01 AL Physics/M.C./P.1
2001 Hong Kong Advanced Level Examination
AL Physics
Multiple Choice Questions (Solution)
1.
2.
3.
E
The weight must pass through the center
of the mass of the block.
The frictional force must lie along the
surface between the block and the inclined
plane.
The reaction force which is perpendicular
to the inclined plane could not pass
through the center of the mass, or else
there would be a net moment of force
about the center of the mass of the block
and the block would rotate. To be in
equilibrium the reaction force vector must
lie on the left of the center of mass. The
two moment of forces due to the reaction
force (clockwise) and the frictional force
(anticlockwise) will then balance each
other.
B
Remark:
Properties of action and reaction pair:
- equal in magnitude but opposite in
direction;
- they act on “different” object
C
Initial k.e. due to the vertical motion = E/2
When the stone is “halfway” up,
k.e. due to the vertical motion = E/4 and
k.e. due to the horizontal motion remains
constant = E/2
Hence, when the stone is halfway up, total
k.e. of the stone = E/4 + E/2 = 3E/4
4.
A
Remark:
Impulse = Ft
= change of momentum (vector)
5.
A
Let w be the speed of the object at point Y.
Since the object undergoes uniform
acceleration motion,
w2 = u2 + 2as
v2 = w2 + 2as
By solving the above two equations,
w=
6.
C
7.
E

u2  v2
2
vmax = A
 = 1/0.5 = 2 rads-1
By conservation of energy,
2
= v2 + (x)2
vmax
12 = v2 + (2  0.3)2
v = 0.8 ms-1
8.
C
Remark: Forces acting on the pendulum
bob are weight of the bob and the tension
in the string
9.
E
Ihollow > Isolid
Hence, at the bottom of the incline plane,
rotational k.e. of the Ihollow > rotational k.e.
of the Isolid.
10. A
1 2
mv
2
Angular momentum l = mvr
1
l
Hence, k.e. = m( ) 2
2 mr
1 l2
=
2 mr 2
Since F is a central force, during the
process, no torque is acting on the subject
 angular momentum conserves, i.e. l
remains constant throughout the process.
1
Hence, k.e.  2
r
as r  r/2, k.e.  4 k.e.
k.e. =
11. D
1 s = 9 192 631 770  period
Hence, frequency f = 9 192 631 770
Wavelength  = c/f = 3.3 cm, microwave
01 AL Physics/M.C./P.2
17. D
P
12. E
13. E
y =
(2) False
The outer rings are packed more
closely than the inner rings because
the slope of the lens surface increases
outwards.
(3) False
Even if liquid is introduced between
the convex lens and the reflector, for
the light ray reflected by the reflector,
it still suffers a phase inversion due to
the reflection from the optically
denser medium. Hence, there is a
phase difference between the two
reflected light rays.
20
2
R
I ’ = I0 cos2 
I = I ’ sin2 
= I0 sin2  cos2 
I
= 0 sin 2 2
4
When  = 0 and 90, I = 0
otherwise I  0
Hence, when Q is being rotated slowly
through 90, I increases and then
decreases.
18. D
min.
frequency
C
15. D
40

I’
D
, d   y 
d
14. A
(1) True
If the lens is with long focal length,
the slope of the lens surface will be
gentle.
60
Q
I0
I
= 10 log 1
I0
I
= 10 log 2
I0
I
I
= 10(log 1 – log 2 )
I0
I0
I1
= log
I2
I1
= 100
I2
max.
frequency
19. D
20. A
(1) True
With the positively charged rod
brought near X, electrons will be
drawn from the earth.
(2) False
16. A
(3) & (4) False
X and Y are in contact with each other
and earthed. Both X and Y are at zero
potential.
I
01 AL Physics/M.C./P.3
21. C
when the 100 V is applied across AD
R
VCD = 100 
= 80
10  10  R
 R = 80 
when the 100 V is applied across CD
80
VAB = 100 
= 40 V
60  60  80
22. D
bulb X : 6 V, 12 W
bulb Y : 6 V, 3 W
If the two bulbs work at their respective
rated values, the voltage across both of the
bulbs must be the same i.e. 6 V, but the
current passing through X should be with
greater value.
A & B: it is not possible that voltage
across X and Y are the same.
C: the current passing through X and Y are
the same.
E: current passing through X is smaller
than that of Y
24. A
(1) True
Q = CV = 100000  10-6  20 = 2 C
To discharge the capacitor in 2000 s,
a mean current of 2/2000 = 1 mA will
be provided.
(2) False
Time constant RC = 10  100000 
10-6 = 1 s. In the discharging process,
which is an exponential decay, the
capacitor is assumed to be completely
discharged in 5RC.
(3) False
1
CV 2
2
1
=  100000  10 6  20 2
2
= 20 J
Energy =
25. A
C = 0
hence,
1
d
1
1

d f di
=
1
di
di  d f
=
df
C 
23. B
C f  Ci
Ci
C = Q/V
=
To determine the charge stored in the
capacitor, imagine that the two terminals
of the capacitor are shorted by a
connecting wire. The net charge flows
through the wire is equal to the charge
stored in the capacitor. With the terminals
being shorted, the charges on the two
metal plates will redistribute until
equilibrium state is reached.
At
equilibrium there should be (900 + 100)/2
= 500 C of charge on one each metal
plates. Hence, there is 400 C of charge
flowing through the connecting wire and
the original charge stored on the capacitor
should equal to 400 C.
400 
C=
= 100 F
4
A
d
di
1
df
1
1
1.2
= –16.7%
=
26. C
(1) True
The induced e.m.f. of an inductor is
given by
dI
d
= L
= 
dt
dt
Hence, henry-ampere is equivalent to
weber.
(2) True
By Faraday’s Law,
d
= 
dt
Hence, volt-second is equivalent to
weber.
01 AL Physics/M.C./P.4
(3) False
magnetic flux = magnetic field  area
1 weber = 1 Teslametre2
27. D
It is not possible to represent an electric
field by field lines of closed loops.
28. E
According to the laws of electromagnetic
induction, the current is induced as to
oppose the change in magnetic flux. In the
question, the magnetic pole P is
approaching and the magnetic flux
confined by the loop will increase. To
oppose the change (i.e. increase in
magnetic flux) the loop C should move
away from the original position and its
area should slightly decrease, no matter
whether P is an N-pole or a S-pole.
29. D
30. B
Average power dissipation
2
= irms
R
2
=2 5
= 20 W
Remark: reactance does not dissipate
power.
31. C
A: circle, sinusoidal signals to X and Y
are with phase difference 90 and equal
amplitude
B: ellipse, sinusoidal signals to X and Y
are with phase difference in between 0
and 90.
C: sinusoidal curve, sinusoidal signal to Y
and X is controlled by the time base.
D: straight line, sinusoidal signals to X
and Y are antiphase.
E. Lissajous figure, sinusoidal signals to
X and Y in frequency ratio of 2:1.
32. B
33. E
The ideal is being compressed, volume of
the gas decreases. Work done on the gas
W is positive. As the pressure remains
constant, by the ideal gas equation PV =
nRT, temperature of the gas decreases.
Hence, internal energy U decreases (i.e.
negative).
34. B
Remark: The concept is required in
explaining the emission of -rays in
radioactive decay. In -rays emission, the
nucleons transit from high to low energy
level in the nucleus. These energy levels
are discrete in nature.
35. E
dU
dr
Hence, the potential energy due to shortrange repulsive and long-range attractive
forces should vary inversely as r5 and r2
respectively. Potential energy due to a
repulsive force is positive while that of an
attractive force is negative.
Interatomic force F = 
36. B
F
e
= E
A
l
4mgl
Fl
mgl
E=
=
=
d 2
Ae
d 2 e
( ) e
2
m g l
d e
E
=


2

m
g
l
d
e
E
m
0.01
=
= 0.1%
m
10
g
0.1
=
= 1.0%
g
9.8
l
0.01
=
= 0.51%
l
1.96
d
0.01
= 2
= 3.3%
2
d
0.61
e
0.1
=
= 2.6%
e
3.9
37. E
(1) True
 = Eu
Slope of a stress-strain graph in the
proportional region is Young modulus
which reflects the stiffness of the
material.
01 AL Physics/M.C./P.5
(2) True
Breaking stress of Y is greater than
that of Z.
(3) True
Maximum strain Z can be stretched to
is 1.5, i.e. e/l = 1.5. Hence, Z can be
stretched to 2.5  original length.
38. C
(1) True
Equation of continuity: Av = constant.
42. B
70
49
=
A0
70
A0 =
70 2
= 100 Bq
49
43. B
(1) False
The nuclear binding energy depends
on the total number of nucleons inside
the nucleus.
(2) False
(2) True
X and Z are with the same crosssectional area A. By equation of
continuity, the water speeds are the
same.
(3) False
Water speed in Y is the highest. By
Bernoulli’s Principle, pressure at Y is
lowest. Hence, the water level in
manometer Y is the lowest.
39. C
c
.

Wavelength of red light is longer than that
of yellow light.
Energy of a photon = hf = h
(3) True
44. C
Binding energy per nucleon
1.0087  1.0078  2.0146
= (
)  931
2
= 0.88 MeV
45. D
(1) False
Penetrating power of a radiation
depends on the type of radiations it is
and its corresponding energy.
(2) True
After 20 minutes, percentage of the
original radioactive isotope remains:
40. A
f
1
1

ni n f
41. B
Vout = 2 V
 voltage across 2 k = 6 – 2 = 4 V
current flow through 2 k = 4/2k = 2 mA
Ib = Ic/ = 2 m/100 = 20 A
6 V – Ib  R – Vbe = 0
6 – 20   R – 0.6 = 0
R = 270 k
Isotope
X
Y
Z
(3) True
Percentage
50
25
75