2014 HKDSE. Physics Paper 1A Suggested Solution
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1 2014 HKDSE 1 Physics Paper 1A Suggested Solution Prepared by C.M. <HKCEE & HKALE A in Physics>
2 2014 HKDSE Physics Paper 1A Suggested Solution 1. D 2. A 3. C 4. A 5. B 6. D 7. C 8. C 9. B 10. B 11. B 12. B 13. B 14. A 15. C 16. A 17. B 18. C 19. A 20. C 21. D 22. D 23. B 24. C 25. D 26. A 27. D 28. D 29. B 30. B 31. D 32. A 33. C HKDSE Physics Paper 1A Question Analysis Number of MCQ Percentage I. Heat & Gas 2 6.1% II. Force & Motion % III. Wave Motion % IV. Electricity & Magnetism % V. Radioactivity & Nuclear Energy 3 9.1% % 40.0% Topic Covered in Physics Paper IA 30.0% 20.0% 10.0% 0.0% Heat & Gas Force & Motion Wave Motion Electricity & Magnetism Radioactivity & Nuclear Energy Paper IA 6.1% 30.3% 21.2% 33.3% 9.1%
3 Q1. D 1) Temp. difference heat transfer (from high T to low T) T of surrounding > T of ice-cream Heat gain by ice-cream from surrounding 2) Vacuum flask (Y) rate of heat gain from surrounding by conduction & convection 3) Rate of heat gain = E t = ml t (identical ice-cream same m) 3 Rate of heat gain 1 t Rate of heat gain in Y < Rate of heat gain in X t y > t x Q2. A 1) Same electric heater same P 2) By Pt = mc T (for changing T) P(2x60) = m(800)(80 20) P = 400m 3) By Pt = ml f (for changing state) 400m(8 2)(60) = ml f l f = 144 kjkg -1 Remarks: CANNOT use the change in T in LIQUID state to calculate Question only give specific heat capacity in SOLID state Q3. C 1) Uniform density Center of mass at the middle 2) Rod suspended at Q Take moment at Q Clockwise moment = Anti-clockwise moment M QR (3) = M PQ (2) M PQ M QR = 3 2 M PQ M QR = 3 : 2
4 Q4. A 1) Two strings (non-identical) Two tensions (T 1 for left, T 2 for right) 2) At equilibrium no F net Considering 30N objects, F = F T 1 = 30N Considering 20N objects, F = F T 2 = 20N 4 3) Considering object W, F = F 30sinθ + 20sinϕ = W F = F 30cosθ = 20cosϕ 4) θ & ϕ < 90 o sinθ & sinϕ < 1 W < W < 50N Q5. B 1) Consider the 1 st phase, By s 1 = ut a(t 1) 2 36 = 4u a(4)2 36 = 4u + 8a (1) 2) Consider the whole motion, By s = ut at2 72 = 6u a(6)2 72 = 6u + 18a (2) 3) By solving (1) & (2) a = 3ms -2
5 Q6. D 1) Identical blocks same mass From rest u = 0 2) By law of conservation of energy Loss in G.P.E. = Gain in K.E. mgh 0 = 1 2 mv2 0 v = 2gH v H same H v 1 = v 2 3) By Newton s 2 nd law, F net = ma mg sin θ = ma a = g sin θ a sin θ θ 5 θ 1 > θ 2 a 1 > a 2 4) By a = v u t a 1 t ( v & u is the same) a 1 t t 1 < t 2 Q7. C Take right as +ve Q can only move to right after collision ( F net acting on Q is to the right side) By Law of conservation of momentum m p u P + m Q u Q = m p v P + m Q v Q 1) (2)(+6) + (1)(0) = (2)( v P ) + (1)(+2) v P = +5ms 1 > v Q ( +2 ms 1 ) It is not possible as Q cannot exceed P 2) (2)(+6) + (1)(0) = (2)( v P ) + (1)(+4) v P = +4ms 1 = v Q It is possible as Q move with P together 3) (2)(+6) + (1)(0) = (2)( v P ) + (1)(+6) v P = +3ms 1 < v Q (+6 ms 1 ) It is possible. Q move faster than P
6 Q8. C Consider whole system By Newton s 2 nd law F net = ma 5g 3g = 8a a = g 4 ms 2 Q9. B By law of conservation of energy Loss in G.P.E. = Gain in K.E. mgh 0 = 1 2 m(v2 u 2 ) 6 (9.81)(h) = 1 2 ( ) h = 6.0m Q10. B Take downwards & right as +ve 1) Consider vertical motion, u y = 0 s y = u y t a yt = 1 2 (9.81)t2 t = 0.4s 2) Consider horizontal motion u x = s x t u x = u x = 2.5ms 1 u = 2.5ms 1 ( u y = 0) Q11. B Option A: Weightless Normal reaction = 0N, not related to G-force Option B: In circular motion, Centripetal force is provided by the F net towards center F net = mv2 r GMm r 2 = mv2 r a c = v2 r Both astronaut and the spacecraft are moving with same a c Option C: Reaction force of the floor = normal reaction = 0N Option D: F c = F net towards center F c is not an extra force!!!
7 Q12. B 1) G-field experienced on the surface of the Earth = g g = GM = acceleration due to gravity R2 2) Consider circular motion (with radius = 2R), F net = mv2 r GMm (2R) 2 = mv2 2R 3) Consider a c v = GM 2R 7 a c = v2 r a c = a c = GM 4R 2 = 1 4 g ( GM 2R )2 2R Q13. B Note that the direction of wave is to the wavefront Q14. A 1) P is moving upwards 2) Q and S are moving in same direction 3) R is moving downwards Only the particles at extreme points (max. / min.) will be momentarily at rest
8 Q15. C By Snell's law & v = fλ, sin θ 1 sin θ 2 = v 1 v 2 = λ 1 λ 2 = n 2 n 1 θ v λ 1 n θ v in this question v III < v I < v II Q16. A Degree of diffraction λ 1 a where a is the gap width 8 Q17. B Let the distance travelled by car be x 1) Consider the speed of ultrasound, By v = d t x 340= ( ) x = 4m 2) Consider the speed of the car, By v = d t 4 v = ( )/2 v = 20ms 1 Q18. C 1) For anti-phase, Constructive interference occur if PD = n λ Destructive interference occur if PD = nλ 2) Path diff. at O = 0λ Destructive Interference Path diff. at P = (3 2.8) 0.1 = 1λ Destructive Interference
9 Q19. A 1) Sound wave are mechanical waves 2) Sound wave mechanical wave need vibration of paricle to transmit cannot transmit in vacuum 3) Sound can form stationary wave 9 Q20. C 1) Two sphere in contact consider as same conductor 2) +ve charged rod near X -ve charged induced in X near rod +ve charged induced in Y 3) When X is earthed, e - flow to Y ( -ve charged is attracted by rod) to neutralize the +ve 4) Y become neutral X become ve charged after separation & rod removal
10 Q21. D By Columb s Law, F = Q AQ B 4πεr 2 Let distance between each particle be r Option A F net acting on Q 1 to left F net acting on Q 2 = 0N F net acting on Q 3 to right Option B F net acting on Q 1 = (+2)( 1) 4πεr 2 + (+2)(+2) 4πε(2r) 2 = F net acting on Q 2 = 0N 1 4πεr 2 N F net acting on Q 3 = (+2)(+2) 4πε(2r) 2 + (+2)( 1) 4πεr 2 = 1 4πεr 2 N 10 Option C F net acting on Q 1 to right F net acting on Q 2 to left F net acting on Q 3 = (+4)( 4) + (+4)(+1) = 0N 4πε(2r) 2 4πεr 2 Option D F net acting on Q 1 = ( 4)(+1) 4πεr 2 F net acting on Q 2 = (+1)( 4) 4πεr 2 + ( 4)( 4) 4πε(2r) 2 = 0N + (+1)( 4) 4πεr 2 = 0N F net acting on Q 3 = ( 4)( 4) + ( 4)(+1) = 0N 4πε(2r) 2 4πεr 2 Q22. D 1) Electron is ve charged F E to left Left plate is +ve E-field point from P to Q 2) By E = F Q E = E = 50NC
11 Q23. B Let the distance between charge +Q & X be 3x By V = +Q 4πεr V X = V Y = +Q 4πε(3r) +Q V 4πε(2r) Y = 3 ( +Q ) 2 4πε(3r) V Y = 3 2 V X 11 Q24. C 1) When S is closed 3Ω resistor is shorted By V = IR e. m. f. = (3)(6 + r) (1) 2) When S is opened, e. m. f. = I( r) (2) 3) Combine (1) & (2) (3)(6 + r) = I( r) r = 9I + Ir (3) By substitution I into (3), internal resistor (r) is calculated Option A Option B Option C Option D When I = 1.6A, r = 2.57Ω (Not possible) When I = 2.0A, r = 0Ω (Not possible) When I = 2.4A, r = 6Ω (Possible) When I = 3.2A, r = 54Ω (Possible)
12 Q25. D 1) For short-circuited, R = 0Ω, V = 0V, i = 2) For open circuit, R =, V = e. m. f. i = 0A 3) The reading of voltmeter = V P (Potential difference across P) Option A When P & Q are shorted V P = 0V 12 Option B When P & Q are in opon circuit V P + V Q = 6V V Q or V P 6V Option C When P is shortrd V P = 0V When Q is in open circuit V Q = e. m. f. = 6V V P = 0V ( V Q + V P = 6V) Option D When P is in open circuit V P = e. m. f. = 6V When Q is shorted V Q = 0V V P = 6V ( V Q + V P = 6V)
13 Q26. A 1) By right hand grip rule, consider force acting at point O Force by charge Q towards P Force by charge P towards S Force by charge S towards P Force by charge R towards Q 2) Force toward Q & S are balanced 13 Force in OP direction Q27. D 1) By right hand grip rule, i B If i = +ve (flow upwards), B into coil PQRS If i = ve (flow downwards), B out of coil PQRS 2) Consider 1 st phase, when i = +ve & B into coil By Lenz s Law, induced i flow in clockwise to generate B coil (to oppose the change B into coil) into 3) Consider 2 nd phase, when i = -ve & B out of coil By Lenz s Law, induced i flow in clockwise to generate B coil (to oppose the change B out of coil) into Q28. D Consider B = μ oni I B N 1 l When l & N B (Independent of cross sectional area)
14 Q29. B 1) Metal block e is charge carrier 2) PQ is higher potential +ve charged e is accumulated at SR F B acting into block (towards SR) 3) By Fleming left hand rule, i to the left, F B into paper B upwards (Q to P) Q30. B 1) In d.c. condition, P = V2 R P = 102 R P = 100 R 14 2) In a.c. condition, 1 2 P = V r.m.s. R (100 R ) = V 2 r.m.s. R V r.m.s. = 5 2 V Q31. D 1) M = Mass no. = no. of p + + n A = Atomic no. = no. of p + 2) Consider the decays equations, M M 4 4 AW A 2X + 2 α M 4 M 4 Y + β A 2X A 1Y A M 4 M 4 Z + β A 1 0 Statement (1) Y has 1 more proton than X Statement (2) No. of n = M A Statement (3) Isotope = atoms of same element of same no. of p +, different no. of n (OR mass number)
15 Q32. A 1) α radiation blocked by paper x < 450 2) Aluminum is used to block β radiation No β radiation no after passing through aluminum no difference in x & y Only A & C are possible answers 3) γ willl be to 1 2 of initial intensity by 25mm lead 15 2mm lead will block a small amount of γ z < y 4) Blackgrond radiation (50) cannot be blocked by any material y > 50 Only A is possible answer Q33. C 1) Consider the decay equation, Ra Rn + α + 4.9MeV 2) Energy difference = 4.9MeV = ) By E = mc = ( m)(3 x10 8 ) 2 m = kg 4) Energy is released in the product mass of product < mass of reactant - The End -
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