Cambridge A Level Physics 9702 — 2014 May/June Paper 4 · Variant 3

1 (a) Define gravitational potential at a point. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) A stone of mass m has gravitational potential energy EP at a point X in a gravitational field. The magnitude of the gravitational potential at X is φ. State the relation between m, EP and φ. .............................................................................................................................................. [1] (c) An isolated spherical planet of radius R may be assumed to have all its mass concentrated at its centre. The gravitational potential at the surface of the planet is − 6.30 × 107 J kg−1. A stone of mass 1.30 kg is travelling towards the planet such that its distance from the centre of the planet changes from 6R to 5R. Calculate the change in gravitational potential energy of the stone. change in energy = ..................................................... J [4]

Mark scheme: 1 (a) work done bringing unit mass M1 from infinity (to the point) A1 [2] (b) EP = –mφ B1 [1] (c) φ ∝ 1/x C1 either at 6R from centre, potential is (6.3 × 107)/6 (= 1.05 × 107 J kg–1) and at 5R from centre, potential is (6.3 × 107)/5 (= 1.26 × 107 J kg–1) C1 change in energy = (1.26 – 1.05) × 107 × 1.3 C1 = 2.7 × 106 J A1 or change in potential = (1/5 – 1/6) × (6.3 × 107) (C1) change in energy = (1/5 – 1/6) × (6.3 × 107) × 1.3 (C1) = 2.7 × 106 J (A1) [4]

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2 (a) Explain what is meant by the Avogadro constant. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) Argon-40 (4018Ar) may be assumed to be an ideal gas. A mass of 3.2 g of argon-40 has a volume of 210 cm3 at a temperature of 37 °C. Determine, for this mass of argon-40 gas, (i) the amount, in mol, amount = ................................................. mol [1] (ii) the pressure, pressure = ................................................... Pa [2] (iii) the root-mean-square (r.m.s.) speed of an argon atom. r.m.s. speed = ............................................... m s−1 [3]

Mark scheme: 2 (a) the number of atoms M1 in 12 g of carbon-12 A1 [2] (b) (i) amount = 3.2/40 = 0.080 mol A1 [1] (ii) pV = nRT p × 210 × 10–6 = 0.080 × 8.31 × 310 C1 p = 9.8 × 105 Pa A1 [2] (do not credit if T in °C not K) (iii) either pV = 1/3 × Nm <c2> N = 0.080 × 6.02 × 1023 (= 4.82 × 1022) and m = 40 × 1.66 × 10–27 (= 6.64 × 10–26) C1 9.8 × 105 × 210 × 10–6 = 1/3 × 4.82 × 1022 × 6.64 × 10–26 × <c2> C1 <c2> = 1.93 × 105 cRMS = 440 m s–1 A1 [3] or Nm = 3.2 × 10–3 (C1) 9.8 × 105 × 210 × 10–6 = 1/3 × 3.2 × 10–3 × <c2> (C1) <c2> = 1.93 × 105 cRMS = 440 m s–1 (A1) or 1/2 m<c2> = 3/2 kT (C1) 1/2 × 40 × 1.66 × 10–27 <c2> = 3/2 × 1.38 × 10–23 × 310 (C1) <c2> = 1.93 × 105 cRMS = 440 m s–1 (A1) (if T in °C not K award max 1/3, unless already penalised in (b)(ii)) GCE AS/A LEVEL – May/June 2014 9702 43 3

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3 The volume of 1.00 kg of water in the liquid state at 100 °C is 1.00 × 10−3 m3. The volume of 1.00 kg of water vapour at 100 °C and atmospheric pressure 1.01 × 105 Pa is 1.69 m3. (a) Show that the work done against the atmosphere when 1.00 kg of liquid water becomes water vapour is 1.71 × 105 J. [2] (b) (i) The first law of thermodynamics may be given by the expression ΔU = + q + w where ΔU is the increase in internal energy of the system. State what is meant by 1. + q, ...................................................................................................................................... [1] 2. + w. ...................................................................................................................................... [1] (ii) The specific latent heat of vaporisation of water at 100 °C is 2.26 × 106 J kg−1. A mass of 1.00 kg of liquid water becomes water vapour at 100 °C. Determine, using your answer in (a), the increase in internal energy of this mass of water during vaporisation. increase in internal energy = ..................................................... J [2]

Mark scheme: 3 (a) either change in volume = (1.69 – 1.00 × 10–3) or liquid volume << volume of vapour M1 work done = 1.01 × 105 × 1.69 = 1.71 × 105 (J) A1 [2] (b) (i) 1. heating of system/thermal energy supplied to the system B1 [1] 2. work done on the system B1 [1] (ii) ∆U = (2.26 × 106) – (1.71 × 105) C1 = 2.09 × 106 J (3 s.f. needed) A1 [2]

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4 A student investigates the energy changes of a mass oscillating on a vertical spring, as shown in Fig. 4.1. spring mass Fig. 4.1 The student draws a graph of the variation with displacement x of energy E of the oscillation, as shown in Fig. 4.2. 2.5 2.0 E / mJ 1.5 1.0 0.5 0 – 1.5 – 1.0 – 0.5 0 0.5 1.0 1.5 x / cm Fig. 4.2 (a) State whether the energy E represents the total energy, the potential energy or the kinetic energy of the oscillations. .............................................................................................................................................. [1] (b) The student repeats the investigation but with a smaller amplitude. The maximum value of E is now found to be 1.8 mJ. Use Fig. 4.2 to determine the change in the amplitude. Explain your working. change in amplitude = .................................................. cm [3]

Mark scheme: 4 (a) kinetic (energy)/KE/EK B1 [1] (b) either change in energy = 0.60 mJ or max E proportional to (amplitude)2/equivalent numerical working B1 new amplitude is 1.3 cm B1 change in amplitude = 0.2 cm B1 [3]

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5 An isolated solid metal sphere of radius r is given a positive charge. The distance from the centre of the sphere is x. (a) The electric potential at the surface of the sphere is V0. On the axes of Fig. 5.1, sketch a graph to show the variation with distance x of the electric potential due to the charged sphere, for values of x from x = 0 to x = 4r. 1.00 V0 0.75 V0 potential 0.50 V0 0.25 V0 0 0 r 2r 3r 4r distance x Fig. 5.1 [3] (b) The electric field strength at the surface of the sphere is E0. On the axes of Fig. 5.2, sketch a graph to show the variation with distance x of the electric field strength due to the charged sphere, for values of x from x = 0 to x = 4r. 1.00 E0 0.75 E0 field strength 0.50 E0 0.25 E0 0 0 r 2r 3r 4r distance x Fig. 5.2 [3]

Mark scheme: 5 (a) graph: straight line at constant potential = V0 from x = 0 to x = r B1 curve with decreasing gradient M1 passing through (2r, 0.50V0) and (4r, 0.25V0) A1 [3] (b) graph: straight line at E = 0 from x = 0 to x = r B1 curve with decreasing gradient from (r, E0) M1 passing through (2r, ¼E0) A1 [3] (for 3rd mark line must be drawn to x = 4r and must not touch x-axis)

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6 An uncharged capacitor is connected in series with a battery, a switch and a resistor, as shown in Fig. 6.1. 9.0 V 4700 +F Fig. 6.1 The battery has e.m.f. 9.0 V and negligible internal resistance. The capacitance of the capacitor is 4700 μF. The switch is closed at time t = 0. During the time interval t = 0 to t = 4.0 s, the charge passing through the resistor is 22 mC. (a) (i) Calculate the energy transfer in the battery during the time interval t = 0 to t = 4.0 s. energy transfer = ..................................................... J [2] (ii) Determine, for the capacitor at time t = 4.0 s, 1. the potential difference V across the capacitor, V = ..................................................... V [2] 2. the energy stored in the capacitor. energy = ..................................................... J [2] (b) Suggest why your answers in (a)(i) and (a)(ii) part 2 are different. ................................................................................................................................................... .............................................................................................................................................. [1]

Mark scheme: 6 (a) (i) energy = EQ C1 = 9.0 × 22 × 10–3 = 0.20 J A1 [2] (ii) 1. C = Q / V V = (22 × 10–3)/(4700 × 10–6) C1 = 4.7 V A1 [2] 2. either E = ½CV 2 C1 = ½ × 4700 × 10–6 × 4.72 = 5.1 × 10–2 J A1 [2] or E = ½QV (C1) = ½ × 22 × 10–3 × 4.7 = 5.1 × 10–2 J (A1) or E = ½Q2/C (C1) = ½ × (22 × 10–3)2/4700 ×10–6 = 5.1 × 10–2 J (A1) GCE AS/A LEVEL – May/June 2014 9702 43 (b) energy lost (as thermal energy) in resistance/wires/battery/resistor B1 [1] (award only if answer in (a)(i) > answer in (a)(ii)2)

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7 A solenoid is connected in series with a battery and a switch. A Hall probe is placed close to one end of the solenoid, as illustrated in Fig. 7.1. solenoid Hall probe Fig. 7.1 The current in the solenoid is switched on. The Hall probe is adjusted in position to give the maximum reading. The current is then switched off. (a) The current in the solenoid is now switched on again. Several seconds later, it is switched off. The Hall probe is not moved. On the axes of Fig. 7.2, sketch a graph to show the variation with time t of the Hall voltage VH. VH 0 t current current switched on switched off Fig. 7.2 [3] (b) The Hall probe is now replaced by a small coil. The plane of the coil is parallel to the end of the solenoid. (i) State Faraday’s law of electromagnetic induction. ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (ii) On the axes of Fig. 7.3, sketch a graph to show the variation with time t of the e.m.f. E induced in the coil when the current in the solenoid is switched on and then switched off. E 0 t current current switched on switched off Fig. 7.3 [3]

Mark scheme: 7 (a) graph: VH increases from zero when current switched on B1 VH then non-zero constant B1 VH returns to zero when current switched off B1 [3] (b) (i) (induced) e.m.f. proportional to rate M1 of change of (magnetic) flux (linkage) A1 [2] (ii) pulse as current is being switched on B1 zero e.m.f. when current in coil B1 pulse in opposite direction when switching off B1 [3]

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8 (a) State what is meant by quantisation of charge. ................................................................................................................................................... .............................................................................................................................................. [1] (b) A student carries out an experiment to determine the elementary charge. A charged oil drop is positioned between two horizontal metal plates, as shown in Fig. 8.1. + 680 V oil drop 7.0 mm Fig. 8.1 The plates are separated by a distance of 7.0 mm. The lower plate is earthed. The potential of the upper plate is gradually increased until the drop is held stationary. The potential for the drop to be stationary is 680 V. The weight of the oil drop, allowing for the upthrust of the air, is 4.8 × 10−14 N. Calculate the value for the charge on the oil drop. charge = ..................................................... C [2] (c) The student repeats the experiment and determines the following values for the charge on oil drops. 3.3 × 10−19 C 4.9 × 10−19 C 9.7 × 10−19 C 3.4 × 10−19 C Use these values to suggest a value for the elementary charge. Explain your working. elementary charge = ..................................................... C [2]

Mark scheme: 8 (a) discrete and equal amounts (of charge) B1 [1] allow: discrete amounts of 1.6 × 10–19C/elementary charge/e integral multiples of 1.6 × 10–19C/elementary charge/e (b) weight = qV / d 4.8 × 10–14 = (q × 680)/(7.0 × 10–3) C1 q = 4.9 × 10–19 C A1 [2] (c) elementary charge = 1.6 × 10–19 C (allow 1.6 × 10–19 C to 1.7 × 10–19 C ) M0 either the values are (approximately) multiples of this or it is a common factor C1 it is the highest common factor A1 [2]

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9 For a particular metal surface, it is observed that there is a minimum frequency of light below which photoelectric emission does not occur. This observation provides evidence for a particulate nature of electromagnetic radiation. (a) State three further observations from photoelectric emission that provide evidence for a particulate nature of electromagnetic radiation. 1. ............................................................................................................................................... ................................................................................................................................................... 2. ............................................................................................................................................... ................................................................................................................................................... 3. ............................................................................................................................................... ................................................................................................................................................... [3] (b) Some data for the variation with frequency f of the maximum kinetic energy EMAX of electrons emitted from a metal surface are shown in Fig. 9.1. 0.6 0.5 E MAX / eV 0.4 0.3 0.2 0.1 0 5.5 6.0 6.5 7.0 7.5 f / 1014 Hz Fig. 9.1 (i) Explain why emitted electrons may have kinetic energy less than the maximum at any particular frequency. ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (ii) Use Fig. 9.1 to determine 1. the threshold frequency, threshold frequency = ................................................... Hz [1] 2. the work function energy, in eV, of the metal surface. work function energy = ................................................... eV [3]

Mark scheme: 9 (a) e.g. no time delay between illumination and emission max. (kinetic) energy of electron dependent on frequency max. (kinetic) energy of electron independent of intensity rate of emission of electrons dependent on/proportional to intensity (any three separate statements, one mark each, maximum 3) B3 [3] (b) (i) (photon) interaction with electron may be below surface B1 energy required to bring electron to surface B1 [2] GCE AS/A LEVEL – May/June 2014 9702 43 (ii) 1. threshold frequency = 5.8 × 1014 Hz A1 [1] 2. Φ = hf0 C1 = 6.63 × 10–34 × 5.8 × 1014 = 3.84 × 10–19 (J) C1 = (3.84 × 10–19)/(1.6 × 10–19) = 2.4 eV A1 [3] or hf = Φ + EMAX (C1) chooses point on line and substitutes values EMAX, f and h into equation with the units of the hf term converted from J to eV (C1) Φ = 2.4 eV (A1)

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10 (a) Explain what is meant by the binding energy of a nucleus. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) Data for the masses of some particles are given in Fig. 10.1. mass / u proton 1.00728 neutron 1.00867 tritium (31H) nucleus 3.01551 polonium (21084Po) nucleus 209.93722 Fig. 10.1 The energy equivalent of 1.0 u is 930 MeV. (i) Calculate the binding energy, in MeV, of a tritium (31H) nucleus. binding energy = ................................................ MeV [3] (ii) The total mass of the separate nucleons that make up a polonium-210 (21084Po) nucleus is 211.70394 u. Calculate the binding energy per nucleon of polonium-210. binding energy per nucleon = ................................................ MeV [3] (c) One possible fission reaction is 23592U + 10n 14156Ba + 9236Kr + 310n . By reference to binding energy, explain, without any calculation, why this fission reaction is energetically possible. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2]

Mark scheme: 10 (a) energy required to separate the nucleons (in a nucleus) M1 to infinity A1 [2] (allow reverse statement) (b) (i) ∆m = (2 × 1.00867) + 1.00728 – 3.01551 C1 = 9.11 × 10–3 u C1 binding energy = 9.11 × 10–3 × 930 = 8.47 MeV A1 [3] (allow 930 to 934 MeV so answer could be in range 8.47 to 8.51 MeV) (allow 2 s.f.) (ii) ∆m = 211.70394 – 209.93722 = 1.76672 u C1 binding energy per nucleon = (1.76672 × 930)/210 C1 = 7.82 MeV A1 [3] (allow 930 to 934 MeV so answer could be in range 7.82 to 7.86 MeV) (allow 2 s.f.) (c) total binding energy of barium and krypton M1 is greater than binding energy of uranium A1 [2] Section B

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11 (a) A circuit incorporating an ideal operational amplifier (op-amp) is shown in Fig. 11.1. P – + V IN V OUT Fig. 11.1 (i) State the name of this circuit. ...................................................................................................................................... [1] (ii) Explain why the point P is referred to as a virtual earth. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] (b) The circuit of Fig. 11.1 is modified, as shown in Fig. 11.2. RC C RB B RA A 1.0 k1 – + V IN V Fig. 11.2 The voltmeter has infinite resistance and its full-scale deflection is 1.0 V. The input potential to the circuit is VIN. The switch position may be changed in order to have different values of resistance in the circuit. (i) The input potential VIN and the switch position are varied. For each switch position, the reading of the voltmeter is 1.0 V. Complete Fig. 11.3 for the switch positions shown. switch position VIN / mV resistance A 10 RA = .............................................. B 100 RB = .............................................. C ............................... RC = 1.0 kΩ Fig. 11.3 [3] (ii) By reference to your answers in (i), suggest a use for the circuit of Fig. 11.2. ........................................................................................................................................... ...................................................................................................................................... [1]

Mark scheme: 11 (a) (i) inverting amplifier B1 [1] (ii) gain is very large/infinite B1 V+ is earthed/zero B1 for amplifier not to saturate, P must be (almost) earth/zero B1 [3] (b) (i) RA = 100 kΩ A1 RB = 10 kΩ A1 VIN = 1000 mV A1 [3] (ii) variable range meter B1 [1] GCE AS/A LEVEL – May/June 2014 9702 43

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12 (a) Outline briefly the principles of CT scanning. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [5] (b) In a model for CT scanning, a section is divided into four voxels. The pixel numbers P, Q, R and S of the voxels are shown in Fig. 12.1. D3 D2 D4 P Q D1 S R Fig. 12.1 The section is viewed from the four directions D1, D2, D3 and D4. The detector readings for each direction are noted. The detector readings are summed as shown in Fig. 12.2. 49 61 73 55 Fig. 12.2 The background reading is 34. Determine the pixel numbers P, Q, R and S as shown in Fig. 12.3. P Q S R Fig. 12.3 P = ............................................................... Q = ............................................................... S = ............................................................... R = ............................................................... [4]

Mark scheme: 12 (a) series of X-ray images (for one section/slice) M1 taken from different angles M1 to give image of the section/slice A1 repeated for many slices M1 to build up three-dimensional image (of whole object) A1 [5] (b) deduction of background from readings C1 division by three C1 P = 5 Q = 9 R = 7 S =13 (four correct 2/2, three correct 1/2) A2 [4]

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13 The signal from a microphone is to be transmitted in digital form. A block diagram of part of the transmission system is shown in Fig. 13.1. parallel-to ADC -serial converter Fig. 13.1 (a) Suggest two advantages of the transmission of a signal in digital form rather than in analogue form. 1. ............................................................................................................................................... ................................................................................................................................................... 2. ............................................................................................................................................... ................................................................................................................................................... [2] (b) State the function of the parallel-to-serial converter. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (c) In a particular telephone system, the sampling frequency is 8 kHz. In the manufacture of a compact disc, the sampling frequency is approximately 44 kHz. Suggest and explain why the sampling frequency is much higher for the compact disc. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3]

Mark scheme: 13 (a) e.g. noise can be eliminated/waveform can be regenerated extra bits of data can be added to check for errors cheaper/more reliable greater rate of transfer of data (1 each, max 2) B2 [2] (b) receives bits all at one time B1 transmits the bits one after another B1 [2] (c) sampling frequency must be higher than/(at least) twice frequency to be sampled M1 either higher (range of) frequencies reproduced on the disc or lower (range of) frequencies on phone A1 either higher quality (of sound) on disc or high quality (of sound) not required for phone B1 [3]

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14 (a) State what is meant by the attenuation of a signal. ................................................................................................................................................... .............................................................................................................................................. [1] (b) A transmission cable has a length of 30 km. The attenuation per unit length of the cable is 2.4 dB km−1. Calculate, for a signal being transmitted along the cable, (i) the total attenuation, in dB, attenuation = ................................................... dB [1] (ii) the ratio input power of signal . output power of signal ratio = ......................................................... [3] (c) By reference to your answers in (b), suggest why the attenuation of transmitted signals is usually expressed in dB. ................................................................................................................................................... .............................................................................................................................................. [1]

Mark scheme: 14 (a) reduction in power (allow intensity/amplitude) B1 [1] (b) (i) attenuation = 2.4 × 30 = 72 dB A1 [1] (ii) gain/attenuation/dB = 10 lg(P2/P1) C1 72 = 10 lg(PIN/POUT) or –72 = 10 lg(POUT /PIN) C1 ratio = 1.6 × 107 A1 [3] (c) e.g. enables smaller/more manageable numbers to be used e.g. gains in dB for series amplifiers are added, not multiplied B1 [1]

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