Download 1. Which of the following pairs of quantities of a moving object must

00 AL Physics/M.C./P.1
2000 Hong Kong Advanced Level Examination
AL Physics
Multiple Choice Questions (Solution)
1.
D
3.
F
A
B
C
The whole system is in equilibrium
 (i) total force acting on the system = 0
(ii) total moment of force acting on the
system = 0
R
(1) False
Consider the two books A and B as a
whole system. The forces acting on
the system are F and the frictional
force acting on B. Because the system
remains stationary, these two forces
must equal in magnitude and opposite
in direction.
T
Q
(1) True
As the weight moves towards Q, d
increases. Tension T increases.
B
(1) False
The normal reaction must be greater
than the weight of the ball or else
linear momentum of ball would not
change from vertically downwards to
upwards.
(3) True
The boy moves with uniform velocity.
Net force acting on the boy must be
zero. This implies normal reaction
acting on the boy equal to the weight.
P
Take moment at point P
Tr - Wd = 0
Tr = Wd
T = Wd/r
(3) True
(2) False
Normal reaction is equal to zero.
Weight of the astronaut in the orbiting
spacecraft is used as centripetal force
for the circular motion.
d
W
(2) True
F acting on A is towards the right.
The frictional force acting on A by B
should then be towards the left.
2.
r
(2) True
With a shorter string, r would
decrease and tension T increases.
(3) False
With a longer string, r would
increase and tension T decreases.
4.
A
T – mg = ma
120 – 8  10 = 8a
a = 5.0 ms-2
5.
C
mg sin 1 – f = ma1
mg sin 2 – f = ma2

mg sin 1  f
ma 1
=
mg sin  2  f
ma 2
5 sin 15   f
1
=
5 sin 20   f
2
10 sin 15 - 2f = 5sin 20 - f
f = 10sin 15 - 5sin 20
= 0.88 N
00 AL Physics/M.C./P.2
6.
A
9.
vu
F = m
t
Ft = mv
1
k.e. = mv 2
2
1 m2v 2
=
2 m
m2r = G
(u = 0)
2 = G
1 F 2t 2
2 m
1
 k.e. 
m
Hence, ratio of gain in k.e. of A to that of
B = 2:1
=
7.
C
(1) True
When the ball eventually comes to
rest, it is in equilibrium. Resultant
force acting on the ball is equal to
zero  ke = mg. Hence, e  m.
(2) True
Final compression of the spring e is
given by the equation: ke = mg. It is
independent of h.
(3) False
The strain energy stored in the spring
1
= ke 2 which is independent of the
2
height h while the gravitational
potential loss is mg(h + e). Hence, it
is not possible these two energies to
be equal.
8.
B
A
Mm
r2
M
r3
To have the same period (i.e. same ), the
two planets must have the same value of
M
, i.e. average density.
r3
10. B
speed at equilibrium position
= max. speed = A
max. acceleration = 2A
=   vmax
= v max
k
m
12
0 .1
= 5.5 ms-1
= 0 .5
11. C
glass
tube

L

T
rubber
bung
m
W
Let m be the mass of the rubber bung.
Tension T = weight W
Consider the motion of the rubber bung,
vertical:
T cos  = mg
i.e. W cos  = mg
(1)
N
r
horizontal:

T sin  = m2r
W sin  = m2L sin 
W = m2L
(2)
mg
vertical:
horizontal:
N cos  - mg = ma = 0
N sin  = ma = m
v2
r
(1) False
According to eq. (1),  depends on
ratio of W and m only.
(2) False
If  is fixed, ratio of W and m is
constant. According to eq. (2), L 
1/2.
00 AL Physics/M.C./P.3
(3) True
According to eq. (1), cos   1/W.
D. True. Definition of moment of inertia,
I = mr2. By pulling his arms and
legs to his chest, r decreases and I
decreases.
12. B
T = 2
(
l
g
l
T 2
) =
2
g
g =
E. True. Coiling up his body, I would
decease. By conservation of angular
momentum, angular speed  would
increase. More turns hence could be
made before reaching the water.
4 2 l
T2
A. False. Air resistance increases the
period of the pendulum. g would be
smaller.
B. True. If the stop watch used for the
experiment runs too slowly, a shorter
time T would be recorded. g would
be higher.
C. False. Length of the string is smaller
than the effective length (i.e. length of
the string + radius of the pendulum
bob). g would be smaller.
D & E. False. g at places below or above
sea-level is smaller than that at the sea
level.
13. A
Simple harmonic motion is isochronous,
i.e. period (or ) is independent of
amplitude
(1) True
maximum velocity = A
15. E
paper cone
pendulums
(loaded with
metal rings)
string
A
heavy bob
B
C
Barton’s pendulum is a phenomenon of
resonance.
(1) True
Pendulum B and the driver pendulum
(i.e. heavy bob) have the same
effective length  they have the same
oscillating frequency.
Resonance
occurs for pendulum B.
(2) True
In force oscillation, oscillating
systems must have the same frequency
as the driving system.
(2) False
maximum elastic p.e. =
C. True. By conservation of energy, the
potential energy converted into
translational k.e. and rotational k.e.
1 2
kA
2
(3) True
Phase difference
(3) False
isochronous oscillation

/2
14. B
A. True. No external torque acts on the
diver about the center of gravity.
B. False. The force “weight” acts on the
diver.
0
f0
frequency of the
driving force, f
Pendulum A has a shorter length than
the driver pendulum  fA > f and
pendulum A is in phase with the driver
pendulum.
00 AL Physics/M.C./P.4
Pendulum C has a longer length than
the driver pendulum  fC < f and
pendulum C is in antiphase with the
driver pendulum.
18. A
Hence, A and C are in antiphase.
16. B
1 cm
17. E
(1) True
Displacement of
air particles
+
_
P
position
As shown in the figure above, air
particles on the left of P have +ve
displacement, i.e. they move towards
right. Air particles on the right of P
have –ve displacement, i.e. they move
towards left. Hence point P is a
centre of compression.
(2) True
Every point in a wave is in simple
harmonic motion. In s.h.m. the mass
moves with the greatest speed and
hence the greatest kinetic energy
when passing through the equilibrium
position.
(3) True
Displacement of
air particles
+
_
As shown in the figure above, by the time
the wavefront move from the center of the
square to 6 cm, the dot vibrator moves
distance of 3 cm. Hence, the velocities of
the wavefront and the dot vibrator should
be in ratio of 2:1. Hence velocity of the
dot vibrator = 10 cms-1.
19. D
(1) False
Color of an object depends on color
of the reflected light. As the lens is
made non-reflecting for yellow light,
it should not appear yellow.
(2) True
Condition for destructive interference:
2nt = (m + 1/2)
where
m = 0, 1, 2, …
Remark: there is no need to consider
phase inversion as in both the upper
and lower surfaces, phase inversion
occurs.
When m = 0, thickness t is minimum.
t = (/n)/4 = ’/4 (i.e. A quarter of the
wavelength of yellow light in the film.)
(3) True
Conservation of energy
P
position
As shown in the above figure, point P
has a positive displacement (i.e.
moving towards right) a short time
later.
20. D
(1) False
A convex lens can form a virtual
image if the object is inside the focal
point.
(2) True
The point object must lie along the
line OS so that the ray that is directed
toward the center of the lens will
emerge from the lens with no change
in its direction.
(3) True
PQ and OS diverge.
00 AL Physics/M.C./P.5
21. E
(1) True
1 Q
4 0 r
Q is +ve,
1
hence, V(r) 
r
V(r) =
(2) True
Potential at infinity = 0.
W.D. in moving a charge q0 from  to
position r
= potential energy of q0 at r –
potential energy of q0 at infinity
= potential energy of q0 at r
q
1 q0 Q
=
 0
4 0 r
r
Hence,
W.D. in moving +q from  to a
distance d from +Q = W.D. in
bringing +2q from  to a distance 2d
1 qQ
from +Q =
4 0 d

VQ  V R > V P  VQ

VQ  V R > VQ
23. A
0.2 A 5 
B
Q
R
Electric field lines point from left to right

VP > V Q > VR
VP = 0, hence, both VQ & VR are negative.
V
r
Electric field lines in region between Q &
R is denser than that between P & Q. i.e.
E (in-between Q & R) > E (in-between P
& Q)
E = 
D
1
R
Current passing through the 1  resistor
= (0.3 – 0.2) A = 0.1 A
Point C is at a potential 1 V (= 0.2  5 V)
lower than point B and point D is at a
potential 0.1 V (= 0.1  1 V) lower than B.
Hence D is 0.9 V higher than C. Current
passing through the 3  resistor = 0.9/3 =
0.3 A (from D to C). Current passing
through R is 0.2 A from right to left
(Figure below).
3
1
0.1 A
1

d
P
3
0.3 A
(3) True
Total potential energy of the system
1 qQ 2q 2 2qQ
=
(


)
4 0 d
d
2d
22. A
C
0.3 A
D
R
0.2 A
24. E
(1) True
Terminal voltage
= p.d. across R
R
= 
Rr
1
= 
1 r / R
If r >> R, terminal voltage becomes
very small.
Power = i2(r + R)
Same current flowing through r and R,
if r >> R, most of the power is
dissipated by the internal resistance in
battery.
(2) True

Rr
If R is very large, I would be very
small.
Current I =
00 AL Physics/M.C./P.6
Power = i2(r + R)
Same current flowing through r and R,
if r << R, most of the power is
dissipated by R.
(3) True
power dissipated by R
 2
) R
= (
Rr
2R
= 2
R  r 2  2 Rr
=
=
28. B
(1) False
The motor stop rotating, i.e.  = 0.
No change of magnetic flux through
the coil of the motor.
(2) False
Induced e.m.f.   
2R
(3) True
No induced back e.m.f. is induced
across the motor to oppose the current.
R  r  2 Rr  4 Rr
2
the particles must have the same velocity
2
2R
( R  r ) 2  4 Rr
2
=
(R  r) 2
 4r
R
Power dissipated by R is max. when R
= r.
29. D
(1) Diode D were reversed in the circuit:
D
D1
R
D2
25. D
S
R
BP
O
Bs
BQ
BR
P
Q
26. E
V = Ed
max. E-field = V/d
4.5  10 3
=
1.5  10 3
= 3  106 N/C
D3
For upper half cycle of input voltage,
current passes diodes D and D3. No
current flows through R. Voltage
drop across R = 0.
For lower half cycle of input voltage,
no current flows into the circuit.
Voltage drop across R = 0.
Hence, trace III would be displayed
on the CRO.
(2) Diode D were removed:
D1
R
max. acceleration = F/m
= qE/m
1.6  10 19  3  10 6
=
9.1 10 31
= 5.3  1017 ms-2
27. B
to be undeflected
FE
qE
E
v
= FB
= qvB
= vB
= E/B
D2
D3
For upper half cycle of input voltage,
current passes diodes D1 and D3
through R.
For lower half cycle of input voltage,
no current flows into the circuit.
Voltage drop across R = 0.
00 AL Physics/M.C./P.7
Hence, trace II would be displayed on
the CRO.
(2) True
No resistance in a pure inductor.
After a sufficient long time, the
inductor is the same as a conducting
wire.
30. C
X
Y
D
B
6.0 V
A
There will be a time delay in the
glowing of B.
E
Z
(3) False
When the switch S is re-open, energy
stored in L will release. Current will
then flow through B and A which are
now connected in series. A and B
should be in equal brightness.
Let C be the capacitance of the capacitors.
The switch is connected to X:
Capacitors A and B are connected in series,
each of the capacitors will have 3.0 V drop
across.
The switch is connected to Y:
Capacitors D and E are connected in series.
Equivalent capacitor of D and E = C/2.
Capacitor B is connected in parallel to the
capacitor system formed by D and E.
Hence total equivalent capacitor = 3C/2.
33. C
(1) True

B. False. Inductance depends on the
geometry of the inductors only. It
does not depend on time.
C. False.
D. False.   L
Vs
VC
(2) True
Z =
For constant charge, V  1/C
Hence, final p.d. across Z and Y = 2.0 V
31. A
A. True. Current passing through the
solenoid attains a finite value.

VR or I
=
R 2  C2
R2  (
1 2
)
C
, Z
(3) False
power factor = cos 
R
=
Z
, Z, cos  
34. D
dI
.
dt
C
dI
E. False. It is true that   L
.
dt
However, no information about what
the e.m.f. depends on can be read
from the graph.
32. C
(1) True
E.m.f. is induced across L to oppose
the change when switch S is closed.
L
As the magnetic flux inside L is pointing
upward, according to right-hand grip rule
the current should flow in the direction as
shown in the above figure. Hence the
00 AL Physics/M.C./P.8
capacitor could be “charged up” with
“lower” plate positive or “discharge” with
“upper” plate positive.
i.e. U is at a minimum at r0. It is
given that U(r0) = 0. Hence, U is nonnegative for all values of r.
B. True. When r < r0, U increases as r
decreases so that slope = -ve and F is
positive, i.e. repulsive force between
the two molecules dominates.
35. C
Sinusoidal waveform:
1
<P> = I 2p R
(Ip: peak value)
2
Square waveform:
<P>
= <I2R>
= <I2> R
C. True. When r > r0, U increases as r
increases so that slope = +ve and F is
negative, i.e. attractive force between
the two molecules dominates.
area under I 2  t curve for one cycle
R
period
1 2
I0  T
= 2
R
T
1
= I 02 R
2
hence, if the average power dissipated is
the same in both cases, I0 = Ip = 1 A
=
D. False. To separate the two molecules
from r = r0 to , work done, which is
equal to energy difference between
the final position and the initial
position (= U0  0) is needed to work
against the attractive force.
E. True. U is very large when r is very
small. For small r, it is the repulsive
force between the two molecules that
dominates.
36. C
Vol .  
F mg
=

A
A
A
Average density  of the body remains the
same. Volume  (dimensions)3 and area
 (dimensions)2.
Hence, stress 
dimensions. The new average stress = 2S.
stress =
Q  R:
constant T and pressure p decreases
 vol. increases (isothermal expansion)
and work is done by the gas on the
surroundings.
37. D
U
U0
r
0
38. E
P  Q:
straight line passing through the origin
 volume of the gas remains constant
(isovolumetric process).
As W = pV, W = 0.
equilibrium
position, r0
A. True. At r = r0, i.e. equilibrium
position, the attractive force between
the two molecules is equal to their
repulsive force.
dU
As F = 
= -ve slope, at the
dr
dU
equilibrium point, F = 0 
= 0,
dr
R  S:
constant p and T increases
 vol. increases (isobaric expansion) and
work is done by the gas on the
surroundings.
39. E
00 AL Physics/M.C./P.9
40. C
A=
Energy
0
Rf
Vo
 1
Vi
Ri
(1) False
When the sliding contact is at P, Ri =
0 and the gain A  .
-4E
(2) True
When the sliding contact is at Q, Rf =
0 and the gain A = 1.
-6E
(1) True
The atom would absorb 2E from the
electron to excite the electron of the
atom from energy level –6E to –4E.
(2) True
The atom would absorb energy of the
photon to excite the electron of the
atom from energy level –6E to –4E.
(3) False
No partial absorption of energy is
allowed for photon. No transfer of
energy states corresponds to change
of energy of 3E.
41. E
Relationships between different physical
quantities in photoelectric emission:
 K = hf - 
(: work function of the metal)
I
 i  no. of photon 
hf
(3) True
A depends on the ratio of Rf to Ri only.
43. E
A. False. p.e.  1/r. At P the electric
p.e. of the -particle is at a maximum.
B. False. Total energy should remain
constant.
C. False. Electrostatic force acting on
the -particle passes through N. No
external torque acting on the particle about N. Hence, angular
momentum of the -particle should
remain constant.
D. False. The larger the initial k.e., the
smaller the nearest distance.
E. True. If the atomic number of the
nucleus is greater, the repulsive force
between the -particle and the nucleus
increases.
The nearest distance
increases.
42. D
Vi
Vo
Rf
Ri
44. A
As source does not emit -radiation, it
makes no difference on the count rate in
inserting the cardboard or 1 mm
aluminium. The  radiation would be
filtered. The recorded count rate should
drop markedly and x ~ y. Inserting 5 mm
of lead, there would be some residual 
radiation. z should be greater than the
background count rate.
45. D