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02 AL Physics/M.C./P.1
2002 Hong Kong Advanced Level Examination
AL Physics
Multiple Choice Questions (Solution)
1.
D
Before the cord is cut:
X is in equilibrium state
 tension (spring) = mg
Y is in equilibrium state
 tension (cord) = mg + tension (spring)
= 2mg
At the instant the cord is cut:
It takes time for the spring to contract; X is
in equilibrium state momentarily, i.e.
acceleration of X = 0.
Force acting on Y = weight + tension
(spring) = 2mg. Hence, acceleration of Y
= 2g.
2.
C
Separation between P and Q is maximum
when P is just hitting the ground.
Time for P to reach the ground is given by:
1
s = ut  gt 2
2
1 2
= gt
2
80 = 5t2
t =4s
Position of Q when P is hitting the ground:
1
s = u(t  1)  g (t  1) 2
2
1
2
= g (t  1)
2
= 5(4-1)2
= 45 m
Hence, max. separation between P and Q
= 80 – 45 = 35 m
3.
B
1m=n
wavelength for orange-red light ~ 650 nm
hence, n ~ 1.5  106
n is of order 106.
4.
B
(1) False
Motion of the mass between B and C
is s.h.m. (vertical spring and mass
oscillation). In s.h.m. kinetic energy
is maximum when the mass is passing
through the equilibrium position, i.e.
the position where mg = kx.
(2) False
C is the extremity of s.h.m.,
acceleration is maximum  resultant
force acting on the mass is maximum.
(3) True
mg
is the equilibrium position. As
k
the mass is passing through the
equilibrium position with k.e., the
spring will be compressed further.
5.
A

T
T

F
mg
F

mg
Fmin
Forces acting the bob includes the weight,
tension and F. As the bob is in static
equilibrium, the resultant force of these
three forces is zero. This implies the
vectors of these three forces form a
triangle as shown in the above figure. As
angle  is fixed, F is minimum when it is
perpendicular to the tension.
Hence,
minimum F = mg sin .
02 AL Physics/M.C./P.2
6.
D
(1) momentum
by conservation of momentum:
momentum (projectile) + momentum
(target) = constant
10. C
A. a = -2x
B. F  -x
C.
final momentum of P = +ve
final momentum of Q = 0
final momentum of R = -ve
hence, momentum of Z has the great
magnitude
(2) kinetic energy
elastic collision
 total k.e. conserved.
i.e. k.e. (P, Q or R) + k.e. (X, Y or Z)
= constant.
k.e. = total energy – p.e.
1
= E0 - kx2
2
D. v =  A 2  x 2
11. B
center of
mass
l/2
k.e. must be positive.
k.e. of the target is maximum when
k.e. of the projectile is zero.
7.
A
impulse = Ft
p
F =
t
 impulse = p
8.
9.
C
Owing to change of direction of the
velocity, an object undergoing circular
motion must have acceleration in the
direction pointing towards the center of
the circular path (centripetal acceleration).
If the motion is a uniform circular motion,
centripetal acceleration will be the only
acceleration of the object (Figure of option
(1)). However, if the motion is nonuniform, there must be a tangential
acceleration which is due to the change of
“magnitude” of the velocity (Figure of
option (2)).
A
By conservation of energy
Total energy = k.e. + p.e.
3E 1 2
Hence,
E =
 kx
4 2
E
1 2
kx =
2
4
1 1 2
= ( kA )
4 2
A
x =
2
by conservation of energy
1
mgh = I2
2
1 2
mgh = I
2
0.1  10  (0.3/2) = 0.5  3  10-3  2
 = 10 rads-1
12. D
A and C. True
For an orbiting satellite
G

v2
Mm
= m
2
r
r
1
k.e. = mv 2
2
1 Mm
= G
2
r
Total mechanical energy E of a satellite
orbiting the earth
= p.e. + k.e.
Mm 1 Mm
= G
 G
r
2
r
1 Mm
=  G
2
r
Due to air resistance, total mechanical
energy E decreases (i.e. more –ve), hence,
r decreases. As r decreases, k.e. increases
and v increases.
02 AL Physics/M.C./P.3
B. True
2
v
Mm
= m
r
r2
angular momentum = mvr
G

= m GMr
Hence, angular momentum decreases as r
decreases.
D. False
Time needed for the satellite to complete
2r
one revolution =
.
v
r decreases and v increases, the time
needed should decrease.
16. A
Remark: The following three figures show
the emerging light rays of a concave lens
for virtual objects located at: (1) positions
within focal length f; (2) focus and (3)
positions farther than focal length f.
L
(1)
f
L
(2)
13. A
P
Q
f
time
(3)
L
1

4
period later than P. Hence P leads Q by

90 or .
2
Q reaches the maximum at a time
f
14. D
Ultrasonic wave is longitudinal wave.
Polarization cannot be observed for
longitudinal wave.
15. D
For diffraction grating, angular position of
the first-order maximum is given by:
d sin  = 
hence, angle  between the first-order
maxima is given by

 = 2 = 2 sin 1
d
Hence, to maximize , long wavelength
light should be used (e.g. red light) and d
should be as small as possible (i.e.
increases the no. of slit per mm of the
grating.)
L1
L2
20 cm
final image
20 cm
The parallel light rays after passing lens L1
will converge towards a point 40 cm on
the right of L1. For L2 these light rays
form a virtual object at a location 20 cm
on the right of L2, i.e. in between L2 and
the focus. The light rays after passing lens
L2 will converge to a point a little further
away. Hence, the final image is real and
formed on the right of L2.
02 AL Physics/M.C./P.4
17. C
Condition for formation of stationary
waves in a fixed ends string:

l = n( ) where n = 1, 2, 3, …
2
c
= n( )
2f
Hence,

c=
fn
400
n
=
n 1
450
n =8
2lf
2  1 400
=
= 100 ms-1
n
8
18. B
(1) False
Number of waves emitted per second
from the source = frequency.
(2) False
Moving observer Doppler effect
(3) True
Moving source Doppler effect
19. D
According to the diagram the following
features about the spectrum are observed:
a. positions of the spectral lines do not
change;
b. the spectral lines widen.
A. False. In an expanding universe, the
stars move away from the earth.
Wavelength should increase and the
observed spectra should shift toward
the red end (red shift).
B. False. If the star is approaching the
earth, wavelength should decrease and
the observed spectra should shift
toward the violet end.
C. False. Same as A.
D. True. Light sources on the spinning
star are moving with different
velocities, which depend on how far
away the sources from the rotating
axis is, towards/away from the earth.
Hence, the width of the spectral lines
spreads out without change of the
mean position.
20. C
A. False.
B. False. As the path taken by the charge
is convex away from Q, the two
charges must be unlike charges and
the force between them is attractive.
C. True. The fact that force between Q
and q is attractive implies potential
energy is negative. As point X is
closer to Q than Y, q has a more
negative potential energy at X than Y.
By conservation of energy, k.e. + p.e.
= constant.
At X, p.e. is more
negative  k.e. at X is greater, i.e.
greater speed.
D. False. Whether X is at a potential
higher or lower than Y depends on
polarity of Q. If Q is positive, X is at
a potential higher than Y. If Q is
negative, X is at a potential more
negative (i.e. lower) than Y.
21. A
A. True. In a magnetic field the electron
can undergo circular motion, i.e. it
can be deflected by more than 90
while in electric field the electron can
be almost deflected by 90.
B. False.
In both the electric and
magnetic field, the force depends on
the magnitude of the charge; they are
qE and qvB.
C. False. If the region of the uniform
magnetic field is sufficiently large, the
electron will undergo uniform circular
motion. The speed of the electron
will then be constant.
D. False. If the field is magnetic, the
direction of the magnetic force, which
is always perpendicular the velocity of
the moving charge, will change
continuously.
22. B
02 AL Physics/M.C./P.5
23. D
Q: As the battery have been disconnected,
the charge stored on the capacitor
must remain constant.
0 A
, if one of the plates of the
d
capacitor is shifted upward slightly, A
decreases and C decreases.
26. C
time constant  = RC
2  10-3 = 1  103  C
C = 2 F
C: C =
Q
, Q remains constant and C
C
decreases, hence, V should increase.
V: V =
24. D
V1 + V2 = 9 V
Hence, V2 = 5 V.
Let I1, I2 and I3 be the currents passing V1,
V2 and V3 respectively.
Since, all the three voltmeters are with the
same internal resistance. Vi  Ii
27. D
By Faraday’s Law,
d
 =
dt
dA
= B
dt
= Blsin 60  v
=
3Blv
2
28. A
induced e.mf.  enclosed area A
resistance  length l
hence, current  A/l
9V
current (circle) : current (square)
R
A
I1
B
V3
V1
4V
S
I3
C I2
( d / 2) 2 d 2
:
d
4d
= 1:1
=
5V
V2
I3 should flow in the direction as shown in
the above figure.
By Kirchoff’s first law: I1 + I3 = I2
hence, V1 + V3 = V2
and
V3 = 1 V.
29. A
P
Error!
BR
B
Q
Starting from the same point A, potential
drop across R should be 3 V so that point
B is at a potential 1 V higher than C.
25. C
Let  and r be the e.m.f. and inernal
resistance of the supply respectively.
Total power P delivered by the supply
= I
=
2
rR
BQ
R
As shown in the figure above, the resultant
magnetic field B due to the two wires Q
and R are vertically downwards. Hence,
by right-hand grip rule the magnetic force
acting on P points to the right horizontally.
30. C
Power supplied by the battery
= I = 5  2 = 10 W
Power loss due to the resistance of the
armature coil = i2R = 22  0.5 = 2 W
10  2
 100 % = 80%
Efficiency =
10
02 AL Physics/M.C./P.6
31. C
37. A
VR, Y1
Y1
E
Y2
h
3.6  10 15
1.6  10 19
= 22.5 kV

1
n 2f
1
1
c
 2  2

ni n f
1 1

'
1
8
1 32
=
= 1 =
1
1
9
9


1 n 2
In an a.c. circuit in which a resistor and a
capacitor are connected in series, the
voltage across the resistor (VR) should lead
that of the capacitor (VC) by 90.
However, in the circuit as shown, the
polarity of the connection to the CRO is
reversed for the capacitor. Hence, Y1 lags
behind Y2 by 90.
32. B
k.e. of the electrons = photon energy
eV = 3.6  10-15
ni2
1
1
1
 2  2

ni n f
VC
V =
1
hf 
Y2
8

9
’ =
38. D
(1) False
The graph does not pass through the
origin.
(2) True
P
Error!
A
B
33. C
0
34. A
35. C
A. False.
Blue light has shorter
wavelength and higher photon energy
c
= hf = h . The photoelectrons

emitted should have a greater
maximum kinetic energy.
B. False.
C. True. No. of photons = intensity 
photon energy.
With the same
intensity, no. of photons decreases as
the photon energy increases.
D. False. Kinetic energy of the emitted
photoelectrons increases, a higher
stopping voltage is required.
36. B
T
1
T
=
slope
of
the
broken line
P
VB > VA, hence, volume of the gas
increases.
V
(3) True
PV = NkT
PV
= Nk
T
Mass of the ideal gas is fixed, N is
constant.
02 AL Physics/M.C./P.7
39. B
43. B
PV = NkT
PV
N =
kT
PV
=
R
T
NA
PVN A
=
RT
10 5  10 6  6.02  10 23
=
8.31  (25  273 )
= 2.4  1019
40. D
1 2
3
mc = kT
2
2
c  T
c'
273  160
=
273  80
c
c’ = 1.1 c

41. B
 kt
A = A0 e
ln A = ln A0 – kt
slope of ln A – t graph = -k
= -0.0125 s-1
hence,
42. A
k = 0.0125 s-1
half-life = (ln 2)/k
= 55.5 s
IC

=0
IB =
6 – IBRB – VBE
I
6  C RB  VBE = 0

6
3  10 3
RB  0.6 = 0
80
RB = 144 k
44. C
V- ~ V+ = VP
no current flows into the op amp
hence,
V  V0
V1  V
= 
50 k
10 k
VP  V0
V1  VP
=
50 k
10 k
VP  (0.9)
2.7  VP
=
10
50
5(2.7 – VP) = VP+0.9
VP = 2.1 V
45. B
m
V
m
=
Al
density  =
m
(d / 2) 2 l
m
d2 
l
d
m l
=
2

d
m
l
=
hence, % error of d
= (% error of m + % error of l)/2
= 3%