2003 Hong Kong Advanced Level Examination
2006 Hong Kong Advanced Level Examination
AL Physics
Multiple Choice Questions (Solution)
1. A
sin ( = [pic]
T = W sin (
2. C
Let s1 be the distance traveled by the stone in the 1st second;
and s2 be the distance traveled by the stone in the 2nd second.
by s = [pic]
s1 = [pic]
s1 + s2 = [pic]
hence, 4 = [pic]
4s1 = s1 + s2
[pic] = [pic]
3. C
A and B: There is no sufficient information to draw such a conclusion.
C: As friction is negligible, there is no change in the horizontal speed during the collision of the ball with the floor.
time of flight = [pic] = 0.5 s
D: The horizontal speed is unchanged. The time of flight should remain the same.
4. A
5. C
6. D
(1) consider the forces acting on the mass along the string:
T1 - mgcos( = [pic]
at the end points of the motion v = 0
hence, T1 = mgcos(
(2) consider the forces acting on the mass in the vertical direction:
T2cos( - mg = 0
hence, T2 = mg/cos(
7. D
8. B
wavelength ( = c/f = [pic]= 4 m
phase difference = [pic]= [pic]
9. A
(1) correct
n =[pic], v ( 1/n
(2) incorrect,
frequency remains constant in refraction
(3) incorrect,
by Snell’s Law: n1sin(1 = n2sin(2
hence, large n ( small angle
10. D
potential V of an isolated charged sphere ( [pic]
As the two metal spheres are connected with a conducting wire, they have the same potential.
Hence, Q ( r
The charge stored on S
= [pic]
= 2Q
11. B
(1) correct
E-field at X due to P points to the right
E-field at X due to Q points to the right
hence, the resultant E-field at X points towards Q
(2) correct
X is at the mid-point of PQ and P & Q are oppositely charged. Total potential at X is hence equal to zero.
(3) incorrect
resultant E-field at any point between P and Q ( [pic]
where r is the distance of the point from P and
d is the separation between P and Q.
At the point near P (i.e. r is very small), the resultant E-field becomes very large (much greater than that at X).
12. B
given: R1 = R2 = R3 = R
R3 has the largest current (= 2I) flowing through. R3 should reach its limitation of power dissipation first.
( (2I)2R = 4
I2R = 1
hence, total power dissipation of the system = 4 + 1 + 1 = 6 W
13. D
Q = CV
for constant Q, C ( 1/V
hence,
[pic] = [pic]
= [pic]
( 3C1 = 2C1 + 2C2
C1 = 2C2
[pic] = [pic]
14. A
The system is equivalent to 7 capacitors connected in parallel.
Capacitance of each capacitor
= 8/4 pF = 2 pF
Hence, the capacitance of the system
= 7 ( 2 = 14 pF
15. D
[pic]
(1) short b and c
have the output at a & d
total no. of turns of the secondary coil = 250
turns ratio Np : Ns = 2 : 1
hence, output voltage = 120 V
(2) have the output at a & b
no. of turns of the secondary coil
= 200
turns ratio Np : Ns = 5 : 2
hence,
= 240 ( 2/5 = 96 V
(3) short b and d
have the output at a & c
equivalent no. of turns of the secondary coil = 200 – 50 = 150
(remark: the 50 turns coil is now wound in direction opposite to that of the 200 turns coil)
turns ratio Np : Ns = 10 : 3
hence,
output voltage = 240 ( 3/10 = 72 V
16. C
[pic] = [pic]
= [pic]
17. A
qvB = [pic]
qB = [pic]
for the same circular path,
B ( [pic]
[pic] = [pic]
= [pic]
B( = 900 ( 1 ( 10-3 = 0.9 T
18. B
B = [pic]
(B = [pic]
r = [pic]
= [pic]
= 0.02 m
19. B
(1) & (2) incorrect.
NO magnetic field lines are cut by the metal rod in both cases.
(3) correct
Magnetic field lines are cut by the metal rod. By law of EM induction, e.m.f. (= vlB) is induced in the rod.
20. B
( = NBA cos (
= NBA cos (t
with the angular velocity doubled, period is halved and amplitude remains constant
21. A
amplitude of the induced e.m.f. across the secondary coil ( NBA(
(1) correct, ( increases
(2) incorrect
Magnetic field outside the solenoid
= 0
Magnetic flux enclosed by the secondary coil = cross-sectional area of the solenoid ( B-field inside the solenoid
With the cross-sectional area of the secondary coil increasing but that of the solenoid unchanged, there is no effect in the magnetic flux enclosed by the secondary coil.
(3) incorrect
magnetic field of a solenoid depends on the number density of the turns
increasing the total number of turns but keeping the density constant has no effect on the magnetic field
22. A
N : total no. of gas molecules
m : mass of a gas molecule
hence, Nm = total mass of the gas
23. C
24. C
(1) 4 ( 0.005 = 0.02 mSv
(2) 12 ( 10 ( 0.001 = 0.12 mSv
(3) 2 ( 0.02 = 0.04 mSv
25. D
T = [pic]
g = [pic]
[pic] = [pic]
= 2(2 + 1 = 5%
26. B
(1) incorrect
Q should move along the line of collision after the collision.
(2) incorrect
Total momentum after collision lines along the line of collision while the initial momentum is in horizontal direction. Principle of conservation of momentum is being violated.
(3) correct
27. C (out of syllabus)
moment of inertia, I = l2m + l2(2m) = 3ml2
change in p.e. = mgl + 2mgl = 3mgl
by conservation of energy
[pic] = 3mgl
[pic] = 3mgl
( = [pic]
28. B
[pic] = [pic]
[pic] = [pic]
MS = [pic]
= [pic]
= 5.92 ( 10-4 r3 kg
29. A
30. D
31. B
Condition for resonance of a closed tube:
( = [pic] (where n = 1, 3, 5, …)
[pic] = [pic]
l = [pic]
l = 16.5 cm (n = 1)
32. C
33. B
(h = [pic]
= [pic]
6 = [pic]
100.6 = [pic]
If = 398 mW
34. D
Intensity = [pic]
Power output
= [pic]
= 4.24 ( 1026 W
35. D
rate of energy supplied to the capacitor
= power supplied to the capacitor
= VI
= [pic]
= [pic]
= [pic]
at t = 0, power = 0
at t ( (, power ( 0
36. A
point X and Y are shorted by a metal wire, hence, potential difference between X and Y = 0
p.d. across 3k( resistor = 6 V
hence, I2 = 2 mA (direction: as indicated in the figure)
p.d. across 6 k( resistor = 3V
hence, I3 = 0.5 mA (direction: as indicated in the figure)
by Kirchhoff’s first law,
I1 + I3 = I2
( I1 = 1.5 mA
37. C
VS0 = [pic]
= [pic]
= 5 V
38. A
in photoelectric effect, the stopping voltage is independent of the light intensity
39. A
(1) energy of a photon of red light ~ 2 eV
(2) average k.e. of an air molecule at room temperature
~ [pic]
= [pic]
= 6 ( 10-21 J
= 0.04 eV
(3) accelerating potential of a cathode-ray tube ~ kV
k.e. of an electron in a cathode-ray tube ~ keV
40. C
For an extension-load graph, it is the area between the curve and y-axis that represents the work done or energy recovered.
41. A
v = [pic]
[pic] =[pic]
E = 1.2 ( 1011 Pa
42. B (out of syllabus)
43. D (out of syllabus)
44. D
Point A and B are at the same height.
Velocity of flow of the liquid at point B
= 0.
Apply Bernoulli’s Principle to point A and B:
[pic] = [pic]
[pic] = [pic]
( (g(h = [pic]
(h ( v2
If the velocity of flow is doubled, the difference in heights of the liquid will be quadruple.
45. C
-----------------------
VL - VC
VC
VL
VR
I3
I2
hB
I1
Y
X
6 V
3 V
6 k(
3 k(
O
B-field = 0
secondary coil
a
solenoid
m
2m
l
VS
line of collision
TOP VIEW
Q
P
d
c
b
v
B
A
l
O
l
l
2m
m
hA
R3
R2
R1
2I
I
I
(3)
(2)
after
before
T
W
(
P0
(
................
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