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03 AL Physics/M.C./P.1
2003 Hong Kong Advanced Level Examination
AL Physics
Multiple Choice Questions
1.
D
3.
F
F




mg
2  F sin  = mg
mg
F =
2 sin 
The vertical speed increases from 0.15f to
0.25f in T.
0.25 f  0.15 f
vertical acceleration g =
T
10 = 0.1f 2
f = 10 Hz.
tension at the lowest point
= F cos 
mg cos 
=
2 sin 
mg
=
2 tan 
2.
4.
A
R
F
45
45
45
C
average vertical speed v1 between the first
and second spots
= (3  0.05)/T
= 0.15f
average vertical speed v2 between the
second and third spots
= (5  0.05)/T
= 0.25f
R
mg
D
Let l1 and l2 be the length of arms of the
beam balance. When the unknown mass
m is weighted first on the left pan the
condition for equilibrium is given by
ml1 = m1l2.
(1)
When the mass is weighted on the right
pan, the condition becomes
ml2 = m2l1.
(2)
(1)  (2) and eliminate l1 and l2, we get
m2 = m1m2
m = m1m2
acceleration of the blocks = F/2m
consider the movement of the block on the
left, it accelerates to the left with
acceleration F/2m,
hence,
R cos 45 = ma
F
= m
2m
= F/2
(1)
In the vertical direction, the block is in
equilibrium,
i.e.
R sin 45 – mg = 0
R sin 45 = mg
(2)
(2)/(1)

mg
F /2
F = 2mg
tan 45 =
5.
A
Think about an ordinary simple harmonic
oscillator, the compression is maximum
when the oscillator is momentarily at rest.
The oscillator does not have kinetic
energy for further compression of the
spring. All the energy of the oscillator is
in the form of potential energy and is
stored in the spring. The situation in this
question is similar, when the two blocks
move with the same velocity, i.e. velocity
of P relative to Q is zero, P cannot
compress further towards Q.
This
implies the compression of the spring is
maximum.
03 AL Physics/M.C./P.2
6.
7.
8.
9.
B
Before the bullet entering the wooden
block, its horizontal speed is constant.
When the bullet passes through the block,
owing to the frictional force, it decelerates
and the speed decreases. The horizontal
speed remains constant after passing
through the block.
A
At the vertical position, besides balancing
the weight, the tension needs to provide
the centripetal force required for the
circular motion. Tension = mg + mv2/r
B
maximum velocity = A
maximum momentum = mA
C
A: unchanged
m
k
mass increases, period increases
C: max acceleration = 2A, with the
mass increases,  decreases and the
acceleration decreases
1 2
kA
D: unchanged, total energy =
2
B: T = 2 
10. A
(1) True
(2) False
The amplitude of oscillation would
become infinite if there are no
damping forces AND the system is at
resonance.
(3) False
Phase difference between the
displacement and the driving force
depends on the difference between
driving force frequency and the
natural frequency. The displacement
of the system may not be in phase
with the driving force.
11. A
g = G
M
R2
M
4
= G
 R
4 3 3
R
3
4
= G R
3
3 g
4 GR
3
10
=
4   6.7 10 11  6.4 10 6
= 5.6  103 kg m-3
 =
12. D
Potential energy of the mass at the earth’s
surface
Mm
= G
R
Mm
= G 2 R
R
= - mgR
Potential energy of the mass at a height of
3R above the earth’s surface (i.e. 4R from
the center of the earth)
Mm
= G
4R
Mm R
= G 2
R 4
= - mgR/4
Hence, potential energy gained = 3mgR/4
13. D
Standard value for g is 9.8 ms-2.
Result of students P has the greatest
deviation from the standard value.
14. A
R
Q
P
03 AL Physics/M.C./P.3
15. B
At boundary N, the angle of refraction in
(III) is greater than the angle of incident in
(II). This implies refractive index nII >
nIII.
At boundary M, total internal
reflection occurs. Hence, nII > nI and
among the three media nI should be the
smallest. Refractive index n of the three
media should be in order of (II) > (III) >
(I). Speed of light in a medium  1/n.
Hence, the speeds of light in the three
media in descending order is (I) > (III) >
(II).
(3) True
18. C
h = 10 log
4I
I0
I
 10 log 4
I0
= 70 + 10 log 4
= 76 dB
= 10 log
19. C
(1) True
16. C
(1) True
Wavelength of red light ~ 700 nm.
According the graph, a higher
proportion of red light is absorbed by
X.
(2) True
Accordingly to the graph, a higher
proportion of light with wavelength
less than 400 nm (i.e. ultra-violet
radiation) is absorbed by Y.
(3) False
Within the visible light spectrum
(400 – 700 nm), transmitted light
intensity is higher for Y. The view is
brighter.
17. B
(1) False
By the principle of reversibility of
light (i.e. if a ray is reversed, it always
travels along its original path.), the
diagram can be redrawn as below:
d=
1 10 2
= 2000 nm
5000
(2) True
d sin  = m
d sin  = 2 
2000 sin  = 2  500
sin  = 0.5
 = 30
(3) False
max order =
d
2000
=
=4
500

20. D
A. False. Eyepiece should be the
with shorter focal length.
B. False.
It is the case for
telescopes
but
not
for
microscopes.
C. False. It is not the case for
telescopes.
D. True.
one
the
the
the
L
21. C
F
Obviously, the above diagram is not
correct. Virtual image is formed
only when the object is located
between the focus and the pole of the
lens.
(2) False
Only incident light rays which are
parallel to the principle axis converge
to the focus.
(1) True
One displacement node in each case.
(2) True
Speed of sound waves must be the
same in both cases.
(3) False
03 AL Physics/M.C./P.4
Wavelength in a closed tube is double
in length of that in an open tube.
22. A
wavelength of red light ~ 1 m
photon energy
= hf
= hc/
= 6.6  10 34 
= 2  10
-19
26. D
Electric field strength, E =
independent of the separation,
unchanged.
Q
,
A
remain
0
Electric potential, V = Ed, d decreased, V
decreased.
3  10 8
27. C
When the high-resistance voltmeter
connects across terminals a and b, it reads
4 V. This implies e.m.f. of the battery =
4  3 V = 12 V (remark: effect of the two
1  resistors are negligible).
1 10 6
J
23. C
F  Q1Q2
When the low-resistance ammeter is
connected across a and b, equivalent
resistance of the system = 5  and the
current flow = 12/5 = 2.4 A. Half of the
current, i.e. 1.2 A will flow through the
ammeter.
case 1: like charges
F1
5 1
=
3 3
F2
= 5:9
case 2: unlike charges
F1
5 1
=
2 2
F2
= 5:4
28. A
5 F +
+
+
+
24. C
(1) True
_
_
_
_
P
K
5 k
(2) True
Points A and C are on the same
equip-potential line.
Potential
energy of the charged particle at point
A and C are the same.
By
conservation of energy, the particle
has the same kinetic energy at point C
and A.
(3) False
2
24
= +2 V
After:
Potential at P =  6 
= +3V
1 k
Q
6V
Before K is closed, net charge inside the
area enclosed by the broken line is zero.
After K is closed and when steady state is
attained:
Voltage across the 5 k resistor
5
= 6
=5V
5 1
Voltage across 1 k resistor = 1 V.
25. D
Before:
Potential at P =  6 
10 F
_
+
2
1 1
2  (  ) 1
4 4
The respective charges stored on the
capacitors are:
Q (5 F) = 5  5 = 25 C
Q (10 F) = 10  1 = 10 C
Hence, there should be 25 C - 10C = 15
C of positive charges flowing away from
the capacitors through K.
03 AL Physics/M.C./P.5
29. D
A. False.
32. A
E-field = 0
V
Initial current =
,
R
R doubled, initial current halved.
B. False.
Energy dissipated in the resistor
= energy stored in the capacitor
1
CV 2 , unchanged.
=
2
C. False.
Total charge stored in the capacitor =
CV, unchanged.
D. True.
Time constant  = RC, halved.
(1) True
As the conductor is spherical in shape
and there is no electric field inside,
the induced charges must be
distributed uniformly on the outer
surface.
(2) False
30. D
Magnetic field at P = 0  I1 is in opposite
direction to I2 and B-field due to I1 =
B-field due to I2
i.e.
 0 I1
 I
= 0 2
2r1
2r2
I1
0.6
=
10
30
I1 = 1.8 A
31. B
The charged particle undergoes circular
motion in the magnetic field

(3) False
Conductor – equip-potential object
33. D
34. C
Left
P
S
Q
R
I
wire
2
v
r
mv
r =
qB
Right
qvB = m
(1) False
If the magnetic field decreases
gradually, radius of the path will
increase.
(2) False
If the charged particle loses its charge
gradually, again the radius of the path
will increase.
(3) True
If the charged particle loses its kinetic
energy gradually, velocity will
decrease and hence radius of the path
decreases.
B-field
B-field
On the left:
The coil approaches the metal wire.
Magnetic flux inside the coil in the
direction of pointing out of paper
increases. By Faraday’s Law, current
will be induced inside the coil and by
Lenz’s Law the induced current is in
clockwise sense so as to reduce the
change.
Moving across the metal wire:
As the coil moving across the metal wire
from left to right, magnetic flux inside the
coil in the direction of pointing out of
paper decreases. By Lenz’s law, the
induced current is in anti-clockwise
direction.
On the right:
03 AL Physics/M.C./P.6
The coil moves away from the metal wire.
Magnetic flux inside the coil in the
direction of pointing into paper decreases.
By Lenz’s Law, the direction of the
induced current is clockwise.
35. B
(1) False
Increasing
the
frequency,
the
reactance of the inductor (= L) will
increase. Current flow through the
circuit decreases and hence the
brightness of the bulb decreases.
38. C
According to the ideal gas equation PV =
nRT, T  PV. The temperatures at the
state X, Y and Z is in descending order of
TY > TZ > TX. When the gas undergoes
expansion from state X to Y, the pressure
is constant (isobaric process).
By the
equation PV = nRT, the temperature
increases linearly with V.
39. A
F
(2) False
Inserting a soft-iron core into the coil
will increase the inductance of the
coil.
(3) True
With a capacitor added in series with
the inductor, the total impedance of
the circuit might decrease and current
flow through the circuit increases.
36. B
Z2 = R2 + X2
132 = R2 + 122
R =5
Average power dissipated in the circuit
2
= irms
R
= 22  5
= 20 W
37. D
(1) True
PV = nRT
P remains constant and V increases 
T increases. Hence, internal energy
increases. Remark: internal energy
depends on temperature of the gas
only.
(2) True
V remains constant and P increases 
T increases.
(3) True
Adiabatic process (Q = 0).
Compression implies work is done on
the gas (i.e. W is +ve). By first
Law of Thermodynamics U = Q +
W = +ve.
equilibrium
position
A1
r
0
A2
A3
W.D. = area under the F-r curve
The atoms are originally at their
equilibrium position, to separate the atoms
to infinity, W.D. = A2 + A3
40. C
41. B
1

1
102
1


102

42. A
43. B
1
1
 2
2
m
n
1
1
= k( 2  2 )
1
3
1
1
= k( 2  2 )
1
2
8/9
=
3/ 4
= 121 nm

03 AL Physics/M.C./P.7
44. B
1 -1
s
10 6
ln 2
=
k
= 106 ln 2
= 6.9  105 s
= 8 days
= 1 week
k=
t1/2
45. B