-----

~ - - - f V I - - ~1""'-&4_ .. _ 1 2 V _ - - - 4 ~

When K1 is closed, the p.d.across 3 n is 12 V whichis not equal to 24 V, sointernal resistance existsin the battery (R).

24( - - ) - 3 = 123 + R

R = 3 n

When both keys, K1 and K2 ,

are closed1

R =------ + 3 = 5 n1 1 '3+6

The current i := := ¥- = 4.8 A

The p.d. across the resistance

=24

-4.8(3)

. = 906 V

22.

(C)

6



h_ io

23. The B-field due to i 2 on PQ -

The force on PQ = i B_ il i2 ~ 1 1 0- 2nr

... Force is proportional to i 1 i 2.

Because the current i 1 and i 2 flows in the same

direction, the wires attract each other. So theforce is in direction 2.

(D)

24. Because the resistance in the rectifier is zero, sothe power dissipated in i t is zero.Before rectification, the power dissipated in thelamps is 50 W ( equal to the shaded area )

.p

But after rectification, only half cycle will the

current flows through the circuit, so the powerdissipated will be half of 50 W= 25 W.p;

(D)



25. The resistance ofAX = 4000 x } = 1000 Q

The total resistance of the circuit ,1

R = 1 1 + (4000 - 1000)

500 + 1000= 3333.3 [2

360 1The p.d. across AX = 3333.3 x ( .1 1)

500 + 1000= 36 V

(B)

26. The e.m.f. induced in the inductord i

s = a t

• The p.d. across i t is maximum when is maximum.

F h h h 1 d i h" h " .rom t e grap , t e s ope = at w lC IS maxImum at

point ell(C)

27. When the circuit is purely resistive, the powerdissipated in Reat in the resistor is equal to thepower supply by the battery. So the energy convertedby unit charge round the circuit is equal to the e.m.f.of the cell .

(E)

28. Because the current i 4 sin w t which is alternating.. . 10 4

l r m s = ; Z = /Z

· The power dissipated = (i r In s) 2R4 ?

(72) ~ 8 R

(B)

29. For the magnetometer,Opporing Torque = moment of inertia x argular

acceleration- i B l e = I e

8



·0 '

10

.. i B e=-,--e

I t represents the equation of a S.H.M.

W= l i By I

f = 27rf = 2TTjiR-

/13f Cl; I

(E)

30. The electrostatic force on both the positive rays andthe positron is the same F = qE

· TIle acceleration of positive ions, al =! ml

The acceleration of positrons, a2 = ! -m2

(B)

31. The electrostatic force,FE qE

Vqa

100q 0.0254000 q

The gravitational force,Fg = m g

= m g

• FE = tan eFg

· tan 100 4000 gmg

9. g tan 100

m 4000The charged particle attracts to the positive plateq _ g tan 10°ID - - 4000

(A)

9



32. The peak voltage of the a.c. source = 10 x 3= 30 V

The root mean square value that measured by thevoltmeter

30 = 21 V=/Z

(A)

33. N = No e-At

tn N - ~ n N o -Attn N = ~ n - At• ~ n a t

· when. Q , l } N ~ ,

is plotted against t , the graph is astraight line.

(B)

34. We can give the direction of motion by theFleming's left I-land rule.

(E)

35. Fa =' BqaVFS = BqSV

Fa = =•. FS qS 1

(B)

36. The error in k

=error

in a+

error inb + }

error in c+ error

incl

(E) is correct because 4% error in d= 2% error in c only.

(E)

37.

(C)

10



38. (1) TRUE

By law of floation, the upthrust on P is equal tothe weight of P. As the lead shots are placed intothe beaker Q from P, the weight of P decreases andso is the upthrust.

(2) TRUE

(3) TRUEBecause as the shots in P.mg = VIPwg

VI = .E!-pW

Shots in beaker Q_ mV2 - pr

p > pw

• V2 < Vl· Volume of water displaced will decrease and

the height H falls.(A)

39 (1) TRUE

Vo + VtThe average speed = t... t -r 00

:. Average speed -+ 0

(2) TRUE

The total distance travelled =Vt= area under the graph

... Area under the graph is finite

:. The total distance is finite(3) FALSE

The body comes to rest (speed = 0 ms-I)When time is infinity.

(B)

40. (1) FALSEThe e A ~ a n s i o n of all fluids are not linear.

(2) FALSENot all thermometers agree at these temperatures,but we calculate that all thermometers agree at

11



these points.

(3) TRUE

(E)

(C)

41. By ideal gas equation,PV = nRT

At the same temperature, pressure, volume, the numberof moles of gas is fixed.

(1) FALSEThe expression does not relate to the mass of thegas.

(2) TRUE

(3) TRUE

The avogadro no. = 6 x 10 23 molecules/mole· For same no.. of moles of gas, the no.. of

molecules is the same.

42. The question is out of syllabus.

(1) TRUE

For a real gas,Cp -C v=R

: . > 1

(2) FALSEFor the idea gas, Y > 1

(3) FALSE

y for a diatomic gas < y for a rnonoatomic gas.Because the internal energy of the diatomic gasinvolves t ranslational kinetic energy, vibrationalenergy and rotational energy.

r",'1

......,r I

(D)

I r

r!.....

I

i I

t 1--..J...jC---------

sin 90°n2 = sin r

. 1Sln r = -n2sin i _ n2sin r -. nl

43.

12



nl - sin isin 90·S ln 1

sin i n2 = n2nl

1

(E)

i ...

_ C}.l1 - Vl

_ C}.l2 - V2

· : V1>V 1:. }.l2 < }.ll

(3) FALSE

The diagram shows the situation, so i t ispossible. (D)

(1) FALSEThe critical angle equal to i .

(2) FALSE

The critical angle is independent of n2 and t .

(3) TRUE

The expression does not involve n2.

44. (1) TRUE

sin i _VIsrnr - V

2. . r > i

•. V2 > VI

(2) FALSE

45. (1) TRUE

Longitudinal waves have already been polarised.

(2) TRUE

(3) FALSEInterference causes the colors seen in oil film.

(B)

13



N.

46. (1) TRUE

(2) TRUE

(3) FALSE

s

.s

Direction is away from the equator because HongKong situates at the north hemisphere.

(D)

(C)

++

B.

/ //

-------..... ++ C.+. +.-----

A

+ + + ' ' - - - j L - - - : - ~ - : - - : - /--47. (1) FALSE

A has a surplus ofpositive charge ..

(2) TRUE

Diagram shows thatB has a surplus ofnegative charge.

(3) TRUE

14



48. (1) TRUE

••• E =

c

e

• u n i t o f E i s Nc- 1

(2) TRUE

c V As VUnit o f cR = (V) (ft) = (V ) (p) = s

(3) TRUE

E = cV2

2Ec = V2

Unit o f c = JV-2

(A)

49. (1) FALSE

A dark li n e o f wavelength 589 nm i s seen inthe spectrum.

(2) & (3) TRUE

This is the phonomenum is th e a bso rptionspectrum.

(C)

1978-AL-Phy MCQ Sol (Lantco)

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