Cambridge A Level Physics 9702 — 2013 Oct/Nov Paper 4 · Variant 3

1 (a) State Newton’s law of gravitation. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (b) A star and a planet are isolated in space. The planet orbits the star in a circular orbit of radius R, as illustrated in Fig. 1.1. t planet star mass M R Fig. 1.1 The angular speed of the planet about the star is ω. By considering the circular motion of the planet about the star of mass M, show that ω and R are related by the expression R 3ω2 = GM where G is the gravitational constant. Explain your working. [3] (c) The Earth orbits the Sun in a circular orbit of radius 1.5 × 108 km. The mass of the Sun For is 2.0 × 1030 kg. Examiner’s A distant star is found to have a planet that has a circular orbit about the star. The radius Use of the orbit is 6.0 × 108 km and the period of the orbit is 2.0 years. Use the expression in (b) to calculate the mass of the star. mass = ........................................... kg [3]

Mark scheme: 1 (a) force proportional to product of the two masses and inversely proportional to the square of their separation M1 either reference to point masses or separation >> ‘size’ of masses A1 [2] (b) gravitational force provides the centripetal force B1 GMm / R2 = mRω2 M1 where m is the mass of the planet A1 GM = R3ω2 A0 [3] (c) ω = 2π / T C1 either Mstar / MSun = (Rstar / RSun)3 × (TSun / Tstar)2 Mstar = 43 × (½)2 × 2.0 × 1030 C1 = 3.2 × 1031 kg A1 [3] or Mstar = (2π)2 Rstar3 / GT2 (C1) = {(2π)2 × (6.0 × 1011)3} / {6.67 × 10–11 × (2 × 365 × 24 × 3600)2} (C1) = 3.2 × 1031 kg (A1)

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2 (a) (i) State what is meant by the internal energy of a system. For Examiner’s .................................................................................................................................. Use .................................................................................................................................. ..............................................................................................................................[2] (ii) Explain why, for an ideal gas, the internal energy is equal to the total kinetic energy of the molecules of the gas. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (b) The mean kinetic energy <EK> of a molecule of an ideal gas is given by the expression <EK> = 32kT where k is the Boltzmann constant and T is the thermodynamic temperature of the gas. A cylinder contains 1.0 mol of an ideal gas. The gas is heated so that its temperature changes from 280 K to 460 K. (i) Calculate the change in total kinetic energy of the gas molecules. change in energy = ............................................. J [2] (ii) During the heating, the gas expands, doing 1.5 × 103 J of work. For State the first law of thermodynamics. Use the law and your answer in (i) to Examiner’s determine the total energy supplied to the gas. Use .................................................................................................................................. .................................................................................................................................. total energy = ............................................. J [3]

Mark scheme: 2 (a) (i) sum of kinetic and potential energies of the molecules M1 reference to random distribution A1 [2] (ii) for ideal gas, no intermolecular forces M1 so no potential energy (only kinetic) A1 [2] (b) (i) either change in kinetic energy = 3/2 × 1.38 × 10–23 × 1.0 × 6.02 × 1023 × 180 C1 = 2240 J A1 [2] or R = kNA energy = 3/2 × 1.0 × 8.31 × 180 (C1) = 2240 J (A1) (ii) increase in internal energy = heat supplied + work done on system B1 2240 = energy supplied – 1500 C1 energy supplied = 3740 J A1 [3]

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3 (a) Define electric potential at a point. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... ......................................................................................................................................[2] (b) Two point charges A and B are separated by a distance of 20 nm in a vacuum, as illustrated in Fig. 3.1. 20 nm P B A x Fig. 3.1 A point P is a distance x from A along the line AB. The variation with distance x of the electric potential VA due to charge A alone is shown in Fig. 3.2. 0.8 potential V / V VA VB 0.6 0.4 0.2 0 2 4 6 8 10 12 14 16 18 x / nm Fig. 3.2 The variation with distance x of the electric potential VB due to charge B alone is also shown in Fig. 3.2. (i) State and explain whether the charges A and B are of the same, or opposite, sign. For Examiner’s .................................................................................................................................. Use .................................................................................................................................. ..............................................................................................................................[2] (ii) By reference to Fig. 3.2, state how the combined electric potential due to both charges may be determined. .................................................................................................................................. ..............................................................................................................................[1] (iii) Without any calculation, use Fig. 3.2 to estimate the distance x at which the combined electric potential of the two charges is a minimum. x = .......................................... nm [1] (iv) The point P is a distance x = 10 nm from A. An α-particle has kinetic energy EK when at infinity. Use Fig. 3.2 to determine the minimum value of EK such that the α-particle may travel from infinity to point P. EK = ............................................. J [3]

Mark scheme: 3 (a) work done bringing unit positive charge M1 from infinity (to the point) A1 [2] (b) (i) either both potentials are positive / same sign M1 so same sign A1 [2] or gradients are positive & negative (so fields in opposite directions) (M1) so same sign (A1) (ii) the individual potentials are summed B1 [1] (iii) allow value of x between 10 nm and 13 nm A1 [1] (iv) V = 0.43 V (allow 0.42 V → 0.44 V) M1 energy = 2 × 1.6 × 10–19 × 0.43 A1 = 1.4 × 10–19 J A1 [3] GCE A LEVEL – October/November 2013 9702 43

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4 (a) State two functions of capacitors connected in electrical circuits. For Examiner’s 1. ..................................................................................................................................... Use .......................................................................................................................................... 2. ..................................................................................................................................... .......................................................................................................................................... [2] (b) Three capacitors are connected in parallel to a power supply as shown in Fig. 4.1. V C1 C2 C3 Fig. 4.1 The capacitors have capacitances C1, C2 and C3. The power supply provides a potential difference V. (i) Explain why the charge on the positive plate of each capacitor is different. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[1] (ii) Use your answer in (i) to show that the combined capacitance C of the three capacitors is given by the expression C = C1 + C2 + C3. [2] (c) A student has available three capacitors, each of capacitance 12 μF. For Draw circuit diagrams, one in each case, to show how the student connects the three Examiner’s capacitors to provide a combined capacitance of Use (i) 8 μF, [1] (ii) 18 μF. [1]

Mark scheme: 4 (a) e.g. store energy (do not allow ‘store charge’) in smoothing circuits blocking d.c. in oscillators any sensible suggestions, one each, max. 2 B2 [2] (b) (i) potential across each capacitor is the same and Q = CV B1 [1] (ii) total charge Q = Q1 + Q2 + Q3 M1 CV = C1V + C2V + C3V M1 (allow Q = CV here or in (i)) so C = C1 + C2 + C3 A0 [2] (c) (i) A1 [1] (ii) A1 [1]

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5 A uniform magnetic field of flux density B makes an angle θ with a flat plane PQRS, as For shown in Fig. 5.1. Examiner’s Use Q P e e magnetic field flux density B e e R S Fig. 5.1 The plane PQRS has area A. (a) State (i) what is meant by a magnetic field, .................................................................................................................................. ..............................................................................................................................[1] (ii) an expression, in terms of A, B and θ, for the magnetic flux Φ through the plane PQRS. ..............................................................................................................................[1] (b) A vertical aluminium window frame DEFG has width 52 cm and length 95 cm, as shown in Fig. 5.2. 52 cm E hinge D 95 cm hinge F G Fig. 5.2 The frame is hinged along the vertical edge DG. The horizontal component BH of the Earth’s magnetic field is 1.8 × 10–5 T. For the closed window, the frame is normal to the horizontal component BH. The window is opened so that the plane of the window rotates through 90°. (i) Explain why, when the window is opened, the change in magnetic flux linkage due For to the vertical component of the Earth’s magnetic field is zero. Examiner’s Use .................................................................................................................................. ..............................................................................................................................[1] (ii) Calculate, for the window opening through an angle of 90°, the change in magnetic flux linkage. change in flux linkage = .......................................... Wb [2] (c) (i) State Faraday’s law of electromagnetic induction. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (ii) The window in (b) is opened in a time of 0.30 s. Use your answer in (b)(ii) to calculate the average e.m.f. induced in the window frame. e.m.f. = ............................................. V [1] (iii) State the sides of the window frame between which the e.m.f. is induced. between side ............ and side .............. [1]

Mark scheme: 5 (a) (i) region (of space) either where a moving charge (may) experience a force or around a magnet where another magnet experiences a force B1 [1] (ii) (Φ =) BA sinθ A1 [1] (b) (i) plane of frame is always parallel to BV / flux linkage always zero B1 [1] (ii) ∆Φ = 1.8 × 10–5 × 52 × 10–2 × 95 × 10–2 C1 = 8.9 × 10–6 Wb A1 [2] (c) (i) (induced) e.m.f. proportional to rate of M1 change of (magnetic) flux (linkage) A1 [2] (allow rate of cutting of flux) (ii) e.m.f. = (8.9 × 10–6) / 0.30 = 3.0 × 10–5 V A1 [1] (iii) This question part was removed from the assessment. All candidates were awarded 1 mark. B1 [1] GCE A LEVEL – October/November 2013 9702 43

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6 A particle has mass m and charge +q and is travelling with speed v through a vacuum. For The initial direction of travel is parallel to the plane of two charged horizontal metal plates, as Examiner’s shown in Fig. 6.1. Use + V metal plate path of particle metal plate Fig. 6.1 The uniform electric field between the plates has magnitude 2.8 × 104 V m–1 and is zero outside the plates. The particle passes between the plates and emerges beyond them, as illustrated in Fig. 6.1. (a) Explain why the path of the particle in the electric field is not an arc of a circle. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[1] (b) A uniform magnetic field is now formed in the region between the metal plates. The magnetic field strength is adjusted so that the positively charged particle passes undeviated between the plates, as shown in Fig. 6.2. + V region of uniform electric and magnetic fields path of particle path of particle Fig. 6.2 (i) State and explain the direction of the magnetic field. For Examiner’s .................................................................................................................................. Use .................................................................................................................................. ..............................................................................................................................[2] (ii) The particle has speed 4.7 × 105 m s–1. Calculate the magnitude of the magnetic flux density. Explain your working. magnetic fl ux density = ............................................. T [3] (c) The particle in (b) has mass m, charge +q and speed v. Without any further calculation, state the effect, if any, on the path of a particle that has (i) mass m, charge –q and speed v, ..............................................................................................................................[1] (ii) mass m, charge +q and speed 2v, ..............................................................................................................................[1] (iii) mass 2m, charge +q and speed v. ..............................................................................................................................[1]

Mark scheme: 6 (a) either constant speed parallel to plate or accelerated motion / force normal to plate / in direction field B1 so not circular A0 [1] (b) (i) direction of force due to magnetic field opposite to that due to electric field B1 magnetic field into plane of page B1 [2] (ii) force due to magnetic field = force due to electric field B1 Bqv = qE B = E / v C1 = (2.8 × 104) / (4.7 × 105) = 6.0 × 10–2 T A1 [3] (c) (i) no change / not deviated B1 [1] (ii) deviated upwards B1 [1] (iii) no change / not deviated B1 [1]

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7 (a) By reference to the photoelectric effect, explain For Examiner’s (i) what is meant by work function energy, Use .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (ii) why, even when the incident light is monochromatic, the emitted electrons have a range of kinetic energy up to a maximum value. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (b) Electromagnetic radiation of frequency f is incident on a metal surface. The variation with frequency f of the maximum kinetic energy EMAX of electrons emitted from the surface is shown in Fig. 7.1. 4 EMAX / 10 –18 J 3 2 1 0 0 1 2 3 4 5 f / 1015 Hz Fig. 7.1 (i) Use Fig. 7.1 to determine the work function energy of the metal surface. For Examiner’s Use work function energy = ............................................. J [3] (ii) A second metal has a greater work function energy than that in (i). On Fig. 7.1, draw a line to show the variation with f of EMAX for this metal. [2] (iii) Explain why the graphs in (i) and (ii) do not depend on the intensity of the incident radiation. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2]

Mark scheme: 7 (a) (i) minimum photon energy B1 minimum energy to remove an electron (from the surface) B1 [2] (ii) either maximum KE is photon energy – work function energy or max KE when electron ejected from the surface B1 energies lower than max because energy required to bring electron to the surface B1 [2] (b) (i) threshold frequency = 1.0 × 1015 Hz (allow ±0.05 × 1015) C1 work function energy = hf0 C1 = 6.63 × 10–34 × 1.0 × 1015 = 6.63 × 10–19 J A1 [3] (allow alternative approaches based on use of co-ordinates of points on the line) (ii) sketch: straight line with same gradient M1 displaced to right A1 [2] (iii) intensity determines number of photons arriving per unit time B1 intensity determines number of electrons per unit time (not energy) B1 [2]

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8 One possible nuclear fission reaction is For Examiner’s 235 1 141 92 1 Use U + n Ba + Kr + 3 n + energy. 92 0 56 36 0 Barium-141 (141 Ba) and krypton-92 (92 Kr) are both β-emitters. 56 36 Barium-141 has a half-life of 18 minutes and a decay constant of 6.4 × 10–4 s–1. The half-life of krypton-92 is 3.0 seconds. (a) State what is meant by decay constant. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (b) A mass of 1.2 g of uranium-235 undergoes this nuclear reaction in a very short time (a few nanoseconds). (i) Calculate the number of barium-141 nuclei that are present immediately after the reaction has been completed. number = ................................................ [2] (ii) Using your answer in (b)(i), calculate the total activity of the barium-141 and the krypton-92 a time of 1.0 hours after the fission reaction has taken place. activity = ........................................... Bq [4]

Mark scheme: 8 (a) probability of decay (of a nucleus) / fraction of number of nuclei in sample that decay M1 per unit time A1 [2] (allow λ =(dN / dt) / N with symbols explained – (M1), (A1) ) (b) (i) number = (1.2 × 6.02 × 1023) / 235 C1 = 3.1 × 1021 A1 [2] GCE A LEVEL – October/November 2013 9702 43 (ii) N = N0 e–λt negligible activity from the krypton B1 for barium, N = (3.1 × 1021) exp(–6.4 × 10–4 × 3600) = 3.1 × 1020 C1 activity = λN = 6.4 × 10–4 × 3.1 × 1020 C1 = 2.0 × 1017 Bq A1 [4] Section B

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9 (a) State three properties of an ideal operational amplifier (op-amp). 1. ...................................................................................................................................... 2. ...................................................................................................................................... 3. ...................................................................................................................................... [3] (b) An amplifier circuit is shown in Fig. 9.1. 10.8 k1 +9 V – + –9 V V OUT V IN 1.2 k1 Fig. 9.1 (i) Calculate the gain of the amplifier circuit. gain = ................................................ [2] (ii) The variation with time t of the input potential VIN is shown in Fig. 9.2. For Examiner’s 10 Use potential / V 5 0 t V IN – 5 –10 Fig. 9.2 On the axes of Fig. 9.2, show the variation with time t of the output potential VOUT . [3]

Mark scheme: 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) gain = 1 + (10.8 / 1.2) C1 = 10 A1 [2] (ii) graph: straight line from (0,0) towards VIN = 1.0 V, VOUT = 10 V B1 horizontal line at VOUT = 9.0 V to VIN = 2.0 V B1 correct +9.0 V → –9.0 V (and correct shape to VIN = 0) B1 [3]

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10 Magnetic resonance imaging (MRI) requires the use of a non-uniform magnetic field For superimposed on a large uniform magnetic field. Examiner’s Use State and explain the purpose of (a) the large uniform magnetic field, .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[3] (b) the non-uniform magnetic field. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[3]

Mark scheme: 10 (a) nuclei spin / precess B1 spin / precess about direction of magnetic field B1 either frequency of precession depends on magnetic field strength or large field means frequency in radio frequency range B1 [3] (b) non-uniform field means frequency of precession different in different regions of subject B1 enables location of precessing nuclei to be determined B1 enables thickness of slice to be varied / location of slice to be changed B1 [3]

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11 Data may be transmitted in either analogue or digital form. For Examiner’s (a) State Use (i) what is meant by a digital signal, .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2] (ii) three advantages of the digital transmission of data when compared to analogue transmission. 1. .............................................................................................................................. 2. .............................................................................................................................. 3. .............................................................................................................................. [3] (b) The block diagram of Fig. 11.1 represents the digital transmission of music. parallel- serial-to- Y ADC to-serial parallel X Y converter converter Fig. 11.1 (i) State the name of 1. the blocks labelled Y, ..............................................................................................................................[1] 2. the block labelled X. ..............................................................................................................................[1] (ii) Describe the function of the parallel-to-serial converter. .................................................................................................................................. .................................................................................................................................. ..............................................................................................................................[2]

Mark scheme: 11 (a) (i) either series of ‘highs’ and ‘lows’ or two discrete values M1 with no intermediate values A1 [2] (ii) e.g. noise can be eliminated (NOT ‘no noise’) signal can be regenerated addition of extra data to check for errors larger data carrying capacity cheaper circuits more reliable circuits (any three, 1 each) B3 [3] GCE A LEVEL – October/November 2013 9702 43 (b) (i) 1. amplifier B1 [1] 2. digital-to-analogue converter (allow DAC) B1 [1] (ii) output of ADC is number of digits all at one time B1 parallel-to-serial sends digits one after another B1 [2]

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12 (a) State two reasons why frequencies in the gigahertz (GHz) range are used in satellite For communication. Examiner’s Use 1. ..................................................................................................................................... .......................................................................................................................................... 2. ..................................................................................................................................... .......................................................................................................................................... [2] (b) In one particular satellite communication system, the frequency of the signal transmitted from Earth to the satellite (the up-link) is 6 GHz. The frequency of the signal transmitted back to Earth from the satellite (the down-link) is 4 GHz. Explain why the two signals are transmitted at different frequencies. .......................................................................................................................................... .......................................................................................................................................... ......................................................................................................................................[2] (c) A signal transmitted from Earth has a power of 3.1 kW. This signal, received by a satellite, has been attenuated by 185 dB. Calculate the power of the signal received by the satellite. power = ............................................ W [3]

Mark scheme: 12 (a) e.g. no / little ionospheric reflection large information carrying capacity (any two sensible suggestions, 1 each) B2 [2] (b) prevents (very) low power signal received at satellite M1 being swamped by high-power transmitted signal A1 [2] (c) attenuation / dB = 10 lg(P2/P1) C1 185 = 10 lg({3.1 × 103}/P) C1 P = 9.8 × 10–16 W A1 [3]

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