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01 AL Physics/M.C./P.1 2001 Hong Kong Advanced Level Examination AL Physics Multiple Choice Questions (Solution) 1. 2. 3. E The weight must pass through the center of the mass of the block. The frictional force must lie along the surface between the block and the inclined plane. The reaction force which is perpendicular to the inclined plane could not pass through the center of the mass, or else there would be a net moment of force about the center of the mass of the block and the block would rotate. To be in equilibrium the reaction force vector must lie on the left of the center of mass. The two moment of forces due to the reaction force (clockwise) and the frictional force (anticlockwise) will then balance each other. B Remark: Properties of action and reaction pair: - equal in magnitude but opposite in direction; - they act on “different” object C Initial k.e. due to the vertical motion = E/2 When the stone is “halfway” up, k.e. due to the vertical motion = E/4 and k.e. due to the horizontal motion remains constant = E/2 Hence, when the stone is halfway up, total k.e. of the stone = E/4 + E/2 = 3E/4 4. A Remark: Impulse = Ft = change of momentum (vector) 5. A Let w be the speed of the object at point Y. Since the object undergoes uniform acceleration motion, w2 = u2 + 2as v2 = w2 + 2as By solving the above two equations, w= 6. C 7. E u2 v2 2 vmax = A = 1/0.5 = 2 rads-1 By conservation of energy, 2 = v2 + (x)2 vmax 12 = v2 + (2 0.3)2 v = 0.8 ms-1 8. C Remark: Forces acting on the pendulum bob are weight of the bob and the tension in the string 9. E Ihollow > Isolid Hence, at the bottom of the incline plane, rotational k.e. of the Ihollow > rotational k.e. of the Isolid. 10. A 1 2 mv 2 Angular momentum l = mvr 1 l Hence, k.e. = m( ) 2 2 mr 1 l2 = 2 mr 2 Since F is a central force, during the process, no torque is acting on the subject angular momentum conserves, i.e. l remains constant throughout the process. 1 Hence, k.e. 2 r as r r/2, k.e. 4 k.e. k.e. = 11. D 1 s = 9 192 631 770 period Hence, frequency f = 9 192 631 770 Wavelength = c/f = 3.3 cm, microwave 01 AL Physics/M.C./P.2 17. D P 12. E 13. E y = (2) False The outer rings are packed more closely than the inner rings because the slope of the lens surface increases outwards. (3) False Even if liquid is introduced between the convex lens and the reflector, for the light ray reflected by the reflector, it still suffers a phase inversion due to the reflection from the optically denser medium. Hence, there is a phase difference between the two reflected light rays. 20 2 R I ’ = I0 cos2 I = I ’ sin2 = I0 sin2 cos2 I = 0 sin 2 2 4 When = 0 and 90, I = 0 otherwise I 0 Hence, when Q is being rotated slowly through 90, I increases and then decreases. 18. D min. frequency C 15. D 40 I’ D , d y d 14. A (1) True If the lens is with long focal length, the slope of the lens surface will be gentle. 60 Q I0 I = 10 log 1 I0 I = 10 log 2 I0 I I = 10(log 1 – log 2 ) I0 I0 I1 = log I2 I1 = 100 I2 max. frequency 19. D 20. A (1) True With the positively charged rod brought near X, electrons will be drawn from the earth. (2) False 16. A (3) & (4) False X and Y are in contact with each other and earthed. Both X and Y are at zero potential. I 01 AL Physics/M.C./P.3 21. C when the 100 V is applied across AD R VCD = 100 = 80 10 10 R R = 80 when the 100 V is applied across CD 80 VAB = 100 = 40 V 60 60 80 22. D bulb X : 6 V, 12 W bulb Y : 6 V, 3 W If the two bulbs work at their respective rated values, the voltage across both of the bulbs must be the same i.e. 6 V, but the current passing through X should be with greater value. A & B: it is not possible that voltage across X and Y are the same. C: the current passing through X and Y are the same. E: current passing through X is smaller than that of Y 24. A (1) True Q = CV = 100000 10-6 20 = 2 C To discharge the capacitor in 2000 s, a mean current of 2/2000 = 1 mA will be provided. (2) False Time constant RC = 10 100000 10-6 = 1 s. In the discharging process, which is an exponential decay, the capacitor is assumed to be completely discharged in 5RC. (3) False 1 CV 2 2 1 = 100000 10 6 20 2 2 = 20 J Energy = 25. A C = 0 hence, 1 d 1 1 d f di = 1 di di d f = df C 23. B C f Ci Ci C = Q/V = To determine the charge stored in the capacitor, imagine that the two terminals of the capacitor are shorted by a connecting wire. The net charge flows through the wire is equal to the charge stored in the capacitor. With the terminals being shorted, the charges on the two metal plates will redistribute until equilibrium state is reached. At equilibrium there should be (900 + 100)/2 = 500 C of charge on one each metal plates. Hence, there is 400 C of charge flowing through the connecting wire and the original charge stored on the capacitor should equal to 400 C. 400 C= = 100 F 4 A d di 1 df 1 1 1.2 = –16.7% = 26. C (1) True The induced e.m.f. of an inductor is given by dI d = L = dt dt Hence, henry-ampere is equivalent to weber. (2) True By Faraday’s Law, d = dt Hence, volt-second is equivalent to weber. 01 AL Physics/M.C./P.4 (3) False magnetic flux = magnetic field area 1 weber = 1 Teslametre2 27. D It is not possible to represent an electric field by field lines of closed loops. 28. E According to the laws of electromagnetic induction, the current is induced as to oppose the change in magnetic flux. In the question, the magnetic pole P is approaching and the magnetic flux confined by the loop will increase. To oppose the change (i.e. increase in magnetic flux) the loop C should move away from the original position and its area should slightly decrease, no matter whether P is an N-pole or a S-pole. 29. D 30. B Average power dissipation 2 = irms R 2 =2 5 = 20 W Remark: reactance does not dissipate power. 31. C A: circle, sinusoidal signals to X and Y are with phase difference 90 and equal amplitude B: ellipse, sinusoidal signals to X and Y are with phase difference in between 0 and 90. C: sinusoidal curve, sinusoidal signal to Y and X is controlled by the time base. D: straight line, sinusoidal signals to X and Y are antiphase. E. Lissajous figure, sinusoidal signals to X and Y in frequency ratio of 2:1. 32. B 33. E The ideal is being compressed, volume of the gas decreases. Work done on the gas W is positive. As the pressure remains constant, by the ideal gas equation PV = nRT, temperature of the gas decreases. Hence, internal energy U decreases (i.e. negative). 34. B Remark: The concept is required in explaining the emission of -rays in radioactive decay. In -rays emission, the nucleons transit from high to low energy level in the nucleus. These energy levels are discrete in nature. 35. E dU dr Hence, the potential energy due to shortrange repulsive and long-range attractive forces should vary inversely as r5 and r2 respectively. Potential energy due to a repulsive force is positive while that of an attractive force is negative. Interatomic force F = 36. B F e = E A l 4mgl Fl mgl E= = = d 2 Ae d 2 e ( ) e 2 m g l d e E = 2 m g l d e E m 0.01 = = 0.1% m 10 g 0.1 = = 1.0% g 9.8 l 0.01 = = 0.51% l 1.96 d 0.01 = 2 = 3.3% 2 d 0.61 e 0.1 = = 2.6% e 3.9 37. E (1) True = Eu Slope of a stress-strain graph in the proportional region is Young modulus which reflects the stiffness of the material. 01 AL Physics/M.C./P.5 (2) True Breaking stress of Y is greater than that of Z. (3) True Maximum strain Z can be stretched to is 1.5, i.e. e/l = 1.5. Hence, Z can be stretched to 2.5 original length. 38. C (1) True Equation of continuity: Av = constant. 42. B 70 49 = A0 70 A0 = 70 2 = 100 Bq 49 43. B (1) False The nuclear binding energy depends on the total number of nucleons inside the nucleus. (2) False (2) True X and Z are with the same crosssectional area A. By equation of continuity, the water speeds are the same. (3) False Water speed in Y is the highest. By Bernoulli’s Principle, pressure at Y is lowest. Hence, the water level in manometer Y is the lowest. 39. C c . Wavelength of red light is longer than that of yellow light. Energy of a photon = hf = h (3) True 44. C Binding energy per nucleon 1.0087 1.0078 2.0146 = ( ) 931 2 = 0.88 MeV 45. D (1) False Penetrating power of a radiation depends on the type of radiations it is and its corresponding energy. (2) True After 20 minutes, percentage of the original radioactive isotope remains: 40. A f 1 1 ni n f 41. B Vout = 2 V voltage across 2 k = 6 – 2 = 4 V current flow through 2 k = 4/2k = 2 mA Ib = Ic/ = 2 m/100 = 20 A 6 V – Ib R – Vbe = 0 6 – 20 R – 0.6 = 0 R = 270 k Isotope X Y Z (3) True Percentage 50 25 75