Cambridge A Level Physics 9702 — 2021 May/June Paper 4 · Variant 1

1 The Earth may be assumed to be an isolated uniform sphere with its mass of 6.0 × 1024 kg concentrated at its centre. A satellite of mass 1200 kg is in a circular orbit about the Earth in the Earth’s gravitational field. The period of the orbit is 94 minutes. (a) Define gravitational field strength. ................................................................................................................................................... ............................................................................................................................................. [1] (b) Calculate the radius of the orbit of the satellite. radius = ..................................................... m [3] (c) Rockets on the satellite are fired so that the satellite enters a different circular orbit that has a period of 150 minutes. The change in the mass of the satellite may be assumed to be negligible. (i) Show that the radius of the new orbit is 9.4 × 106 m. [2] (ii) State, with a reason, whether the gravitational potential energy of the satellite increases or decreases. ........................................................................................................................................... ..................................................................................................................................... [1] (iii) Determine the magnitude of the change in the gravitational potential energy of the satellite. change in potential energy = ...................................................... J [3] [Total: 10]

Mark scheme: 1(a) force per unit mass B1 1(b) GMm / r 2 = mrω 2 and ω = 2π/T or GMm / r 2 = mv2 / r and v = 2πr / T C1 6.67 × 10–11 × 6.0 × 1024 = r3 × [2π / (94 × 60)]2 C1 r = 6.9 × 106 m A1 1(c)(i) r3ω2 = constant or r3 / T2 = constant C1 r3 / (6.9 × 106)3 = (150 / 94)2 so r = 9.4 × 106 m A1 or GMT2/4π2 = r3 and clear that M is 6.0 × 1024 (C1) 6.67 × 10–11 × 6.0 × 1024 = r3 × [2π / (150 × 60)]2 so r = 9.4 × 106 m (A1) 1(c)(ii) separation increases so (potential energy) increases or movement is against gravitational force so (potential energy) increases B1 1(c)(iii) potential energy = (–)GMm / r C1 ΔEP = 6.67 × 10–11 × 6.0 × 1024 × 1200 × [(6.9 × 106)–1 – (9.4 × 106)–1] C1 = 1.9 × 1010 J A1

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2 An ideal gas is contained in a cylinder by means of a movable frictionless piston, as illustrated in Fig. 2.1. cylinder movement of piston piston gas molecule Fig. 2.1 Initially, the gas has a volume of 1.8 × 10−3 m3 at a pressure of 3.3 × 105 Pa and a temperature of 310 K. (a) Show that the number of gas molecules in the cylinder is 1.4 × 1023. [2] (b) Use kinetic theory to explain why, when the piston is moved so that the gas expands, this causes a decrease in the temperature of the gas. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (c) The gas expands so that its volume increases to 2.4 × 10−3 m3 at a pressure of 2.3 × 105 Pa and a temperature of 288 K, as shown in Fig. 2.2. 1.8 × 10−3 m3 2.4 × 10−3 m3 3.3 × 105 Pa 2.3 × 105 Pa 310 K 288 K Fig. 2.2 (i) The average translational kinetic energy EK of a molecule of an ideal gas is given by 3 EK = kT 2 where k is the Boltzmann constant and T is the thermodynamic temperature. Calculate the increase in internal energy ΔU of the gas during the expansion. ΔU = ...................................................... J [3] (ii) The work done by the gas during the expansion is 76 J. Use your answer in (i) to explain whether thermal energy is transferred to or from the gas during the expansion. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 10]

Mark scheme: 2(a) pV = NkT C1 N = (1.8 × 10–3 × 3.3 × 105) / (1.38 × 10–23 × 310) = 1.4 × 1023 A1 or pV = nRT and nNA = N (C1) N = (1.8 × 10–3 × 3.3 × 105 × 6.02 × 1023) / (8.31 × 310) = 1.4 × 1023 (A1) 2(b) speed of molecule decreases on impact with moving piston B1 mean square speed (directly) proportional to (thermodynamic) temperature or mean square speed (directly) proportional to kinetic energy (of molecules) or kinetic energy (of molecules) (directly) proportional to (thermodynamic) temperature B1 kinetic energy (of molecules) decreases (so temperature decreases) B1 2(c)(i) ΔU = 3/2 × k × ΔT × N C1 = 3/2 × 1.38 × 10–23 × (288 – 310) × 1.4 × 1023 C1 = – 64 J A1 2(c)(ii) decrease in internal energy is less than work done by gas M1 (thermal energy is) transferred to the gas (during the expansion) A1

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3 (a) State what is meant by simple harmonic motion. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A trolley of mass m is held on a horizontal surface by means of two springs. One spring is attached to a fixed point P. The other spring is connected to an oscillator, as shown in Fig. 3.1. spring trolley spring oscillator P Fig. 3.1 The springs, each having spring constant k of 130 N m−1, are always extended. The oscillator is switched off. The trolley is displaced along the line of the springs and then released. The resulting oscillations of the trolley are simple harmonic. The acceleration a of the trolley is given by the expression ⎛ ⎞2k a = − x ⎝ ⎠m where x is the displacement of the trolley from its equilibrium position. The mass of the trolley is 840 g. Calculate the frequency f of oscillation of the trolley. f = .................................................... Hz [3] (c) The oscillator in (b) is switched on. The frequency of oscillation of the oscillator is varied, keeping its amplitude of oscillation constant. The amplitude of oscillation of the trolley is seen to vary. The amplitude is a maximum at the frequency calculated in (b). (i) State the name of the effect giving rise to this maximum. ..................................................................................................................................... [1] (ii) At any given frequency, the amplitude of oscillation of the trolley is constant. Explain how this indicates that there are resistive forces opposing the motion of the trolley. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 8]

Mark scheme: 3(a) acceleration (directly) proportional to displacement B1 acceleration is in opposite direction to displacement B1 3(b) ω2 = 2k / m and ω = 2πf C1 (2πf)2 = (2 × 130) / 0.84 C1 f = 2.8 Hz A1 3(c)(i) resonance B1 3(c)(ii) oscillator supplies energy (continuously) B1 energy of trolley constant so energy must be dissipated or without loss of energy the amplitude would continuously increase B1

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4 Outline the use of ultrasound to obtain diagnostic information about internal body structures. .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .................................................................................................................................................... [5]

Mark scheme: 4 (ultrasound) pulse B1 reflected at boundaries B1 gel is used to minimise reflection at skin or generated and detected by quartz crystal B1 time delay between generation and detection gives information about depth B1 intensity (of reflected wave) gives information about nature of boundary B1

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5 (a) State what is meant by the amplitude modulation (AM) of a radio wave. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A radio wave is modulated by an audio signal. The variation with frequency f of the amplitude of the modulated wave is shown in Fig. 5.1. amplitude 0 292 300 308 f / kHz Fig. 5.1 Determine: (i) the wavelength of the carrier wave wavelength = ..................................................... m [1] (ii) the bandwidth of the modulated wave bandwidth = .................................................. kHz [1] (iii) the maximum frequency of the audio signal. maximum frequency = .................................................. kHz [1] (c) The power of a radio signal at a transmitter is PT. At a receiver, the received power PR is given by the expression 0.082 PT PR = x2 where x is the distance, in metres, between the transmitter and the receiver. For the transmission of this signal, the attenuation is 73 dB. Determine the distance x. x = ..................................................... m [3] [Total: 8]

Mark scheme: 5(a) amplitude of the carrier wave varies M1 in synchrony with the displacement of the (information) signal A1 5(b)(i) wavelength = (3.0 × 108) / (300 × 103) = 1000 m A1 5(b)(ii) bandwidth = 16 kHz A1 5(b)(iii) frequency = 8 kHz A1 5(c) attenuation = 10 lg (P1 / P2) C1 73 = 10 lg (PT / PR) 73 = 10 lg (PT x2 / 0.082 PT) or x2 / 0.082 = 107.3 C1 x = 1300 m A1

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6 (a) An isolated metal sphere of radius r is charged so that the electric field strength at its surface is E0. On Fig. 6.1, sketch the variation of the electric field strength E with distance x from the centre of the sphere. Your sketch should extend from x = 0 to x = 3r. E0 field strength E 0 0 r 2r 3r distance x Fig. 6.1 [3] (b) The de Broglie wavelength of a particle is λ 0 when its momentum is p0. On Fig. 6.2, sketch the variation with momentum p of the de Broglie wavelength λ of the p0 particle for values of momentum from to p0. 2 2λ0 wavelength λ λ 0 0 0 p0 p0 2 momentum p Fig. 6.2 [2] (c) A radioactive isotope decays with a half-life of 15 s to form a stable product. A fresh sample of the radioactive isotope at time t = 0 contains N0 nuclei and no nuclei of the stable product. On Fig. 6.3, sketch the variation with t of the number n of nuclei of the stable product for time t = 0 to time t = 45 s. N0 number n 0.5 N0 0 0 15 30 45 time t / s Fig. 6.3 [3] [Total: 8]

Mark scheme: 6(a) from x = 0 to x = r: E = 0 B1 from x = r to x = 3r: curve with negative gradient of decreasing magnitude passing through (r, E0) B1 line passing through (2r, E0 / 4) and (3r, E0 / 9) B1 6(b) from p = p0 / 2 to p = p0: curve with negative gradient of decreasing magnitude passing through (p0, λ0) B1 line passing through (½p0, 2λ0) B1 6(c) from t = 0 to t = 45 s: curve with positive gradient of decreasing magnitude starting at (0, 0) B1 line passing through (15, ½N0) B1 line passing through (30, 0.75N0) and (45, 0.88N0) B1

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7 (a) State what is meant by the capacitance of a parallel plate capacitor. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A capacitor of capacitance C is connected into the circuit shown in Fig. 7.1. A B sensitive + ammeter V A – C Fig. 7.1 When the two-way switch is in position A, the capacitor is charged so that the potential difference across it is V. The switch moves to position B and the capacitor fully discharges through the sensitive ammeter. The switch moves repeatedly between A and B so that the capacitor charges and then discharges with frequency f. (i) Show that the average current I in the ammeter is given by the expression I = fCV. [2] (ii) For a potential difference V of 150 V and a frequency f of 60 Hz, the average current in the ammeter is 4.8 μA. Calculate the capacitance, in pF, of the capacitor. capacitance = .................................................... pF [2] (c) A second capacitor, having the same capacitance as the capacitor in (b), is connected into the circuit of Fig. 7.1. The two capacitors are connected in series. State and explain the new reading on the ammeter. new reading = ......................................................... μA ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] [Total: 9]

Mark scheme: 7(a) charge / potential M1 charge is on one plate, potential is p.d. between the plates A1 7(b)(i) I = Q / t M1 charge = CV and time = 1 / f leading to I = fCV A1 7(b)(ii) 4.8 × 10–6 = 150 × 60 × C C1 C = 530 pF A1 7(c) (total) capacitance is halved B1 charge (for each cycle/discharge) is halved or since f and V are constant, current is proportional to capacitance B1 current = 2.4 μA B1

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8 The variation with temperature of the resistance of a thermistor is shown in Fig. 8.1. 4.0 3.0 resistance / kΩ 2.0 1.0 0 0 10 20 30 temperature / °C Fig. 8.1 A student includes the thermistor and an ideal operational amplifier (op-amp) in the circuit of Fig. 8.2. +3.0 V 2.5 kΩ + – + – 3.0 kΩ 5.0 kΩ Fig. 8.2 (a) Calculate the potential V + at the non-inverting input of the op-amp. V + = ...................................................... V [2] (b) At 10 °C, the resistance of the thermistor is 2.5 kΩ. State and explain whether the light-emitting diode (LED) is emitting light. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (c) Explain why the student’s circuit will not indicate any change in temperature above 0 °C. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (d) The resistor of resistance 5.0 kΩ is changed to a resistor of resistance R so that the LED switches on or off at a temperature of 20 °C. Determine R in kΩ. R = .................................................... kΩ [3] [Total: 9]

Mark scheme: 8(a) V+ = 3.0 × 3.0 / (2.5 + 3.0) = 1.6 V A1 8(b) V – is +2.0 V or V – > V + B1 output is negative so (LED) does not emit light B1 8(c) at 0 °C, V – = 1.7 V or for all temperatures above 0 °C, resistance of thermistor < 4.2 kΩ B1 V – always greater than V + (so no switching) B1 8(d) (at 20 °C,) RT = 1.8 kΩ C1 2.5 / 3.0 = 1.8 / R or [R / (R + 1.8)] × 3.0 = 1.6 C1 R = 2.2 kΩ A1

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9 (a) State what is meant by a magnetic field. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A rectangular piece of aluminium foil is situated in a uniform magnetic field of flux density B, as shown in Fig. 9.1. magnetic field, flux density B Q R T aluminium movement foil of electrons P S V W Fig. 9.1 The magnetic field is normal to the face PQRS of the foil. Electrons, each of charge −q, enter the foil at right angles to the face PQTV. (i) On Fig. 9.1, shade the face of the foil on which electrons initially accumulate. [1] (ii) Explain why electrons do not continuously accumulate on the face you have shaded. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (c) The Hall voltage VH developed across the foil in (b) is given by the expression BI VH = ntq where I is the current in the foil. (i) State the meaning of the quantity n. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) Using the letters on Fig. 9.1, identify the distance t. ..................................................................................................................................... [1] (d) Suggest why, in practice, Hall probes are usually made using a semiconductor material rather than a metal. ................................................................................................................................................... ............................................................................................................................................. [1] [Total: 9]

Mark scheme: 9(a) region where there is a force exerted on M1 a current-carrying conductor or a moving charge or a magnetic material/magnetic pole A1 9(b)(i) face PSWV shaded B1 9(b)(ii) accumulating electrons cause an electric field (between the faces) B1 force due to electric field opposes force due to magnetic field B1 accumulation stops when magnetic force equals electric force B1 9(c)(i) number density of charge carriers B1 9(c)(ii) PV or QT or SW B1 9(d) (for semiconductor,) n is (much) smaller so VH (much) larger B1

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10 (a) State Lenz’s law. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A metal ring is suspended from a fixed point P by means of a thread, as shown in Fig. 10.1. P P metal magnet ring pole piece metal ring N S Fig. 10.1 Fig. 10.2 The ring is displaced a distance d and then released. The ring completes many oscillations before coming to rest. The poles of a magnet are now placed near to the ring so that the ring hangs midway between the poles of the magnet, as shown in Fig. 10.2. The ring is again displaced a distance d and then released. Explain why the ring completes fewer oscillations before coming to rest. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [4] (c) The ring in (b) is now cut so that it has the shape shown in Fig. 10.3. Fig. 10.3 Explain why, when the procedure in (b) is repeated, the cut ring completes more oscillations than the complete ring when oscillating between the poles of the magnet. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] [Total: 9]

Mark scheme: 10(a) direction of (induced) e.m.f. M1 is such as to oppose the change causing it A1 10(b) ring cuts (magnetic) flux and causes induced e.m.f. in ring B1 (induced) e.m.f. causes (eddy/induced) currents (in ring) B1 currents (in ring) cause magnetic field (around ring) M1 two fields interact to cause resistive/opposing force A1 or current (in ring) is in a magnetic field (M1) which causes resistive force (A1) or currents (in ring) dissipate thermal energy (M1) (thermal) energy comes from energy of oscillations (A1) 10(c) current cannot pass all the way around the ring B1 (induced) currents smaller B1 smaller resistive force (so more oscillations) or smaller rate of dissipation of energy (so more oscillations) B1

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11 (a) State how, in a modern X-ray tube, the intensity of the X-ray beam and its hardness are controlled. intensity: ................................................................................................................................... ................................................................................................................................................... hardness: .................................................................................................................................. ................................................................................................................................................... [2] (b) A model of a limb consists of soft tissue and bone, as illustrated in Fig. 11.1. 3.0 cm I0 IC incident transmitted intensity intensity I0 IS bone soft tissue 9.0 cm Fig. 11.1 The soft tissue has a thickness of 9.0 cm. The bone within the soft tissue has a thickness of 3.0 cm. Data for the linear attenuation (absorption) coefficient μ of X-rays in soft tissue and in bone are shown in Table 11.1. Table 11.1 μ / cm−1 soft tissue 0.92 bone 2.90 A parallel beam of X-rays of intensity I0 is incident normally on the model. Calculate, in terms of I0: (i) the transmitted intensity IS through soft tissue alone IS = ..................................................... I0 [2] (ii) the transmitted intensity IC through soft tissue and bone. IC = ..................................................... I0 [2] (c) By reference to your answers in (b), suggest, with a reason, whether good contrast on an X-ray image would be obtained. ................................................................................................................................................... ............................................................................................................................................. [1] [Total: 7]

Mark scheme: 11(a) intensity: vary filament current/p.d. across filament B1 hardness: vary accelerating potential difference B1 11(b)(i) I = I0e –μx C1 IS = I0 exp(–0.92 × 9.0) = 2.5 × 10–4 I0 A1 11(b)(ii) IC = [exp(–0.92 × 6.0) × exp(–2.9 × 3.0)] I0 C1 = 6.7 × 10–7 I0 A1 11(c) conclusion consistent with values in (b)(i) and (b)(ii) e.g. IS ≫ IC so good contrast B1

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12 (a) Electromagnetic radiation of a single constant frequency is incident on a metal surface. This causes an electron to be emitted. Explain why the maximum kinetic energy of the electron is independent of the intensity of the incident radiation. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (b) Ultraviolet radiation of wavelength 250 nm is incident on the surface of a sheet of zinc. The maximum kinetic energy of the emitted electrons is 1.4 eV. Determine, in eV: (i) the energy of a photon of the ultraviolet radiation energy = .................................................... eV [3] (ii) the work function energy of the surface of the zinc. energy = .................................................... eV [2] [Total: 8]

Mark scheme: 12(a) • frequency determines energy of photon • intensity determines number of photons (per unit time) • intensity does not determine energy of a photon Any two points, 1 mark each B2 kinetic energy (of the electron) depends on the energy of one photon B1 12(b)(i) E = hc / λ or E = hf and c = fλ C1 E = (6.63 × 10–34 × 3.00 × 108) / (250 × 10–9) C1 (= 7.96 × 10–19 J) = 5.0 eV A1 12(b)(ii) EMAX = photon energy – work function C1 work function = 5.0 – 1.4 = 3.6 eV A1

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