# 9702_w11_ms_41

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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2011 question paper…
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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2011 question paper for the guidance of teachers 9702 PHYSICS 9702/41 Paper 4 (A2 Structured Questions), maximum raw mark 100 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes must be read in conjunction with the question papers and the report on the examination. ã Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the October/November 2011 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9702 41 Section A 1 (a) gravitational force provides the centripetal force B1 GMm/r 2 = mrω2 (must be in terms of ω) B1 r 3ω2 = GM and GM is a constant B1 [3] (b) (i) 1. for Phobos, ω = 2π/(7.65 × 3600) C1 = 2.28 × 10–4 rad s–1 (9.39 × 10 ) × (2.28 × 10–4)2 = 6.67 × 10–11 × M 6 3 C1 M = 6.46 × 1023 kg A1 [3] 2. (9.39 × 106)3 × (2.28 × 10–4)2 = (1.99 × 107)3 × ω2 C1 ω = 7.30 × 10–5 rad s–1 C1 T = 2π/ω = 2π/(7.30 × 10–5) = 8.6 × 104 s = 23.6 hours A1 [3] (ii) either almost ‘geostationary’ or satellite would take a long time to cross the sky B1 [1] 2 (a) e.g. moving in random (rapid) motion of molecules/atoms/particles no intermolecular forces of attraction/repulsion volume of molecules/atoms/particles negligible compared to volume of container time of collision negligible to time between collisions (1 each, max 2) B2 [2] (b) (i) 1. number of (gas) molecules B1 [1] 2. mean square speed/velocity (of gas molecules) B1 [1] (ii) either pV = NkT or pV = nRT and links n and k and EK = ½m c2 M1 3 clear algebra leading to EK = kT A1 [2] 2 (c) (i) sum of potential energy and kinetic energy of molecules/atoms/particles M1 reference to random (distribution) A1 [2] (ii) no intermolecular forces so no potential energy B1 (change in) internal energy is (change in) kinetic energy and this is proportional to (change in ) T B1 [2] © University of Cambridge International Examinations 2011 Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9702 41 3 (a) (i) amplitude remains constant B1 [1] (ii) amplitude decreases gradually M1 light damping A1 [2] (iii) period = 0.80 s C1 frequency = 1.25 Hz (period not 0.8 s, then 0/2) A1 [2] (b) (i) (induced) e.m.f. is proportional to M1 rate of change/cutting of (magnetic) flux (linkage) A1 [2] (ii) a current is induced in the coil M1 as magnet moves in coil A1 current in resistor gives rise to a heating effect M1 thermal energy is derived from energy of oscillation of the magnet A1 [4] 4 (a) (i) zero field (strength) inside spheres B1 [1] (ii) either field strength is zero or the fields are in opposite directions M1 at a point between the spheres A1 [2] (b) (i) field strength is (–) potential gradient (not V/x) B1 [1] (ii) 1. field strength has maximum value B1 at x = 11.4 cm B1 [2] 2. field strength is zero B1 either at x = 7.9 cm (allow ±0.3 cm) or at 0 to 1.4 cm or 11.4 cm to 12 cm B1 [2] 5 (a) (i) Bqv(sinθ) or Bqv(cosθ) B1 [1] (ii) qE B1 [1] (b) FB must be opposite in direction to FE B1 so magnetic field into plane of paper B1 [2] © University of Cambridge International Examinations 2011 Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9702 41 6 (a) (i) period = 1/50 C1 t1 = 0.03 s A1 [2] (ii) peak voltage = 17.0 V A1 [1] (iii) r.m.s. voltage = 17.0/√2 = 12.0 V A1 [1] (iv) mean voltage = 0 A1 [1] (b) power = V 2/R C1 = 122/2.4 = 60 W A1 [2] 7 (a) each line represents photon of specific energy M1 photon emitted as a result of energy change of electron M1 specific energy changes so discrete levels A1 [3] (b) (i) arrow from –0.85 eV level to –1.5 eV level B1 [1] (ii) ∆E = hc /λ C1 = (1.5 – 0.85) × 1.6 × 10–19 C1 = 1.04 × 10–19 J λ = (6.63 × 10–34 × 3.0 × 108)/(1.04 × 10–19) = 1.9 × 10–6 m A1 [3] (c) spectrum appears as continuous spectrum crossed by dark lines B1 two dark lines B1 electrons in gas absorb photons with energies equal to the excitation energies M1 light photons re-emitted in all directions A1 [4] 8 (a) (i) time for initial number of nuclei/activity M1 to reduce to one half of its initial value A1 [2] (ii) λ = ln 2/(24.8 × 24 × 3600) M1 = 3.23 × 10–7 s–1 A0 [1] (b) (i) A = λN C1 3.76 × 106 = 3.23 × 10–7 × N N = 1.15 × 1013 A1 [2] (ii) N = N0 e–λt = 1.15 × 1013 × exp(–{ln 2 × 30}/24.8) C1 = 4.97 × 1012 A1 [2] (c) ratio = (4.97 × 1012)/(1.15 × 1013 – 4.97 × 1012) C1 = 0.76 A1 [2] © University of Cambridge International Examinations 2011 Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9702 41 Section B 9 (a) e.g. reduced gain increased stability greater bandwidth or less distortion (allow any two sensible suggestions, 1 each, max 2) B2 [2] (b) (i) V – connected to midpoint between resistors B1 VOUT clear and input to V+ clear B1 [2] (ii) gain = 1 + RF/R 15 = 1 + 12000/R C1 R = 860 Ω A1 [2] (c) graph: straight line from (0,0) to (0.6,9.0) B1 straight line from (0.6,9.0) to (1.0,9.0) B1 [2] (d) either relay can be used to switch a large current/voltage M1 output current of op-amp is a few mA/very small A1 [2] or relay can be used as a remote switch (M1) for inhospitable region/avoids using long heavy cables (A1) 10 (a) e.g. large bandwidth/carries more information low attenuation of signal low cost smaller diameter, easier handling, easier storage, less weight high security/no crosstalk low noise/no EM interference (allow any four sensible suggestions, 1 each, max 4) B4 [4] (b) (i) infra-red B1 [1] (ii) lower attenuation than for visible light B1 [1] (c) (i) gain/dB = 10 lg(P2/P1) C1 26 = 10 lg(P2/9.3 × 10–6) P2 = 3.7 × 10–3 W A1 [2] (ii) power loss along fibre = 30 × 0.2 = 6.0 dB C1 either 6 = 10 lg(P/3.7 × 10–3) or 6 dB = 4 × 3.7 × 10–3 or 32 = 10 lg(P/9.3 × 10–6) input power = 1.5 × 10–2 W A1 [2] © University of Cambridge International Examinations 2011 Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – October/November 2011 9702 41 11 (a) (i) switch M1 so that one aerial can be used for transmission and reception A1 [2] (ii) tuning circuit M1 to select (one) carrier frequency (and reject others) A1 [2] (iii) analogue-to-digital converter/ADC M1 converts microphone output to a digital signal A1 [2] (iv) (a.f.) amplifier (not r.f. amplifier) M1 to increase (power of) signal to drive the loudspeaker A1 [2] (b) e.g. short aerial so easy to handle short range so less interference between base stations larger waveband so more carrier frequencies (any two sensible suggestions, 1 each, max 2) B2 [2] © University of Cambridge International Examinations 2011
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