Cambridge A Level Physics 9702 — 2019 Oct/Nov Paper 4 · Variant 1

1 (a) State Newton’s law of gravitation. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A geostationary satellite orbits the Earth. The orbit of the satellite is circular and the period of the orbit is 24 hours. (i) State two other features of this orbit. 1. ....................................................................................................................................... ........................................................................................................................................... 2. ....................................................................................................................................... ........................................................................................................................................... [2] (ii) The radius of the orbit of the satellite is 4.23 × 104 km. Determine a value for the mass of the Earth. Explain your working. mass = ..................................................... kg [4] [Total: 8]

Mark scheme: 1(a) force proportional to product of masses and inversely proportional to square of separation B1 idea of (gravitational) force between point masses B1 1(b)(i) above the equator B1 from west to east B1 1(b)(ii) gravitational force provides/is the centripetal force B1 GM / r2 = r (2π / T)2 C1 (6.67 × 10–11 × M) = {(4.23 × 107)3 × 4π2} / (24 × 3600)2 C1 M = 6.0 × 1024 kg A1

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2 (a) The kinetic theory of gases is based on a number of assumptions about the molecules of a gas. State the assumption that is related to the volume of the molecules of the gas. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) An ideal gas occupies a volume of 2.40 × 10–2 m3 at a pressure of 4.60 × 105 Pa and a temperature of 23 °C. (i) Calculate the number of molecules in the gas. number = ......................................................... [3] (ii) Each molecule has a diameter of approximately 3 × 10–10 m. Estimate the total volume of the gas molecules. volume = .................................................... m3 [3] (c) By reference to your answer in (b)(ii), suggest why the assumption in (a) is justified. ................................................................................................................................................... ............................................................................................................................................. [1] [Total: 9]

Mark scheme: 2(a) (total volume of molecules is) negligible M1 compared with volume occupied by the gas A1 2(b)(i) pV = NkT C1 4.60 × 105 × 2.40 × 10–2 = N × 1.38 × 10–23 × (273 + 23) C1 or pV = nRT (C1) 4.60 × 105 × 2.40 × 10–2 = n × 8.31 × (273 + 23) n = 4.49 (mol) N = nNA = 4.49 × 6.02 × 1023 (C1) N = 2.7 × 1024 A1 Question Answer Marks 2(b)(ii) volume of one atom = d3 (= 2.7 × 10–29 m3) C1 volume of all atoms = 2.7 × 10–29 × 2.7 × 1024 C1 = 7 × 10–5 m3 A1 or volume of one atom = (4 / 3)πr3 (= 1.41 × 10–29 m3) (C1) volume of all atoms = 2.7 × 1024 × 1.41 × 10–29 (C1) = 4 × 10–5 m3 (A1) 2(c) numerical comparison between answer to (b)(ii) and 2.4 × 10–2 (m3) showing (b)(ii) is much less than 2.4 × 10–2 (m3) B1

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3 (a) State what is meant by specific latent heat. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A student determines the specific latent heat of vaporisation of a liquid using the apparatus illustrated in Fig. 3.1. + V liquid A – heater pan of balance Fig. 3.1 The heater is switched on. When the liquid is boiling at a constant rate, the balance reading is noted at 2.0 minute intervals. After 10 minutes, the current in the heater is reduced and the balance readings are taken for a further 12 minutes. The readings of the ammeter and of the voltmeter are given in Fig. 3.2. ammeter reading voltmeter reading / A / V from time 0 to time 10 minutes 1.2 230 after time 10 minutes 1.0 190 Fig. 3.2 The variation with time of the balance reading is shown in Fig. 3.3. 500 480 balance reading / g 460 440 420 400 380 0 4 8 12 16 20 24 time / minutes Fig. 3.3 (i) From time 0 to time 10.0 minutes, the mass of liquid evaporated is 56 g. Use Fig. 3.3 to determine the mass of liquid evaporated from time 12.0 minutes to time 22.0 minutes. mass = .......................................................g [1] (ii) Explain why, although the power of the heater is changed, the rate of loss of thermal energy to the surroundings may be assumed to be constant. ........................................................................................................................................... ..................................................................................................................................... [1] (iii) Determine a value for the specific latent heat of vaporisation L of the liquid. L = ................................................. J g–1 [4] (iv) Calculate the rate at which thermal energy is transferred to the surroundings. rate = ..................................................... W [2] [Total: 10]

Mark scheme: 3(a) (thermal) energy per (unit) mass (to change state) B1 (heat transfer during) change of state at constant temperature B1 3(b)(i) 32 g A1 3(b)(ii) temperature difference (between liquid and surroundings) does not change B1 3(b)(iii) VIt = mL C1 230 × 1.2 × 60 × 10 = (56 × L) + H or 190 × 1.0 × 60 × 10 = (32 × L) + H C1 86 × 600 = (56 – 32) × L C1 or 230 × 1.2 = (56 × L) / (60 × 10) + P or 190 × 1.0 = (32 × L) / (60 × 10) + P (C1) 276 – 190 = (24 × L) / 600 (C1) L = 2200 J g–1 A1 Question Answer Marks 3(b)(iv) 230 × 1.2 × 600 = (56 × 2150) + H or 190 × 1.0 × 600 = (32 × 2150) + H C1 H = 45 200 rate = 45 200 / 600 = 75 W A1 or 230 × 1.2 = (56 × 2150) / (60 × 10) + P or 190 × 1.0 = (32 × 2150) / (60 × 10) + P (C1) rate (= P) = 75 W (A1)

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4 A mass is suspended vertically from a fixed point by means of a spring, as illustrated in Fig. 4.1. spring mass Fig. 4.1 The mass is oscillating vertically. The variation with displacement x of the acceleration a of the mass is shown in Fig. 4.2. 1.5 a / m s–2 1.0 0.5 0 –1.5 –1.0 –0.5 0 0.5 1.0 1.5 x / cm –0.5 –1.0 –1.5 Fig. 4.2 (a) (i) State what is meant by the displacement of the mass on the spring. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) Suggest how Fig. 4.2 shows that the mass is not performing simple harmonic motion. ........................................................................................................................................... ..................................................................................................................................... [1] (b) (i) The amplitude of oscillation of the mass may be changed. State the maximum amplitude x0 for which the oscillations are simple harmonic. x0 = .................................................... cm [1] (ii) For the simple harmonic oscillations of the mass, use Fig. 4.2 to determine the frequency of the oscillations. frequency = .................................................... Hz [3] (c) The maximum speed of the mass when oscillating with simple harmonic motion of amplitude x0 is v0. On Fig. 4.3, show the variation with displacement x of the velocity v of the mass for displacements from +x0 to –x0. v v0 0 –x0 0 x0 x –v0 Fig. 4.3 [2] [Total: 8]

Mark scheme: 4(a)(i) distance from a (reference) point in a given direction B1 4(a)(ii) line is not straight or gradient is not constant B1 4(b)(i) 0.85–0.90 cm A1 4(b)(ii) a = – (2πf )2 x C1 e.g. 1.2 = 4π2 × f 2 × (0.90 × 10–2) C1 f = 1.8 Hz A1 4(c) complete circle/ellipse enclosing the origin B1 closed shape passing through (0, ±v0) and (±x0, 0) B1

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5 (a) A section of a coaxial cable is shown in Fig. 5.1. copper braid insulation copper wire plastic covering Fig. 5.1 (i) Suggest two functions of the copper braid. 1. ....................................................................................................................................... ........................................................................................................................................... 2. ....................................................................................................................................... ........................................................................................................................................... [2] (ii) Suggest one application of a coaxial cable for the transmission of electrical signals. ........................................................................................................................................... ..................................................................................................................................... [1] (b) (i) The constant noise power in a transmission cable is 7.6 μW. The minimum acceptable signal-to-noise ratio is 32 dB. Calculate the minimum acceptable signal power PMIN in the cable. PMIN = ......................................................W [2] (ii) The input power of the signal to the transmission cable is 2.6 W. The attenuation per unit length of the cable is 6.3 dB km–1. Use your answer in (i) to determine the maximum uninterrupted length L of cable along which the signal may be transmitted. L = .................................................... km [2] [Total: 7]

Mark scheme: 5(a)(i) provides return for the signal B1 shields signal from noise B1 5(a)(ii) e.g. connection between aerial and TV set B1 5(b)(i) gain / dB = 10 lg (P1 / P2) C1 32 = 10 lg {PMIN / (7.6 × 10–6)} PMIN = 0.012 W A1 5(b)(ii) attenuation per unit length = (1 / L) × 10 lg (P1 / P2) 6.3 = (1 / L) × 10 lg (2.6 / 0.012) C1 L = 3.7 km A1

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6 (a) State an expression for the electric field strength E at a distance r from a point charge Q in a vacuum. State the name of any other symbol used. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Two point charges A and B are situated a distance 10.0 cm apart in a vacuum, as illustrated in Fig. 6.1. charge A charge B P x 10.0 cm Fig. 6.1 A point P lies on the line joining the charges A and B. Point P is a distance x from A. The variation with distance x of the electric field strength E at point P is shown in Fig. 6.2. 2.5 E / 10–2 N C–1 2.0 1.5 1.0 0 2 4 6 8 10 x / cm Fig. 6.2 State and explain whether the charges A and B: (i) have the same, or opposite, signs ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) have the same, or different, magnitudes. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (c) An electron is situated at point P. Without calculation, state and explain the variation in the magnitude of the acceleration of the electron as it moves from the position where x = 3 cm to the position where x = 7 cm. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [4] [Total: 10]

Mark scheme: 6(a) M1 where ε0 is permittivity (of free space) A1 6(b)(i) field does not change direction/field does not become zero M1 so (charges have) opposite (sign) A1 6(b)(ii) minimum is at the midpoint (between the charges) M1 so (magnitudes are the) same A1 6(c) force = field strength × charge and force = mass × acceleration or acceleration is proportional to field strength B1 (from x = 3.0 cm) to x = 5.0 cm: acceleration decreases B1 at x = 5.0 cm: acceleration is a minimum B1 from x = 5.0 cm (to x = 7.0 cm): acceleration increases B1

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7 (a) An ideal operational amplifier (op-amp) has infinite bandwidth and zero output impedance. State what is meant by: (i) infinite bandwidth ........................................................................................................................................... ..................................................................................................................................... [1] (ii) zero output impedance. ........................................................................................................................................... ..................................................................................................................................... [1] (b) The circuit for a non-inverting amplifier incorporating an ideal op-amp is shown in Fig. 7.1. 4.0 kΩ +5.0 V – + –5.0 V V OUT V IN 800 Ω R Fig. 7.1 The light-emitting diode (LED) emits light when the potential difference across it is at least 2.0 V. The current in the LED must not be greater than 20 mA. (i) Calculate the gain of the amplifier circuit. gain = ......................................................... [2] (ii) Determine the value of VIN for which the value of VOUT is +2.0 V. VIN = ...................................................... V [1] (iii) State the maximum value of the output potential VOUT. maximum potential = ...................................................... V [1] (iv) When the op-amp is saturated, the potential difference across the LED is 2.2 V. Calculate the minimum resistance of resistor R so that the current in the LED is limited to 20 mA. resistance = ...................................................... Ω [2] [Total: 8]

Mark scheme: 7(a)(i) all frequencies are amplified equally B1 7(a)(ii) no drop in output voltage (when there is a current) B1 7(b)(i) gain = 1 + RF / RIN gain = 1 + (4000 / 800) C1 gain = 6.0 A1 7(b)(ii) 2.0 / VIN = 6.0 VIN = (+)0.33 V A1 7(b)(iii) 5.0 V A1 7(b)(iv) V = 5.0 – 2.2 (= 2.8 V) C1 R = V / I = 2.8 / 0.020 = 140 Ω A1

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8 (a) A long straight vertical wire carries a current I. The wire passes through a horizontal card EFGH, as shown in Fig. 8.1 and Fig. 8.2. current out of plane of paper H G I wire H G E F E F Fig. 8.1 Fig. 8.2 (view from above) On Fig. 8.2, draw the pattern of the magnetic field produced by the current-carrying wire on the plane EFGH. [3] (b) Two long straight parallel wires P and Q are situated a distance 3.1 cm apart, as illustrated in Fig. 8.3. 6.2 A 8.5 A wire P wire Q 3.1cm Fig. 8.3 The current in wire P is 6.2 A. The current in wire Q is 8.5 A. The magnetic flux density B at a distance x from a long straight wire carrying current I is given by the expression μ0I B = 2πx where μ0 is the permeability of free space. Calculate: (i) the magnetic flux density at wire Q due to the current in wire P flux density = ...................................................... T [2] (ii) the force per unit length, in N m–1, acting on wire Q due to the current in wire P. force per unit length = ............................................... N m–1 [2] (c) The currents in wires P and Q are different in magnitude. State and explain whether the forces per unit length on the two wires will be different. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 9]

Mark scheme: 8(a) concentric circles (around the wire) M1 at least 3 circles shown, all with increasing separation A1 direction anticlockwise B1 8(b)(i) B = (4π × 10–7 × 6.2) / (2π × 3.1 × 10–2) C1 B = 4.0 × 10–5 T A1 8(b)(ii) F = BIL or F / L = BI C1 F / L = 4.0 × 10–5 × 8.5F / L = 3.4 × 10–4 N m–1 A1 8(c) correct application of Newton’s 3rd law to the forces or F / L is proportional to the product of the two currents M1 so same magnitude A1

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9 Diagnosis using nuclear magnetic resonance imaging (NMRI) requires the use of a non-uniform magnetic field superimposed on a constant magnetic field of large magnitude. Explain the purpose of: (a) the large constant magnetic field ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) the non-uniform magnetic field. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 4]

Mark scheme: 9(a) nuclei precess B1 precession is about (direction of magnetic) field or frequency of precession is in radio-frequency range B1 9(b) • frequency (of precession) depends on field strength • to locate/find position of (spinning) nuclei • to change region where nuclei are detected any two points, one mark each B2

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10 A bridge rectifier using four ideal diodes is shown in Fig. 10.1. A B R Fig. 10.1 The sinusoidal alternating electromotive force (e.m.f.) applied between points A and B has a root- mean-square (r.m.s.) value of 7.0 V. (a) (i) On Fig. 10.1, circle the diodes that conduct when point B is positive with respect to point A. [1] (ii) Calculate the maximum potential difference VMAX across resistor R. VMAX = ...................................................... V [1] (b) A capacitor is connected into the circuit to produce smoothing of the potential difference across resistor R. The variation with time t of the potential difference V across resistor R is shown in Fig. 10.2. V magnitude of ripple 0 t Fig. 10.2 (i) On Fig. 10.1, draw the symbol for a capacitor, connected so as to produce smoothing. [1] (ii) State the effect, if any, on the magnitude of the ripple on V when, separately: 1. a capacitor of larger capacitance is used ........................................................................................................................................... 2. the resistor R has a smaller resistance. ........................................................................................................................................... [2] [Total: 5]

Mark scheme: 10(a)(i) lower right and upper left diodes circled B1 10(a)(ii) maximum = 7.0√2 maximum = 9.9 V A1 10(b)(i) correct symbol for capacitor, shown connected in parallel with R B1 10(b)(ii) 1. (ripple) decreases B1 2. (ripple) increases B1

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11 (a) With reference to the photoelectric effect, state what is meant by work function energy. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) The work function energy of a clean metal surface is 5.5 × 10–19 J. Electromagnetic radiation of wavelength 280 nm is incident on the metal surface. The metal is in a vacuum. (i) Calculate: 1. the photon energy photon energy = ....................................................... J [2] 2. the maximum speed vMAX of the electrons emitted from the surface. vMAX = ................................................ m s–1 [3] (ii) Explain why most of the emitted electrons will have a speed lower than vMAX. ........................................................................................................................................... ..................................................................................................................................... [1] (c) The electromagnetic radiation incident on the metal surface may change in intensity or in frequency. Complete Fig. 11.1 by inserting either ‘increases’ or ‘decreases’ or ‘no change’ to describe the effects of the changes shown on the maximum speed and on the rate of emission of electrons. maximum speed of rate of emission change electrons of electrons reduced intensity at constant frequency ................................... ................................... increased frequency at constant intensity ................................... ................................... Fig. 11.1 [4] [Total: 12]

Mark scheme: 11(a) energy (of photon) required to remove electron M1 from a surface or reference to minimum energy or reference to zero kinetic energy A1 11(b)(i) 1. photon energy = hc / λ C1 = (6.63 × 10–34 × 3.00 × 108) / (280 × 10–9) = 7.1 × 10–19 J A1 2. electron energy = (7.1 – 5.5) × 10–19 J C1 ½ × 9.11 × 10–31 × v2 = (7.1 – 5.5) × 10–19 C1 v = 5.9 × 105 m s–1 A1 11(b)(ii) energy is required to bring electron to the surface B1 11(c) no change decreases increases decreases B4

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12 One possible nuclear reaction that takes place is 23 9 52U + 10n 9452Mo + 1357La9 + 210n + 7–10e Data for nuclei in this reaction are given in Fig. 12.1. total mass of separate binding energy nucleus mass / u mass defect / u nucleons / u per nucleon / MeV 9 4 52Mo 94.906 95.765 0.859 8.443 13 57La9 138.906 140.125 1.219 8.189 23 9 52U 235.044 236.909 1.865 .................................. Fig. 12.1 (a) Show that the energy equivalent to a mass of 1.00 u is 934 MeV. [2] (b) (i) Use data from Fig. 12.1 to calculate the binding energy per nucleon of a nucleus of uranium-235 (23952U). Complete Fig. 12.1. [2] (ii) The nucleon number of an isotope of the element rutherfordium is 267. State whether the binding energy per nucleon of this isotope will be greater than, equal to or less than the binding energy per nucleon of uranium-235. ..................................................................................................................................... [1] (c) Calculate the total energy, in MeV, released in this nuclear reaction. energy = ................................................. MeV [2] (d) The nuclei in 1.2 × 10–7 mol of uranium-235 all undergo this reaction in a time of 25 ms. Calculate the average power release during the time of 25 ms. power = ..................................................... W [3] [Total: 10]

Mark scheme: 12(a) M1 E = (1.49 × 10–10) / (1.60 × 10–19) = 9.34 × 108 = 934 MeV A1 or binding energy = 8.443 × 95 [or equivalent using La-139 nucleus] (M1) binding energy / mass defect = (8.443 × 95) / 0.859 = 934 MeV (A1) 12(b)(i) binding energy = 1.865 × 934 (= 1741.91 MeV) C1 binding energy per nucleon = 1741.91 / 235 = 7.41 (MeV) A1 12(b)(ii) less (than) B1 12(c) energy = {(1.219 + 0.859) – 1.865)} × 934 or energy = (95 × 8.443) + (139 × 8.189) – (235 × 7.412) C1 = 199 MeV A1 12(d) number of reactions = 1.2 × 10–7 × 6.02 × 1023 = 7.22 × 1016 C1 energy release (for one reaction) = 199 × 1.60 × 10–13 (= 3.18 × 10–11 J) C1 power = (7.22 × 1016 × 3.18 × 10–11) / (25 × 10–3) = 9.2 × 107 W A1

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