Cambridge A Level Physics 9702 — 2015 May/June Paper 4 · Variant 3

1 (a) State Newton’s law of gravitation. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) The planet Neptune has eight moons (satellites). Each moon orbits Neptune in a circular path of radius r with a period T. Assuming that Neptune and each moon behave as point masses, show that r and T are related by the expression 4π2r3 GMN = 2 T where G is the gravitational constant and MN is the mass of Neptune. [3] (c) Data for the moon Triton that orbits Neptune and for the moon Oberon that orbits the planet Uranus are given in Fig. 1.1. planet moon radius of orbit period of orbit r /105 km T / days Neptune Triton 3.55 5.9 Uranus Oberon 5.83 13.5 Fig. 1.1 Use the expression in (b) to determine the ratio mass of Neptune . mass of Uranus ratio = ......................................................... [3]

Mark scheme: 1 (a) (gravitational) force proportional to product of masses and inversely proportional to square of separation M1 reference to either point masses or particles or ‘size’ much less than separation A1 [2] (b) gravitational force provides/is the centripetal force B1 GMNm / r 2 = mrω2 (or mv 2 / r) M1 2π / T (or v = 2πr / T) leading to GMN = 4π2r 3 / T 2 A1 [3] (c) MN / MU = (3.55 / 5.83)3 × (13.5 / 5.9)2 x3 factor correct C1 T2 factor correct C1 ratio = 1.18 (allow 1.2) A1 alternative method: mass of Neptune = 1.019 × 1026 kg (C1) mass of Uranus = 8.621 × 1025 kg (C1) ratio = 1.18 (A1) [3]

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2 (a) State what is meant by internal energy. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) The variation with volume V of the pressure p of an ideal gas as it undergoes a cycle ABCA of changes is shown in Fig. 2.1. 4.0 p / 105 Pa B 3.5 3.0 2.5 2.0 1.5 A C 1.0 3.0 4.0 5.0 6.0 7.0 8.0 V / 10ï m3 Fig. 2.1 The temperature of the gas at A is 290 K. The temperature at B is 870 K. Determine (i) the amount, in mol, of gas, amount = .................................................. mol [2] (ii) the temperature of the gas at C. temperature = ..................................................... K [2] (c) Explain why the change from C to A involves external work and a change in internal energy. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2]

Mark scheme: 2 (a) (sum of) potential energy and kinetic energy of molecules/atoms/particles M1 mention of random motion/distribution A1 [2] (b) (i) pV = nRT either at A, 1.2 × 105 × 4.0 × 10−3 = n × 8.31 × 290 or at B, 3.6 × 105 × 4.0 × 10−3 = n × 8.31 × 870 C1 n = 0.20 mol A1 [2] (ii) 1.2 × 105 × 7.75 × 10–3 = 0.20 × 8.31 × T or T = (7.75 / 4.0) × 290 C1 T = 560 K A1 [2] (Allow tolerance from graph: 7.7–7.8 × 10–3 m3) (c) temperature changes/decreases so internal energy changes/decreases B1 volume changes (at constant pressure) so work is done B1 [2]

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3 (a) Define specific latent heat. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) An electrical heater is immersed in some melting ice that is contained in a funnel, as shown in Fig. 3.1. heater melting ice water Fig. 3.1 The heater is switched on and, when the ice is melting at a constant rate, the mass m of ice melted in 5.0 minutes is noted, together with the power P of the heater. The power P of the heater is then increased. A new reading for the mass m of ice melted in 5.0 minutes is recorded when the ice is melting at a constant rate. Data for the power P and the mass m are shown in Fig. 3.2. power of heater mass m melted in mass m melted P / W 5.0 minutes / g per second / g s−1 70 78 ................................. 110 114 ................................. Fig. 3.2 (i) Complete Fig. 3.2 to determine the mass melted per second for each power of the heater. [2] (ii) Use the data in the completed Fig. 3.2 to determine 1. a value for the specific latent heat of fusion L of ice, L = ................................................ J g−1 [3] 2. the rate h of thermal energy gained by the ice from the surroundings. h = .................................................... W [2]

Mark scheme: 3 (a) (numerically equal to) quantity of (thermal) energy/heat to change state/phase of unit mass M1 at constant temperature A1 [2] (allow 1/2 for definition restricted to fusion or vaporisation) (b) (i) at 70 W, mass s–1 = 0.26 g s–1 A1 at 110 W, mass s–1 = 0.38 g s–1 A1 [2] (ii) 1. P + h = mL or substitution of one set of values C1 (110 – 70) = (0.38 – 0.26)L C1 L = 330 J g–1 A1 [3] 2. either 70 + h = 0.26 × 330 or 110 + h = 0.38 × 330 C1 h = 17 / 16 / 15 W A1 [2]

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4 (a) For an oscillating body, state what is meant by (i) forced frequency, ........................................................................................................................................... ...................................................................................................................................... [1] (ii) natural frequency of vibration, ........................................................................................................................................... ...................................................................................................................................... [1] (iii) resonance. ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (b) State and explain one situation where resonance is useful. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (c) In some situations, resonance should be avoided. State one such situation and suggest how the effects of resonance are reduced. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2]

Mark scheme: 4 (a) (i) frequency at which object is made to vibrate/oscillate B1 [1] (ii) frequency at which object vibrates when free to do so B1 [1] (iii) maximum amplitude of vibration of oscillating body B1 when forced frequency equals natural frequency (of vibration) B1 [2] (b) e.g. vibration of quartz/piezoelectric crystal (what is vibrating) M1 either for accurate timing or maximise amplitude of ultrasound waves (why it is useful) A1 [2] (c) e.g. vibrating metal panels (what is vibrating) M1 either place strengthening struts across the panel or change shape/area of panel (how it is reduced) A1 [2]

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5 A charged metal sphere is isolated in space. Measurements of the electric potential V are made for different distances x from the centre of the sphere. The variation with distance x of the potential V is shown in Fig. 5.1. 4.0 3.0 V / 103 V 2.0 1.0 0 0 2.0 4.0 6.0 8.0 10.0 x / cm Fig. 5.1 (a) Use Fig. 5.1 to determine the electric field strength, in N C−1, at a point where x = 4.0 cm. Explain your working. electric field strength = ............................................... N C−1 [3] (b) The charge on the sphere is 8.0 × 10−9 C. (i) Use Fig. 5.1 to state the electric potential at the surface of the sphere. potential = ..................................................... V [1] (ii) The sphere acts as a capacitor. Determine the capacitance of the sphere. capacitance = ..................................................... F [2]

Mark scheme: 5 (a) (magnitude of electric field strength is the potential gradient B1 use of gradient at x = 4.0 cm M1 gradient = 4.5 × 104 N C–1 (allow ± 0.3 × 10 4) A1 or Q Q V V = and E = 2 leading to E = (B1) 4πε 0 x 4 πε 0 x x E = 1.8 × 103 / 0.04 (M1) = 4.5 × 104 N C–1 (A1) [3] (b) (i) 3.6 × 103 V A1 [1] (ii) capacitance = Q / V C1 = (8.0 × 10–9) / (3.6 × 103) = 2.2 × 10–12 F A1 [2]

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6 (a) State the type of field, or fields, that may cause a force to be exerted on a particle that is (i) uncharged and moving, ...................................................................................................................................... [1] (ii) charged and stationary, ...................................................................................................................................... [1] (iii) charged and moving at right-angles to the field. ...................................................................................................................................... [2] (b) A particle X has mass 3.32 × 10−26 kg and charge +1.60 × 10−19 C. The particle is travelling in a vacuum with speed 7.60 × 104 m s−1. It enters a region of uniform magnetic field that is normal to the direction of travel of the particle. The particle travels in a semicircle of diameter 12.2 cm, as shown in Fig. 6.1. region of uniform magnetic field 12.2 cm path of particle X Fig. 6.1 For the uniform magnetic field, (i) state its direction, ........................................................................................................................................... ...................................................................................................................................... [1] (ii) calculate the magnetic flux density. magnetic flux density = ..................................................... T [3] (c) A second particle Y has mass less than that of particle X in (b) and the same charge. It enters the region of uniform magnetic field in (b) with the same speed and along the same initial path as particle X. On Fig. 6.1, draw the path of particle Y in the region of the magnetic field. [1]

Mark scheme: 6 (a) (i) gravitational B1 [1] (ii) gravitational and electric B1 [1] (iii) magnetic and one other field given B1 magnetic, graviational and electric B1 [2] (b) (i) out of (plane of) paper/page (not “upwards”) B1 [1] (ii) B = mv / qr C1 = (3.32 × 10–26 × 7.6 × 104) / (1.6 × 10–19 × 6.1 × 10–2) C1 = 0.26 T A1 [3] (c) sketch: semicircle with diameter < 12.2 cm B1 [1]

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7 In many distribution systems for electrical energy, the energy is transmitted using alternating current at high voltages. Suggest and explain an advantage, one in each case, for the use of (a) alternating voltages, ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) high voltages. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2]

Mark scheme: 7 (a) can change (output) voltage efficiently or to suit different consumers/appliances B1 by using transformers B1 [2] (b) for same power, current is smaller B1 less heating in cables/wires or thinner cables possible or less voltage loss in cables B1 [2]

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8 A photon of wavelength 6.50 × 10−12 m is incident on an isolated stationary electron, as illustrated in Fig. 8.1. deflected photon wavelength 6.84 × 10–12 m incident photon e wavelength 6.50 × 10–12 m electron mass me Fig. 8.1 The photon is deflected elastically by the electron of mass me. The wavelength of the deflected photon is 6.84 × 10−12 m. (a) Calculate, for the incident photon, (i) its momentum, momentum = .................................................. N s [2] (ii) its energy. energy = ...................................................... J [2] (b) The angle θ through which the photon is deflected is given by the expression h Δλ = (1 – cos θ) mec where Δλ is the change in wavelength of the photon, h is the Planck constant and c is the speed of light in free space. (i) Calculate the angle θ. θ = ...................................................... ° [2] (ii) Use energy considerations to suggest why Δλ must always be positive. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3]

Mark scheme: 8 (a) (i) p = h / λ = (6.63 × 10–34) / (6.50 × 10–12) C1 = 1.02 × 10–22 N s A1 [2] (ii) E = hc / λ or E = pc = (6.63 × 10–34 × 3.00 × 108) / (6.50 × 10–12) C1 = 3.06 × 10–14 J A1 [2] (b) (i) 0.34 × 10–12 = (6.63 × 10–34) / (9.11 × 10–31 × 3.0 × 108) × (1 – cos θ) C1 θ = 30.7° A1 [2] (ii) deflected electron has energy M1 this energy is derived from the incident photon A1 deflected photon has less energy, longer wavelength (so ∆λ always positive) B1 [3]

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9 (a) An isotope of an element is radioactive. Explain what is meant by radioactive decay. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3] (b) At time t, a sample of a radioactive isotope contains N nuclei. In a short time Δt, the number of nuclei that decay is ΔN. State expressions, in terms of the symbols t, Δt, N and ΔN for (i) the number of undecayed nuclei at time (t + Δt), number = ......................................................... [1] (ii) the mean activity of the sample during the time interval Δt, mean activity = ......................................................... [1] (iii) the probability of decay of a nucleus during the time interval Δt, probability = ......................................................... [1] (iv) the decay constant. decay constant = ......................................................... [1] (c) The variation with time t of the activity A of a sample of a radioactive isotope is shown in Fig. 9.1. A 0 0 t ½ 2t ½ 3t ½ t Fig. 9.1 The radioactive isotope decays to form a stable isotope S. At time t = 0, there are no nuclei of S in the sample. On the axes of Fig. 9.2, sketch a graph to show the variation with time t of the number n of nuclei of S in the sample. n 0 0 t ½ 2t ½ 3t ½ t Fig. 9.2 [2]

Mark scheme: 9 (a) nucleus/nuclei emits M1 spontaneously/randomly A1 α-particles, β-particles, γ-ray photons A1 [3] (b) (i) N – ∆N A1 [1] (ii) ∆N / ∆t A1 [1] (iii) ∆N / N A1 [1] (iv) ∆N / N∆t A1 [1] (c) graph: smooth curve in correct direction starting at (0,0) M1 n at 2t½ is 1.5 times that at t½ (± 2 mm) A1 [2] Section B

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10 An operational amplifier (op-amp) is used in the comparator circuit of Fig. 10.1. +4.5 V 4.2 k1 +5 V + – –5 V V IN V OUT 1.2 k1 R Fig. 10.1 (a) (i) Show that the potential at the inverting input of the op-amp is +1.0 V. [1] (ii) Explain why the potential difference across resistor R is + 5 V when VIN is greater than 1.0 V and is zero when VIN is less than 1.0 V. VIN > 1.0 V: ......................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... VIN < 1.0 V: ......................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... [4] (b) The variation with time t of the input voltage VIN is shown in Fig. 10.2. 6 5 4 voltage / V 3 V IN 2 1 +1 V 0 0 time t Fig. 10.2 (i) On the axes of Fig. 10.2, draw the variation with time t of the output potential VOUT. [2] (ii) Suggest a use for this type of circuit. ........................................................................................................................................... ...................................................................................................................................... [1]

Mark scheme: 10 (a) (i) (potential =) 1.2 / (1.2 + 4.2) × 4.5 = +1.0 V A1 [1] (ii) (for VIN > 1.0 V) V+ > V– B1 output (of op-amp) is +5 V or positive M1 diode conducts giving +5 V across R or Vout is +5 V A1 (for VIN < 1.0 V) output of op-amp –5 V / negative so diode does not conduct, giving Vout = 0 or 0 V across R A1 [4] (b) (i) square wave with maximum value +5 V and minimum value 0 M1 vertical sides in correct positions and correct phase A1 [2] (ii) re-shaping (digital) signals/regenerator (amplifier) B1 [1]

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11 (a) State and explain how, in an X-ray tube, the hardness of the X-ray beam is controlled. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3] (b) A parallel beam of X-rays has intensity I0 and is incident on a medium having a linear absorption (attenuation) coefficient μ. (i) State an equation for the variation of the intensity I with the thickness x of the medium. ...................................................................................................................................... [1] (ii) Data for the linear absorption (attenuation) coefficient μ for an X-ray beam in blood and in muscle is shown in Fig. 11.1. μ/ cm−1 blood 0.23 muscle 0.22 Fig. 11.1 Suggest why, if this X-ray beam is used to image blood vessels in muscle, contrast on the image would be poor. ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2]

Mark scheme: 11 (a) change/increase/decrease anode/tube voltage B1 electrons striking anode have changed (kinetic) energy/speed B1 X-ray/photons/beam have different wavelength/frequency B1 [3] (b) (i) I = I0 e–µx B1 [1] (ii) contrast is difference in degree of blackening (of regions of the image) B1 µ (very) similar so similar absorption of radiation (for same thickness) so little contrast A1 [2]

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12 (a) Information may be carried by means of various channels of communication. Name examples, one in each case, of devices where information is carried to the device using (i) a wire pair, ...................................................................................................................................... [1] (ii) a coaxial cable, ...................................................................................................................................... [1] (iii) microwaves. ...................................................................................................................................... [1] (b) State two advantages of optic fibres as compared with coaxial cables for long-range communication. 1. .............................................................................................................................................. 2. .............................................................................................................................................. [2] (c) An optic fibre has length 62 km and an attenuation per unit length of 0.21 dB km−1. The input power to the fibre is P. At the receiver, the noise power is 9.2 μW. The signal-to-noise ratio at the receiver is 25 dB. (i) Calculate the ratio, in dB, of the input power P to the noise power at the receiver. ratio = ................................................... dB [2] (ii) Use your answer in (i) to determine the input power P. P = .................................................... W [2]

Mark scheme: 12 (a) (i) loudspeaker/doorbell/telephone etc. B1 [1] (ii) television set/audio amplifier etc. B1 [1] (iii) satellite/satellite dish/mobile phone etc. B1 [1] (b) e.g. lower attenuation/fewer repeaters more secure less prone to noise/interference physically smaller/less weight lower cost greater bandwidth (any two sensible suggestions, 1 each) B2 [2] (c) (i) ratio = 25 + (62 × 0.21) C1 = 38 dB A1 [2] (ii) ratio / dB = 10 lg (P2 / P1) C1 38 = 10 lg (P / {9.2 × 10–6}) P = 58 mW or 5.8 × 10–2 W A1 [2] (allow 1/2 for missing 10 in equation)

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13 During magnetic resonance imaging to obtain information about internal body structures, a large constant magnetic field is used with a calibrated non-uniform magnetic field superimposed on it. (a) State and explain the purpose of (i) the large constant magnetic field, ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (ii) the non-uniform magnetic field. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] (b) The de-excitation energy E (measured in joule) of a proton in magnetic resonance imaging is given by the expression E = 2.82 × 10−26 B where B is the magnetic flux density measured in tesla. The energy E is emitted as a photon of electromagnetic radiation in the radio-frequency range. Calculate the magnetic flux density required for the radio frequency to be 42 MHz. magnetic flux density = ..................................................... T [2]

Mark scheme: 13 (a) (i) to align nuclei/protons B1 to cause Larmor/precessional frequency to be in r.f. region B1 [2] (ii) Larmor/precessional frequency depends on (applied magnetic) field strength B1 knowing field strength enables (region of precessing) nuclei to be located M1 by knowing the frequency A1 [3] (b) E = 2.82 × 10–26 × B 6.63 × 10–34 × 42 × 106 = 2.82 × 10–26 × B C1 B = 0.99 T A1 [2]

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