03 AL Physics/M.C./P.6
2003 Hong Kong Advanced Level Examination
AL Physics
Multiple Choice Questions
03 AL Physics/M.C./P.6
1. D
2 ´ F sin q = mg
F =
tension at the lowest point
= F cos q
=
=
2. A
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/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) Þ tan 45° =
F = 2mg
3. 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
The vertical speed increases from 0.15f to 0.25f in T.
vertical acceleration g =
10 = 0.1f 2
f = 10 Hz.
4. 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 =
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.
6. 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.
7. 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
8. B
maximum velocity = wA
maximum momentum = mwA
9. C
A: unchanged
B: T =
mass increases, period increases
C: max acceleration = w2A, with the mass increases, w decreases and the acceleration decreases
D: unchanged, total energy =
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 =
=
=
r =
=
= 5.6 ´ 103 kg m-3
12. D
Potential energy of the mass at the earth’s surface
=
=
= - 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)
=
=
= - 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
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 µ 1/n. Hence, the speeds of light in the three media in descending order is (I) > (III) > (II).
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:
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.
(3) True
18. C
h =
=
= 70 + 10 log 4
= 76 dB
19. C
(1) True
d = = 2000 nm
(2) True
d sin q = ml
d sin q = 2 l
2000 sin q = 2 ´ 500
sin q = 0.5
q = 30°
(3) False
max order = = = 4
20. D
A. False. Eyepiece should be the one with shorter focal length.
B. False. It is the case for the telescopes but not for the microscopes.
C. False. It is not the case for the telescopes.
D. True.
21. C
(1) True
One displacement node in each case.
(2) True
Speed of sound waves must be the same in both cases.
(3) False
Wavelength in a closed tube is double in length of that in an open tube.
22. A
wavelength of red light ~ 1 mm
photon energy
= hf
= hc/l
=
= 2 ´ 10-19 J
23. C
F µ Q1Q2
case 1: like charges
=
= 5:9
case 2: unlike charges
=
= 5:4
24. C
(1) True
(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
25. D
Before:
Potential at P =
= +2 V
After:
Potential at P =
= +3V
26. D
Electric field strength, E = , independent of the separation, remain unchanged.
Electric potential, V = Ed, d decreased, V decreased.
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 W resistors are negligible).
When the low-resistance ammeter is connected across a and b, equivalent resistance of the system = 5 W and the current flow = 12/5 = 2.4 A. Half of the current, i.e. 1.2 A will flow through the ammeter.
28. A
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 kW resistor
= = 5 V
Voltage across 1 kW resistor = 1 V.
The respective charges stored on the capacitors are:
Q (5 mF) = 5 ´ 5 = 25 mC
Q (10 mF) = 10 ´ 1 = 10 mC
Hence, there should be 25 mC - 10mC = 15 mC of positive charges flowing away from the capacitors through K.
29. D
A. False.
Initial current = ,
R doubled, initial current halved.
B. False.
Energy dissipated in the resistor
= energy stored in the capacitor
= , unchanged.
C. False.
Total charge stored in the capacitor = CV, unchanged.
D. True.
Time constant t = RC, halved.
30. D
Magnetic field at P = 0 Þ I1 is in opposite direction to I2 and B-field due to I1 = B-field due to I2
i.e. =
=
I1 = 1.8 A
31. B
The charged particle undergoes circular motion in the magnetic field
Þ qvB =
r =
(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.
32. A
(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
(3) False
Conductor – equip-potential object
33. D
34. C
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:
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 (= wL) will increase. Current flow through the circuit decreases and hence the brightness of the bulb decreases.
(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 W
Average power dissipated in the circuit
=
= 22 ´ 5
= 20 W
37. D
(1) True
PV = nRT
P remains constant and V increases Þ T increases. Hence, internal energy increases. Remark: internal energy depends on temperature of the gas only.
(2) True
V remains constant and P increases Þ T increases.
(3) True
Adiabatic process (DQ = 0). Compression implies work is done on the gas (i.e. DW is +ve). By first Law of Thermodynamics DU = DQ + DW = +ve.
38. C
According to the ideal gas equation PV = nRT, T µ 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
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
µ
=
=
=
l = 121 nm
42. A
43. B
44. B
k = s-1
t1/2 =
= 106 ln 2
= 6.9 ´ 105 s
= 8 days
= 1 week
45. B