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CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2012 series 9702 PHYSICS 9702/41…
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CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2012 series 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 should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 2012 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components. Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL – October/November 2012 9702 41 Section A 1 (a) force is proportional to the product of the masses and inversely proportional to the square of the separation M1 either point masses or separation size of masses A1 [2] (b) (i) gravitational force provides the centripetal force B1 mv2/r = GMm/r2 and EK = ½mv2 M1 hence EK = GMm/2r A0 [2] (ii) 1. ∆EK = ½ × 4.00 × 1014 × 620 × ({7.30 × 106}–1 – {7.34 × 106}–1) C1 = 9.26 × 107 J (ignore any sign in answer) A1 [2] (allow 1.0 × 108 J if evidence that EK evaluated separately for each r) 2. ∆EP = 4.00 × 1014 × 620 × ({7.30 × 106}–1 – {7.34 × 106}–1) C1 = 1.85 × 108 J (ignore any sign in answer) A1 [2] (allow 1.8 or 1.9 × 108 J) (iii) either (7.30 × 106)–1 – (7.34 × 106)–1 or ∆EK is positive / EK increased M1 speed has increased A1 [2] 2 (a) (i) sum of potential energy and kinetic energy of atoms / molecules / particles M1 reference to random A1 [2] (ii) no intermolecular forces B1 no potential energy B1 internal energy is kinetic energy (of random motion) of molecules B1 [3] (reference to random motion here then allow back credit to (i) if M1 scored) (b) kinetic energy ∝ thermodynamic temperature B1 either temperature in Celsius, not kelvin so incorrect or temperature in kelvin is not doubled B1 [2] 3 (a) temperature of the spheres is the same B1 no (net) transfer of energy between the spheres B1 [2] (b) (i) power = m × c × ∆θ where m is mass per second C1 3800 = m × 4.2 × (42 – 18) C1 m = 38 g s–1 A1 [3] (ii) some thermal energy is lost to the surroundings M1 so rate is an overestimate A1 [2] 4 (a) straight line through origin M1 shows acceleration proportional to displacement A1 negative gradient M1 shows acceleration and displacement in opposite directions A1 [4] © Cambridge International Examinations 2012 Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL – October/November 2012 9702 41 (b) (i) 2.8 cm A1 [1] (ii) either gradient = ω2 and ω = 2πf or a = –ω2x and ω = 2πf C1 gradient = 13.5 / (2.8 × 10–2) = 482 ω = 22 rad s–1 C1 frequency = (22/2π =) 3.5 Hz A1 [3] (c) e.g. lower spring may not be extended e.g. upper spring may exceed limit of proportionality / elastic limit (any sensible suggestion) B1 [1] 5 (a) (i) ratio of charge and potential (difference) / voltage (ratio must be clear) B1 [1] (ii) capacitor has equal magnitudes of (+)ve and (-)ve charge B1 total charge on capacitor is zero (so does not store charge) B1 (+)ve and (-)ve charges to be separated M1 work done to achieve this so stores energy A1 [4] (b) (i) capacitance of Y and Z together is 24 µF C1 1 / C = 1 / 24 + 1 / 12 C = 8.0 µF (allow 1 s.f.) A1 [2] (ii) some discussion as to why all charge of one sign on one plate of X B1 Q = (CV =) 8.0 × 10–6 × 9.0 M1 = 72 µC A0 [2] (iii) 1. V = (72 × 10–6) / (12 × 10–6) = 6.0 V (allow 1 s.f.) (allow 72/12) A1 [1] 2. either Q = 12 × 10–6 × 3.0 or charge is shared between Y and Z C1 charge = 36 µC A1 [2] Must have correct voltage in (iii)1 if just quote of 36 µC in (iii)2. 6 (a) (i) particle must be moving M1 with component of velocity normal to magnetic field A1 [2] (ii) F = Bqv sin θ M1 q, v and θ explained A1 [2] (b) (i) face BCGF shaded A1 [1] (ii) between face BCGF and face ADHE A1 [1] (c) potential difference gives rise to an electric field M1 either FE = qE (no need to explain symbols) or electric field gives rise to force (on an electron) A1 [2] © Cambridge International Examinations 2012 Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL – October/November 2012 9702 41 7 (a) induced e.m.f./current produces effects / acts in such a direction / tends M1 to oppose the change causing it A1 [2] (b) (i) 1. to reduce flux losses / increase flux linkage / easily magnetised and demagnetised B1 [1] 2. to reduce energy / heat losses (do not allow ‘to prevent energy losses’) M1 caused by eddy currents A1 [2] (allow 1 mark for ‘reduce eddy currents’) (ii) alternating current / voltage B1 gives rise to (changing) flux in core B1 flux links the secondary coil M1 (by Faraday’s law) changing flux induces e.m.f. (in secondary coil) A1 [4] 8 (a) discrete quantity / packet / quantum of energy of electromagnetic radiation B1 energy of photon = Planck constant × frequency B1 [2] (b) threshold frequency (1) rate of emission is proportional to intensity (1) max. kinetic energy of electron dependent on frequency (1) max. kinetic energy independent of intensity (1) (any three, 1 each, max 3) B3 [3] (c) either E = hc/λ or hc/λ = eV C1 λ = 450 nm to give work function of 3.5 eV energy = 4.4 × 10–19 or 2.8 eV to give λ = 355 nm M1 2.8 eV 3.5 eV so no emission 355 nm 450 nm so no A1 [3] or work function = 3.5 eV threshold frequency = 8.45×1014 Hz C1 450 nm = 6.67×1014 Hz M1 6.67 × 1014 Hz 8.45 × 1014 Hz A1 © Cambridge International Examinations 2012 Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL – October/November 2012 9702 41 Section B 9 (a) e.g. zero output impedance / resistance infinite input impedance / resistance infinite (open loop) gain infinite bandwidth infinite slew rate 1 each, max. 3 B3 [3] (b) (i) graph: square wave M1 correct cross-over points where V2 = V1 A1 amplitude 5 V A1 correct polarity (positive at t = 0) A1 [4] (ii) correct symbol for LED M1 diodes connected correctly between VOUT and earth A1 correct polarity consistent with graph in (i) A1 [3] (R points ‘down’ if (i) correct) 10 X-ray images taken from different angles / X-rays directed from different angles B1 of one section / slice (1) all images in the same plane (1) images combined to give image of section / slice B1 images of successive sections / slices combined B1 image formed using a computer B1 image formed is 3D image (1) that can be rotated / viewed from different angles (1) (four B-marks plus any two additional marks) B2 [6] 11 (a) e.g. noise can be eliminated / filtered / signal can be regenerated extra bits can be added to check for errors multiplexing possible digital circuits are more reliable / cheaper data can be encrypted for security any sensible advantages, 1 each, max. 3 B3 [3] (b) (i) 1. higher frequencies can be reproduced B1 [1] 2. smaller changes in loudness / amplitude can be detected B1 [1] (ii) bit rate = 44.1 × 103 × 16 C1 = 7.06 × 105 s–1 number = 7.06 × 106 × 340 = 2.4 × 108 A1 [2] 12 (a) (i) signal in one wire (pair) is picked up by a neighbouring wire (pair) B1 [1] (ii) outer of coaxial cable is earthed B1 outer shields the core from noise / external signals B1 [2] © Cambridge International Examinations 2012 Page 6 Mark Scheme Syllabus Paper GCE AS/A LEVEL – October/November 2012 9702 41 (b) attenuation per unit length = 1/L × 10 lg(P2/P1) C1 signal power at receiver = 102.5 × 3.8 × 10–8 = 1.2 × 10–5 W C1 attenuation in wire pair = 10 lg({3.0 × 10–3} / {1.2 × 10–5}) = 24 dB C1 attenuation per unit length = 24 / 1.4 = 17 dB km–1 A1 [4] (other correct methods of calculation are possible) © Cambridge International Examinations 2012
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