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02 AL Physics/M.C./P.1 2002 Hong Kong Advanced Level Examination AL Physics Multiple Choice Questions (Solution) 1. D Before the cord is cut: X is in equilibrium state tension (spring) = mg Y is in equilibrium state tension (cord) = mg + tension (spring) = 2mg At the instant the cord is cut: It takes time for the spring to contract; X is in equilibrium state momentarily, i.e. acceleration of X = 0. Force acting on Y = weight + tension (spring) = 2mg. Hence, acceleration of Y = 2g. 2. C Separation between P and Q is maximum when P is just hitting the ground. Time for P to reach the ground is given by: 1 s = ut gt 2 2 1 2 = gt 2 80 = 5t2 t =4s Position of Q when P is hitting the ground: 1 s = u(t 1) g (t 1) 2 2 1 2 = g (t 1) 2 = 5(4-1)2 = 45 m Hence, max. separation between P and Q = 80 – 45 = 35 m 3. B 1m=n wavelength for orange-red light ~ 650 nm hence, n ~ 1.5 106 n is of order 106. 4. B (1) False Motion of the mass between B and C is s.h.m. (vertical spring and mass oscillation). In s.h.m. kinetic energy is maximum when the mass is passing through the equilibrium position, i.e. the position where mg = kx. (2) False C is the extremity of s.h.m., acceleration is maximum resultant force acting on the mass is maximum. (3) True mg is the equilibrium position. As k the mass is passing through the equilibrium position with k.e., the spring will be compressed further. 5. A T T F mg F mg Fmin Forces acting the bob includes the weight, tension and F. As the bob is in static equilibrium, the resultant force of these three forces is zero. This implies the vectors of these three forces form a triangle as shown in the above figure. As angle is fixed, F is minimum when it is perpendicular to the tension. Hence, minimum F = mg sin . 02 AL Physics/M.C./P.2 6. D (1) momentum by conservation of momentum: momentum (projectile) + momentum (target) = constant 10. C A. a = -2x B. F -x C. final momentum of P = +ve final momentum of Q = 0 final momentum of R = -ve hence, momentum of Z has the great magnitude (2) kinetic energy elastic collision total k.e. conserved. i.e. k.e. (P, Q or R) + k.e. (X, Y or Z) = constant. k.e. = total energy – p.e. 1 = E0 - kx2 2 D. v = A 2 x 2 11. B center of mass l/2 k.e. must be positive. k.e. of the target is maximum when k.e. of the projectile is zero. 7. A impulse = Ft p F = t impulse = p 8. 9. C Owing to change of direction of the velocity, an object undergoing circular motion must have acceleration in the direction pointing towards the center of the circular path (centripetal acceleration). If the motion is a uniform circular motion, centripetal acceleration will be the only acceleration of the object (Figure of option (1)). However, if the motion is nonuniform, there must be a tangential acceleration which is due to the change of “magnitude” of the velocity (Figure of option (2)). A By conservation of energy Total energy = k.e. + p.e. 3E 1 2 Hence, E = kx 4 2 E 1 2 kx = 2 4 1 1 2 = ( kA ) 4 2 A x = 2 by conservation of energy 1 mgh = I2 2 1 2 mgh = I 2 0.1 10 (0.3/2) = 0.5 3 10-3 2 = 10 rads-1 12. D A and C. True For an orbiting satellite G v2 Mm = m 2 r r 1 k.e. = mv 2 2 1 Mm = G 2 r Total mechanical energy E of a satellite orbiting the earth = p.e. + k.e. Mm 1 Mm = G G r 2 r 1 Mm = G 2 r Due to air resistance, total mechanical energy E decreases (i.e. more –ve), hence, r decreases. As r decreases, k.e. increases and v increases. 02 AL Physics/M.C./P.3 B. True 2 v Mm = m r r2 angular momentum = mvr G = m GMr Hence, angular momentum decreases as r decreases. D. False Time needed for the satellite to complete 2r one revolution = . v r decreases and v increases, the time needed should decrease. 16. A Remark: The following three figures show the emerging light rays of a concave lens for virtual objects located at: (1) positions within focal length f; (2) focus and (3) positions farther than focal length f. L (1) f L (2) 13. A P Q f time (3) L 1 4 period later than P. Hence P leads Q by 90 or . 2 Q reaches the maximum at a time f 14. D Ultrasonic wave is longitudinal wave. Polarization cannot be observed for longitudinal wave. 15. D For diffraction grating, angular position of the first-order maximum is given by: d sin = hence, angle between the first-order maxima is given by = 2 = 2 sin 1 d Hence, to maximize , long wavelength light should be used (e.g. red light) and d should be as small as possible (i.e. increases the no. of slit per mm of the grating.) L1 L2 20 cm final image 20 cm The parallel light rays after passing lens L1 will converge towards a point 40 cm on the right of L1. For L2 these light rays form a virtual object at a location 20 cm on the right of L2, i.e. in between L2 and the focus. The light rays after passing lens L2 will converge to a point a little further away. Hence, the final image is real and formed on the right of L2. 02 AL Physics/M.C./P.4 17. C Condition for formation of stationary waves in a fixed ends string: l = n( ) where n = 1, 2, 3, … 2 c = n( ) 2f Hence, c= fn 400 n = n 1 450 n =8 2lf 2 1 400 = = 100 ms-1 n 8 18. B (1) False Number of waves emitted per second from the source = frequency. (2) False Moving observer Doppler effect (3) True Moving source Doppler effect 19. D According to the diagram the following features about the spectrum are observed: a. positions of the spectral lines do not change; b. the spectral lines widen. A. False. In an expanding universe, the stars move away from the earth. Wavelength should increase and the observed spectra should shift toward the red end (red shift). B. False. If the star is approaching the earth, wavelength should decrease and the observed spectra should shift toward the violet end. C. False. Same as A. D. True. Light sources on the spinning star are moving with different velocities, which depend on how far away the sources from the rotating axis is, towards/away from the earth. Hence, the width of the spectral lines spreads out without change of the mean position. 20. C A. False. B. False. As the path taken by the charge is convex away from Q, the two charges must be unlike charges and the force between them is attractive. C. True. The fact that force between Q and q is attractive implies potential energy is negative. As point X is closer to Q than Y, q has a more negative potential energy at X than Y. By conservation of energy, k.e. + p.e. = constant. At X, p.e. is more negative k.e. at X is greater, i.e. greater speed. D. False. Whether X is at a potential higher or lower than Y depends on polarity of Q. If Q is positive, X is at a potential higher than Y. If Q is negative, X is at a potential more negative (i.e. lower) than Y. 21. A A. True. In a magnetic field the electron can undergo circular motion, i.e. it can be deflected by more than 90 while in electric field the electron can be almost deflected by 90. B. False. In both the electric and magnetic field, the force depends on the magnitude of the charge; they are qE and qvB. C. False. If the region of the uniform magnetic field is sufficiently large, the electron will undergo uniform circular motion. The speed of the electron will then be constant. D. False. If the field is magnetic, the direction of the magnetic force, which is always perpendicular the velocity of the moving charge, will change continuously. 22. B 02 AL Physics/M.C./P.5 23. D Q: As the battery have been disconnected, the charge stored on the capacitor must remain constant. 0 A , if one of the plates of the d capacitor is shifted upward slightly, A decreases and C decreases. 26. C time constant = RC 2 10-3 = 1 103 C C = 2 F C: C = Q , Q remains constant and C C decreases, hence, V should increase. V: V = 24. D V1 + V2 = 9 V Hence, V2 = 5 V. Let I1, I2 and I3 be the currents passing V1, V2 and V3 respectively. Since, all the three voltmeters are with the same internal resistance. Vi Ii 27. D By Faraday’s Law, d = dt dA = B dt = Blsin 60 v = 3Blv 2 28. A induced e.mf. enclosed area A resistance length l hence, current A/l 9V current (circle) : current (square) R A I1 B V3 V1 4V S I3 C I2 ( d / 2) 2 d 2 : d 4d = 1:1 = 5V V2 I3 should flow in the direction as shown in the above figure. By Kirchoff’s first law: I1 + I3 = I2 hence, V1 + V3 = V2 and V3 = 1 V. 29. A P Error! BR B Q Starting from the same point A, potential drop across R should be 3 V so that point B is at a potential 1 V higher than C. 25. C Let and r be the e.m.f. and inernal resistance of the supply respectively. Total power P delivered by the supply = I = 2 rR BQ R As shown in the figure above, the resultant magnetic field B due to the two wires Q and R are vertically downwards. Hence, by right-hand grip rule the magnetic force acting on P points to the right horizontally. 30. C Power supplied by the battery = I = 5 2 = 10 W Power loss due to the resistance of the armature coil = i2R = 22 0.5 = 2 W 10 2 100 % = 80% Efficiency = 10 02 AL Physics/M.C./P.6 31. C 37. A VR, Y1 Y1 E Y2 h 3.6 10 15 1.6 10 19 = 22.5 kV 1 n 2f 1 1 c 2 2 ni n f 1 1 ' 1 8 1 32 = = 1 = 1 1 9 9 1 n 2 In an a.c. circuit in which a resistor and a capacitor are connected in series, the voltage across the resistor (VR) should lead that of the capacitor (VC) by 90. However, in the circuit as shown, the polarity of the connection to the CRO is reversed for the capacitor. Hence, Y1 lags behind Y2 by 90. 32. B k.e. of the electrons = photon energy eV = 3.6 10-15 ni2 1 1 1 2 2 ni n f VC V = 1 hf Y2 8 9 ’ = 38. D (1) False The graph does not pass through the origin. (2) True P Error! A B 33. C 0 34. A 35. C A. False. Blue light has shorter wavelength and higher photon energy c = hf = h . The photoelectrons emitted should have a greater maximum kinetic energy. B. False. C. True. No. of photons = intensity photon energy. With the same intensity, no. of photons decreases as the photon energy increases. D. False. Kinetic energy of the emitted photoelectrons increases, a higher stopping voltage is required. 36. B T 1 T = slope of the broken line P VB > VA, hence, volume of the gas increases. V (3) True PV = NkT PV = Nk T Mass of the ideal gas is fixed, N is constant. 02 AL Physics/M.C./P.7 39. B 43. B PV = NkT PV N = kT PV = R T NA PVN A = RT 10 5 10 6 6.02 10 23 = 8.31 (25 273 ) = 2.4 1019 40. D 1 2 3 mc = kT 2 2 c T c' 273 160 = 273 80 c c’ = 1.1 c 41. B kt A = A0 e ln A = ln A0 – kt slope of ln A – t graph = -k = -0.0125 s-1 hence, 42. A k = 0.0125 s-1 half-life = (ln 2)/k = 55.5 s IC =0 IB = 6 – IBRB – VBE I 6 C RB VBE = 0 6 3 10 3 RB 0.6 = 0 80 RB = 144 k 44. C V- ~ V+ = VP no current flows into the op amp hence, V V0 V1 V = 50 k 10 k VP V0 V1 VP = 50 k 10 k VP (0.9) 2.7 VP = 10 50 5(2.7 – VP) = VP+0.9 VP = 2.1 V 45. B m V m = Al density = m (d / 2) 2 l m d2 l d m l = 2 d m l = hence, % error of d = (% error of m + % error of l)/2 = 3%