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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2010 question paper for the…

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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2010 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 CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2010 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 – May/June 2010 9702 41 Section A 1 (a) angle (subtended) at centre of circle B1 (by) arc equal in length to radius B1 [2] (b) (i) point S shown below C B1 [1] (ii) (max) force / tension = weight + centripetal force C1 centripetal force = mrω2 C1 15 = 3.0/9.8 × 0.85 × ω2 C1 ω = 7.6 rad s–1 A1 [4] 2 (a) (i) 27.2 + 273.15 or 27.2 + 273.2 C1 300.4 K A1 [2] (ii) 11.6 K A1 [1] (b) (i) ( c2 is the) mean / average square speed B1 [1] (ii) ρ = Nm/V with N explained B1 so, pV = 1/3 Nm c2 B1 and pV = NkT with k explained B1 so mean kinetic energy / EK = ½m c2 = 3/2 kT B1 [4] (c) (i) pV = nRT 2.1 × 107 × 7.8 × 10–3 = n × 8.3 × 290 C1 n = 68 mol A1 [2] (ii) mean kinetic energy = 3/2 kT = 3/2 × 1.38 × 10–23 × 290 C1 = 6.0 × 10–21 J A1 [2] (iii) realisation that total internal energy is the total kinetic energy C1 energy = 6.0 × 10–21 × 68 × 6.02 × 1023 C1 = 2.46 × 105 J A1 [3] 3 (a) (i) to-and-fro / backward and forward motion (between two limits) B1 [1] (ii) no energy loss or gain / no external force acting / constant energy / constant amplitude B1 [1] (iii) acceleration directed towards a fixed point B1 acceleration proportional to distance from the fixed point / displacement B1 [2] (b) acceleration is constant (magnitude) M1 so cannot be s.h.m. A1 [2] © UCLES 2010 Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9702 41 4 (a) ability to do work B1 as a result of the position/shape, etc. of an object B1 [2] (b) (i) 1 ∆Egpe = GMm / r C1 = (6.67 × 10–11 × {2 × 1.66 × 10–27}2) / (3.8 × 10–15) C1 = 1.93 × 10–49 J A1 [3] 2 ∆Eepe = Qq / 4πε0r C1 = (1.6 × 10–19)2 / (4π × 8.85 × 10–12 × 3.8 × 10–15) C1 = 6.06 × 10–14 J A1 [3] (ii) idea that 2EK = ∆Eepe – ∆Egpe B1 EK = 3.03 × 10–14 J = (3.03 × 10–14) / 1.6 × 10–13 M1 = 0.19 MeV A0 [2] (iii) fusion may occur / may break into sub-nuclear particles B1 [1] 5 (a) (i) VH depends on angle between (plane of) probe and B-field B1 either VH max when plane and B-field are normal to each other or VH zero when plane and B-field are parallel or VH depends on sine of angle between plane and B-field B1 [2] (ii) 1 calculates VHr at least three times M1 to 1 s.f. constant so valid or approx constant so valid or to 2 s.f., not constant so invalid A1 [2] 2 straight line passes through origin B1 [1] (b) (i) e.m.f. induced is proportional / equal to M1 rate of change of (magnetic) flux (linkage) A1 constant field in coil / flux (linkage) of coil does not change B1 [3] (ii) e.g. vary current (in wire) / switch current on or off / use a.c. current rotate coil move coil towards / away from wire (1 mark each, max 3) B3 [3] 6 (a) all four diodes correct to give output, regardless of polarity M1 connected for correct polarity A1 [2] (b) NS / NP = VS / VP C1 V0 = √2 × Vrms C1 ratio = 9.0 / (√2 × 240) = 1/38 or 1/37 or 0.027 A1 [3] © UCLES 2010 Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9702 41 7 (a) arrow pointing up the page B1 [1] (b) (i) Eq = Bqv C1 v = (12 × 103) / (930 × 10–6) C1 = 1.3 × 107 m s–1 A1 [3] (ii) Bqv = mv 2 / r C1 q/m = (1.3 × 107) / (7.9 × 10–2 × 930 × 10–6) C1 = 1.8 × 1011 C kg–1 A1 [3] 8 (a) momentum conservation hence momenta of photons are equal (but opposite) M1 same momentum so same energy A1 [2] (b) (i) (∆)E = (∆)mc2 C1 = 1.2 × 10–28 × (3.0 × 108)2 = 1.08 × 10–11 J A1 [2] (ii) E = hc / λ λ = (6.63 × 10–34 × 3.0 × 108) / (1.08 × 10–11) C1 = 1.84 × 10–14 m A1 [2] (iii) λ = h/p p = (6.63 × 10–34) / (1.84 × 10–14) C1 = 3.6 × 10–20 N s A1 [2] Section B 9 (a) (i) point X shown correctly B1 [1] (ii) op-amp has very large / infinite gain M1 non-inverting input is at earth (potential) / earthed / at 0 V M1 if amplifier is not to saturate, inverting input must be (almost) at earth potential / 0 (V) same potential as inverting input A1 [3] (b) (i) total input resistance = 1.2 kΩ C1 (amplifier) gain (= –4.2 / 1.2) = –3.5 C1 (voltmeter) reading = –3.5 × –1.5 = 5.25 V A1 [3] (total disregard of signs or incorrect sign in answer, max 2 marks) (ii) (less bright so) resistance of LDR increases M1 (amplifier) gain decreases M1 (voltmeter) reading decreases A1 [3] © UCLES 2010 Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9702 41 10 (a) X-ray taken of slice / plane / section B1 repeated at different angles B1 images / data is processed B1 combined / added to give (2-D) image of slice B1 repeated for successive slices B1 to build up a 3-D image B1 image can be viewed from different angles / rotated B1 max 6 [6] (b) (i) 16 A1 [1] (ii) evidence of deducting 16 then dividing by 3 C1 to give A1 [2] 3 2 6 5 11 (a) frequency of carrier wave varies (in synchrony) with signal M1 (in synchrony) with displacement of signal A1 [2] (b) advantages e.g. less noise / less interference greater bandwidth / better quality (1 each, max 2) disadvantages e.g. short range / more transmitters / line of sight more complex circuitry greater expense (1 each, max 2) B4 [4] 12 (a) gain / loss/dB = 10 lg(P1/P2) C1 190 = 10 lg(18 × 103 / P2) or –190 = 10 lg P2 / 18 × 103) C1 power = 1.8 × 10–15 W A1 [3] (b) (i) 11 GHz / 12 GHz B1 [1] (ii) e.g. so that input signal to satellite will not be ‘swamped’ to avoid interference of uplink with / by downlink B1 [1] © UCLES 2010

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