04 AL Physics/M.C./P.7
2004 Hong Kong Advanced Level Examination
AL Physics
Multiple Choice Questions (Solution)
04 AL Physics/M.C./P.7
1. D
As the object is in static equilibrium, the resultant force of the three external forces acting on the object is equal to zero.
i.e. = 0
and =
vertical: F cos q = W = mg
horizontal: F sin q = T
Þ T = mg tan q
At the instant the thread is suddenly cut, the force is removed. Resultant force acting on the object =
=
Hence, acceleration of the object is direction II and magnitude of acceleration = g tan q
2. C
(1) If M = m, after collision A will stop moving and all the initial k.e. of A will transfer to B. Kinetic energy of B is greatest.
(2) Let v and V be the final velocity of A and B respectively.
By conservation of momentum
mu = mv + MV
MV = mu – mv
If M > m, A will rebound with velocity –u and the magnitude of momentum attained by B (i.e., MV) will be greatest and equal to 2mu.
(3) If m > M, B will move forwards with 2u, i.e. the greatest speed.
3. A
Let t be the time taken by the bullet to a point under P.
t = =
Vertical distance traveled by the bullet
=
=
=
Hence, separation between the bullet and point P =
=
= = 3.2 m
4. C
To complete looping the loop without leaving the track, minimum speed at the top of the loop is given by
mg =
Þ =
by conservation of energy,
k.e. (bottom) = k.e. + p.e. (top)
=
=
u2 = 5gr
= 5 (1.6) (0.5)
u = 2 ms-1
5. C
(1) mg = ke
(2) force constant of the short springs
= 2k
Hence, 2mg = 2(2ke’)
e’ = e/2
6. B
F µ -x
7. A
g =
µ
=
g =
= 7.5 Nkg-1
8. B
(1) If the ring is a rigid body, the whole body must rotate with the same angular velocity w. Linear speed v = wr. Hence, v µ r.
(2) If the ring is a group of satellites revolving around Saturn, then for each satellite
=
i.e. v µ
9. B
10. D
The figure above shows the position of the two wave pulses at a short time later. P moves downwards and R moves upwards while Q is still at rest.
11. A
separation Dy =
where
L: separation between the screen and the double slit;
l: wavelength and
d: slits separation.
12. B
Remark: Refractive index of blue light in glass is greater than that of red light.
(1) True
P is less deflected. Þ P is red light.
(2) True
The parallel incident light rays is diverged.
(3) False
v = c/n. Q is blue light which has a higher refractive index. Hence, it would travel slower than P.
13. D
A. True. In resonance, amplitude of vibration of the oscillator could be greater than that of the driving force.
C. True. The closed end of the tube is a displacement node. Hence, it is a pressure antinode.
D. False. If f is the fundamental frequency of the air column in the tube, for resonance to occur, frequency of the tuning fork are f, 3f, 5f, … i.e. any odd numbers of f.
14. C
Atmospheric pressure P
=
=
m =
=
= 5 ´ 1018 kg
15. A
Consider the component of forces which are perpendicular to the dotted line as shown in the above figure.
As A and B are in equilibrium state, we have
F sin 45° = mAg sin 30°
and F sin 45° = mBg sin 60°
Þ mAg sin 30° = mBg sin 60°
=
=
16. B
potential V = Þ V µ
hence, potential V’ due the point charge –2Q is given by
=
= –4
V’ = –4V
Resulting potential at B = V – 4V = –3 V
17. B
E: the electric field strength at P
As the battery remains connected to the capacitor, V across the capacitor remains constant.
E = V/d
d , E ¯
V: the electric potential at P
As shown in the above figure, potential at P ¯ as the separation of the capacitor increases.
W: the energy stored in the capacitor
d , C (=)¯, V remains const.
W (=)¯
18. D
A. False. Depending on the initial velocity, the trajectory could be a part of an ellipse, circle, parabola or hyperbola.
B. False. As B is further away from O than A, Coulomb’s force acting on the particle at A is greater and, hence, acceleration at A is greater than that at B.
C. False. Potential energy of the particle at A is more negative. Hence, kinetic energy of the particle at A is greater than that at B.
D. True. Potential energy of the particle at A is more negative.
19. D
The resistance of a reverse-biased diode is very large and is comparable to the internal resistance of the voltmeter. Hence, the voltmeter should not be connected directly in parallel with the reverse-biased diode. The ammeter should be connected in series with the reverse-biased diode and the voltmeter is connected across the ammeter and the diode, i.e. answer D.
20. B
Redraw the resistors network as shown above. Since the equivalent resistance across CD is 2.5 kW, R = 5 kW. Equivalent resistance across AC = 3.5 kW.
21. A
As the light bulbs are connected in series, VX + VY = 200 V. According to the graph, the distribution of the voltage between the two light bulbs should be VX = 50 V and VY = 150 V. The corresponding current flows through the light bulbs is approximately 0.24 A. Hence, the power dissipated in each light bulb is
PX = 50(0.24) = 12 W
and PY = 150(0.24) = 36 W
22. A
Power delivered to the load
=
=
For maximum power delivered to the load, R should be as small as possible.
23. B
When the voltmeter is full-scale deflected, current flowing through the voltmeter is 100 mA and voltage across the voltmeter is 5 V. Hence, internal resistance of this voltmeter = 5/100 m = 50 kW.
24. C
A. True. V µ
B. True. C = 4pe0r
C. False.
Energy stored in a capacitor
= µ
D. True. If the two spheres are connected by a wire, they will have the same potential. Charge on sphere µ r. Final charge on A will be lesser than that of B, i.e. the charge on A decreases.
25. D
(1): equivalent capacitor = C/2
(2): equivalent capacitor = 2C
Energy store =
In both cases, the final steady voltages dropped across the system of capacitor are the same. Hence, energy stored µ C. The ratio of the total electrical energy stored in the capacitors in circuit (1) to that in circuit (2) = 1:4
26. B
(1) True
Polarity of the Hall voltage is related to the polarity of the charge carriers.
(2) True
VH =
where
VH : Hall voltage
I: current
B: magnetic flux density
n: no. density of the charge carriers
q: charge of the charge carriers
t: thickness of the thin film
(3) False
It is can be used for the determination of the magnetic flux density for both d.c. and a.c. field.
27. D
28. D
A. & B.
e0 µ NAwB0
where
e0: amplitude of the induced voltage in the secondary coil;
N: no. of the turns of the secondary coil;
A: cross-sectional area of the secondary coil;
w: angular frequency of the a.c. magnetic field inside the solenoid;
B0: amplitude of the a.c. magnetic field inside the solenoid.
C. Amplitude of the magnetic field at one end of the solenoid is half of that in the middle.
D. Magnetic field in a solenoid is given by B = m0ni, which is independent of the cross-sectional area.
29. D
= µ
Force per unit length acting on R by P = F (vertically upwards)
Force per unit length acting on R by Q = 4F (vertically downwards)
Hence, resultant force per unit length acting on R = 3F (downwards)
30. D
In a LCR series circuit, same current flows through C and L.
31. C
For a sinusoidal a.c., the average power
=
Average power of a half-wave rectified sinusoidal a.c. = =
Hence, the average value of a steady d.c. that gives the same effect as the rectified a.c. is given by
I2R =
Þ I = = 1 A
32. C
Energy of a photon = hf
As blue light has a shorter wavelength than that of yellow light, energy of a blue light photon is larger than that of yellow light. With the blue light replaced by yellow, maximum k.e. of the electrons emitted should hence decrease.
With the same light intensity, no. of photons (µ I/hf) are larger for yellow light and a larger no. of electrons is emitted. Hence, the photoelectric current should increase.
33. A
ionizing energy = E¥ - E2
=
= 3.4 eV
34. A
(1) True
(2) & (3) False
The dark lines in the solar spectrum are due to the solar radiation being absorbed by the elements in the sun’s atmosphere corresponding to these dark lines.
35. D
(1) True
C = C0ekt
log C = log C0 + kt
log C vs. t is a straight line
(2) & (3) True
C1 =
C2 =
= = =
Hence, it takes the same period for value of C at any time to double and ratio of C at equal time interval should be the same at any time.
36. C
37. A
PV = NkT
T µ PV
Hence according to the graph, the temperatures of X, Y and Z are in the ratio of
TX : TY : TZ = 4 ´ 1 : 2 ´ 3 : 1 ´ 5
= 4 : 6 : 5
38. B
39. C
Voltage across the 1kW resistor = 3 V
IC = 3/1k = 3 mA
IB = IC/b = 3 m/50 = 60 mA
6 – IBRB – VBE = 0
6 – 60 ´ 10-6 RB – 0.6 = 0
RB = 90 kW
40. B
Voltage gain = .
To produce the best square wave output, the voltage gain of the op. amp. circuit should be as large as possible so that saturation occurs.
41. B
(1) True
In RC circuit, the time constant t = RC. Hence, 1 F = 1 sW-1
(2) True
In RL circuit, the time constant t = . Hence, 1 H = 1 Ws
(3) False
Faraday’s Law,
e.m.f. =
Hence, 1 Wb = 1 Vs
42. C
A =
DN = ADt
= 1.0 ´ 106 ´ 24 ´ 3600
= 8.6 ´ 1010
43. C
(1) False
It is a-particles which cannot pass through the mica end-window of the GM tube.
(2) True
The b-particles are emitted in all direction.
(3) True
44. A
45. C
(1) True
No voltmeter is ideal and has infinitely large internal resistance. With the voltmeter connected across the resistor, the equivalent resistance of the resistor and voltmeter is smaller than the original resistance. Voltage across the resistor should be smaller.
(2) True
With the mercury-in-glass thermometer inserted into the hot water, heat energy will be absorbed until thermal equilibrium is attained by the thermometer and the beaker of water. Thus the temperature of the water will be lowered.
(3) False
04 AL Physics/M.C./P.7