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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2012 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 2012 question paper for the guidance of teachers 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 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 May/June 2012 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 2012 9702 43 Section A 1 (a) work done in bringing unit mass from infinity (to the point) B1 [1] (b) gravitational force is (always) attractive B1 either as r decreases, object/mass/body does work or work is done by masses as they come together B1 [2] (c) either force on mass = mg (where g is the acceleration of free fall /gravitational field strength) B1 g = GM/r2 B1 if r @ h, g is constant B1 ∆EP = force × distance moved M1 = mgh A0 or ∆EP = m∆φ (C1) = GMm(1/r1 – 1/r2) = GMm(r2 – r1)/r1r2 (B1) if r2 ≈ r1, then (r2 – r1) = h and r1r2 = r2 (B1) g = GM/r2 (B1) ∆EP = mgh (A0) [4] (d) ½mv2 = m∆φ v2 = 2 × GM/r C1 = (2 × 4.3 × 1013) / (3.4 × 106) C1 v = 5.0 × 103 m s–1 A1 [3] (Use of diameter instead of radius to give v = 3.6 × 103 m s–1 scores 2 marks) 2 (a) (i) either random motion or constant velocity until hits wall/other molecule B1 [1] (ii) (total) volume of molecules is negligible M1 compared to volume of containing vessel A1 or radius/diameter of a molecule is negligible (M1) compared to the average intermolecular distance (A1) [2] (b) either molecule has component of velocity in three directions or c2 = cX2 + cY2 + cZ2 M1 random motion and averaging, so cX2 = cY2 = cZ2 M1 c2 = 3 cX2 A1 so, pV = ⅓Nm c2 A0 [3] (c) c2 ∝ T or crms ∝ T C1 temperatures are 300 K and 373 K C1 crms = 580 m s–1 A1 [3] (Do not allow any marks for use of temperature in units of ºC instead of K) © University of Cambridge International Examinations 2012 Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 43 3 (a) (numerically equal to) quantity of (thermal) energy required to change the state of unit mass of a substance M1 without any change of temperature A1 [2] (Allow 1 mark for definition of specific latent heat of fusion/vaporisation) (b) either energy supplied = 2400 × 2 × 60 = 288000 J C1 energy required for evaporation = 106 × 2260 = 240000 J C1 difference = 48000 J rate of loss = 48000 / 120 = 400 W A1 or energy required for evaporation = 106 × 2260 = 240000 J (C1) power required for evaporation = 240000 / (2 × 60) = 2000 W (C1) rate of loss = 2400 – 2000 = 400 W (A1) [3] 4 (a) a = (–)ω2x and ω = 2π/T C1 T = 0.60 s C1 a = (4π2 × 2.0 × 10–2) / (0.6)2 = 2.2 m s–2 A1 [3] (b) sinusoidal wave with all values positive B1 all values positive, all peaks at EK and energy = 0 at t = 0 B1 period = 0.30 s B1 [3] 5 (a) force per unit positive charge acting on a stationary charge B1 [1] (b) (i) E = Q / 4πε0r2 C1 Q = 1.8 × 104 × 102 × 4π × 8.85 × 10–12 × (25 × 10–2)2 M1 Q = 1.25 × 10–5 C = 12.5 µC A0 [2] (ii) V = Q / 4πε0r = (1.25 × 10–5) / (4π × 8.85 × 10–12 × 25 × 10–2) C1 = 4.5 × 105 V A1 [2] (Do not allow use of V = Er unless explained) © University of Cambridge International Examinations 2012 Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 43 6 (a) (i) peak voltage = 4.0 V A1 [1] (ii) r.m.s. voltage (= 4.0/√2) = 2.8 V A1 [1] (iii) period T = 20 ms M1 frequency = 1 / (20 × 10–3) M1 frequency = 50 Hz A0 [2] (b) (i) change = 4.0 – 2.4 = 1.6 V A1 [1] (ii) ∆Q = C∆V or Q = CV C1 = 5.0 × 10–6 × 1.6 = 8.0 × 10–6 C A1 [2] (iii) discharge time = 7 ms C1 current = (8.0 × 10–6) / (7.0 × 10–3) M1 = 1.1(4) × 10–3 A A0 [2] (c) average p.d. = 3.2 V C1 resistance = 3.2 / (1.1 × 10–3) = 2900 Ω (allow 2800 Ω) A1 [2] 7 (a) sketch: concentric circles (minimum of 3 circles) M1 separation increasing with distance from wire A1 correct direction B1 [3] (b) (i) arrow direction from wire B towards wire A B1 [1] (ii) either reference to Newton’s third law or force on each wire proportional to product of the two currents M1 so forces are equal A1 [2] (c) force always towards wire A/always in same direction B1 varies from zero (to a maximum value) (1) variation is sinusoidal / sin2 (1) (at) twice frequency of current (1) (any two, one each) B2 [3] 8 (a) packet/quantum/discrete amount of energy M1 of electromagnetic radiation A1 (allow 1 mark for ‘packet of electromagnetic radiation’) energy = Planck constant × frequency (seen here or in b) B1 [3] (b) each (coloured) line corresponds to one wavelength/frequency B1 energy = Planck constant × frequency implies specific energy change between energy levels B1 so discrete levels A0 [2] © University of Cambridge International Examinations 2012 Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 43 9 (a) (i) either probability of decay (of a nucleus) M1 per unit time A1 [2] or λ = (–)(dN/dt) / N (M1) (–)dN/dt and N explained (A1) (ii) in time t½, number of nuclei changes from N0 to ½N0 B1 ½ = exp(–λ t½) or 2 = exp (λ t½) B1 ln (½) = –λ t½ and ln (½) = –0.693 or ln 2 = λ t½ and ln 2 = 0.693 B1 0.693 = λ t½ A0 [3] (b) 228 = 538 exp(–8λ) C1 λ = 0.107 (hours–1) C1 t½ = 6.5 hours (do not allow 3 or more SF) A1 [3] (c) e.g. random nature of decay background radiation daughter product is radioactive (any two sensible suggestions, 1 each) B2 [2] © University of Cambridge International Examinations 2012 Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 43 Section B 10 (a) light-dependent resistor (allow LDR) B1 [1] (b) (i) two resistors in series between +5 V line and earth M1 midpoint connected to inverting input of op-amp A1 [2] (ii) relay coil between diode and earth M1 switch between lamp and earth A1 [2] (c) (i) switch on/off mains supply using a low voltage/current output B1 [1] (allow ‘isolates circuit from mains supply’) (ii) relay will switch on for one polarity of output (voltage) C1 switches on when output (voltage) is negative A1 [2] 11 (a) (i) e.m. radiation produced whenever charged particle is accelerated M1 electrons hitting target have distribution of accelerations A1 [2] (ii) either wavelength shorter/shortest for greater/greatest acceleration or λmin = hc/ Emax or minimum wavelength for maximum energy B1 all electron energy given up in one collision/converted to single photon B1 [2] (b) (i) hardness measures the penetration of the beam C1 greater hardness, greater penetration A1 [2] (ii) controlled by changing the anode voltage C1 higher anode voltage, greater penetration/hardness A1 [2] (c) (i) long-wavelength radiation more likely to be absorbed in the body/less likely to penetrate through body B1 [1] (ii) (aluminium) filter/metal foil placed in the X-ray beam B1 [1] 12 (a) strong uniform (magnetic) field M1 either aligns nuclei or gives rise to Larmor/resonant frequency in r.f. region A1 non-uniform (magnetic) field M1 either enables nuclei to be located or changes the Larmor/resonant frequency A1 [4] (b) (i) difference in flux density = 2.0 × 10–2 × 3.0 × 10–3 = 6.0 × 10–5 T A1 [1] (ii) ∆f = 2 × c × ∆B C1 = 2 × 1.34 × 108 × 6.0 × 10–5 = 1.6 × 104 Hz A1 [2] © University of Cambridge International Examinations 2012 Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9702 43 13 (a) (i) no interference (between signals) near boundaries (of cells) B1 [1] (ii) for large area, signal strength would have to be greater and this could be hazardous to health B1 [1] (b) mobile phone is sending out an (identifying) signal M1 computer/cellular exchange continuously selects cell/base station with strongest signal A1 computer/cellular exchange allocates (carrier) frequency (and slot) A1 [3] © University of Cambridge International Examinations 2012
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