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CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2013 series 9702 PHYSICS 9702/43 Paper 4 (A2…

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CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2013 series 9702 PHYSICS 9702/43 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 May/June 2013 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 – May/June 2013 9702 43 Section A 1 (a) region of space area / volume B1 where a mass experiences a force B1 [2] (b) (i) force proportional to product of two masses M1 force inversely proportional to the square of their separation M1 either reference to point masses or separation ‘size’ of masses A1 [3] (ii) field strength = GM / x2 or field strength ∝ 1 / x2 C1 ratio = (7.78 × 108)2 / (1.5 × 108)2 C1 = 27 A1 [3] (c) (i) either centripetal force = mRω2 and ω = 2π / T or centripetal force = mv2 / R and v = 2πR /T B1 gravitational force provides the centripetal force B1 either GMm / R2 = mRω2 or GMm / R2 = mv2 / R M1 M = 4π2R3 / GT2 A0 [3] (allow working to be given in terms of acceleration) (ii) M = {4π2 × (1.5 × 1011)3} / {6.67 × 10–11 × (3.16 × 107)2} C1 = 2.0 × 1030 kg A1 [2] 2 (a) obeys the equation pV = constant × T or pV = nRT M1 p, V and T explained A1 at all values of p, V and T/fixed mass/n is constant A1 [3] (b) (i) 3.4 × 105 × 2.5 × 103 × 10–6 = n × 8.31 × 300 M1 n = 0.34 mol A0 [1] (ii) for total mass/amount of gas 3.9 × 105 × (2.5 + 1.6) × 103 × 10–6 = (0.34 + 0.20) × 8.31 × T C1 T = 360 K A1 [2] (c) when tap opened gas passed (from cylinder B) to cylinder A B1 work done on gas in cylinder A (and no heating) M1 so internal energy and hence temperature increase A1 [3] © Cambridge International Examinations 2013 Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9702 43 3 (a) (i) 1. amplitude = 1.7 cm A1 [1] 2. period = 0.36 cm C1 frequency = 1/0.36 frequency = 2.8 Hz A1 [2] (ii) a = (–)ω2x and ω = 2π/T C1 acceleration = (2π/0.36)2 × 1.7 × 10–2 M1 = 5.2 m s–2 A0 [2] (b) graph: straight line, through origin, with negative gradient M1 from (–1.7 × 10–2, 5.2) to (1.7 × 10–2, –5.2) A1 [2] (if scale not reasonable, do not allow second mark) (c) either kinetic energy = ½mω2(x02 – x2) or potential energy = ½mω2x2 and potential energy = kinetic energy B1 ½mω (x0 – x2) = ½ × ½mω2x02 or ½mω2x2 = ½ × ½mω2x02 2 C1 x02 = 2x2 x = x0 / √2 = 1.7 / √2 = 1.2 cm A1 [3] 4 (a) work done moving unit positive charge M1 from infinity (to the point) A1 [2] (b) (gain in) kinetic energy = change in potential energy B1 ½mv2 = qV leading to v = (2Vq/m)½ B1 [2] (c) either (2.5 × 105)2 = 2 × V × 9.58 × 107 C1 V = 330 V M1 this is less than 470 V and so ‘no’ A1 [3] or v = (2 × 470 × 9.58 × 107) (C1) v = 3.0 × 105 m s–1 (M1) this is greater than 2.5 × 105 m s–1 and so ‘no’ (A1) or (2.5 × 105)2 = 2 × 470 × (q/m) (C1) (q/m) = 6.6 × 107 C kg–1 (M1) this is less than 9.58 × 107 C kg–1 and so ‘no’ (A1) © Cambridge International Examinations 2013 Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9702 43 5 (a) (uniform magnetic) flux normal to long (straight) wire carrying a current of 1 A M1 (creates) force per unit length of 1 N m–1 A1 [2] (b) (i) flux density = 4π × 10–7 × 1.5 × 103 × 3.5 C1 = 6.6 × 10–3 T A1 [2] (ii) flux linkage = 6.6 × 10–3 × 28 × 10–4 × 160 C1 = 3.0 × 10–3 Wb A1 [2] (c) (i) (induced) e.m.f. proportional to rate of M1 change of (magnetic) flux (linkage) A1 [2] (ii) e.m.f. = (2 × 3.0 × 10–3) / 0.80 C1 = 7.4 × 10–3 V A1 [2] 6 (a) (i) to reduce power loss in the core B1 due to eddy currents/induced currents B1 [2] (ii) either no power loss in transformer or input power = output power B1 [1] (b) either r.m.s. voltage across load = 9.0 × (8100 / 300) C1 peak voltage across load = √2 × 243 = 340 V A1 [2] or peak voltage across primary coil = 9.0 × √2 (C1) peak voltage across load = 12.7 × (8100/300) = 340 V (A1) 7 (a) (i) lowest frequency of e.m. radiation M1 giving rise to emission of electrons (from the surface) A1 [2] (ii) E = hf C1 threshold frequency = (9.0 × 10–19) / (6.63 × 10–34) = 1.4 × 1015 Hz A1 [2] (b) either 300 nm ≡ 10 × 1015 Hz (and 600 nm ≡ 5.0 × 1014 Hz) or 300 nm ≡ 6.6 × 10–19 J (and 600 nm ≡ 3.3 × 10–19 J) or zinc λ0 = 340 nm, platinum λ0 = 220 nm (and sodium λ0 = 520 nm) M1 emission from sodium and zinc A1 [2] (c) each photon has larger energy M1 fewer photons per unit time M1 fewer electrons emitted per unit time A1 [3] © Cambridge International Examinations 2013 Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9702 43 8 (a) two (light) nuclei combine M1 to form a more massive nucleus A1 [2] (b) (i) ∆m = (2.01410 u + 1.00728 u) – 3.01605 u = 5.33 × 10–3 u C1 energy = c2 × ∆m C1 = 5.33 × 10–3 × 1.66 × 10–27 × (3.00 × 108)2 = 8.0 × 10–13 J A1 [3] (ii) speed/kinetic energy of proton and deuterium must be very large B1 so that the nuclei can overcome electrostatic repulsion B1 [2] Section B 9 (a) (i) light-dependent resistor/LDR B1 [1] (ii) strain gauge B1 [1] (iii) quartz/piezo-electric crystal B1 [1] (b) (i) resistance of thermistor decreases as temperature increses M1 etiher VOUT = V × R / (R + RT) or current increases and VOUT = I R A1 VOUT increases A1 [3] (ii) either change in RT with temperature is non-linear or VOUT is not proportional to RT/ change in VOUT with RT is non-linear M1 so change is non-linear A1 [2] 10 (a) sharpness: how well the edges (of structures) are defined B1 contrast: difference in (degree of) blackening between structures B1 [2] (b) e.g. scattering of photos in tissue/no use of a collimator/no use of lead grid large penumbra on shadow/large area anode/wide beam large pixel size (any two sensible suggestions, 1 each) B2 [2] (c) (i) I = I0e–µx C1 ratio = exp(–2.85 × 3.5) / exp(–0.95 × 8.0) C1 = (4.65 × 10–5) / (5.00 × 10–4) = 0.093 A1 [3] (ii) either large difference (in intensities) or ratio much less than 1.0 M1 so good contrast A1 [2] (answer given in (c)(ii) must be consistent with ratio given in (c)(i)) © Cambridge International Examinations 2013 Page 6 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9702 43 11 (a) (i) amplitude of the carrier wave varies M1 (in synchrony) with the displacement of the information signal A1 [2] (ii) e.g. more than one radio station can operate in same region/less interference enables shorter aerial increased range/less power required/less attenuation less distortion (any two sensible answers, 1 each) B2 [2] (b) (i) frequency = 909 kHz C1 wavelength = (3.0 × 108) / (909 × 103) = 330 m A1 [2] (ii) bandwidth = 18 kHz A1 [1] (iii) frequency = 9000 Hz A1 [1] 12 (a) for received signal, 28 = 10 lg(P / {0.36 × 10–6}) C1 P = 2.3 × 10–4 W A1 [2] (b) loss in fibre = 10 lg({9.8 × 10–3} / {2.27 × 10–4}) C1 = 16 dB A1 [2] (c) attenuation per unit length = 16 / 85 = 0.19 dB km–1 A1 [1] © Cambridge International Examinations 2013

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