Cambridge A Level Physics 9702 — 2017 Oct/Nov Paper 4 · Variant 2

1 (a) State Newton’s law of gravitation. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) The planet Jupiter and one of its moons, Io, may be considered to be uniform spheres that are isolated in space. Jupiter has radius R and mean density ρ. Io has mass m and is in a circular orbit about Jupiter with radius nR, as illustrated in Fig. 1.1. Jupiter radius R density ρ Io nR Fig. 1.1 The time for Io to complete one orbit of Jupiter is T. Show that the time T is related to the mean density ρ of Jupiter by the expression 2 3πn3 ρT = G where G is the gravitational constant. [4] (c) (i) The radius R of Jupiter is 7.15 × 104 km and the distance between the centres of Jupiter and Io is 4.32 × 105 km. The period T of the orbit of Io is 42.5 hours. Calculate the mean density ρ of Jupiter. ρ = ............................................... kg m–3 [3] (ii) The Earth has a mean density of 5.5 × 103 kg m–3. It is said to be a planet made of rock. By reference to your answer in (i), comment on the possible composition of Jupiter. ........................................................................................................................................... .......................................................................................................................................[1] [Total: 10]

Mark scheme: 1(a) force proportional to product of masses and inversely proportional to square of separation B1 idea of force between point masses B1 1(b) mass of Jupiter (M) = (4 / 3) πR3ρ B1 ω = 2π / T or v = 2πnR / T B1 (m)ω2x = GM(m) / x2 or (m)v2 / x = GM(m) / x2 M1 substitution and correct algebra leading to ρT2 = 3πn3 / G A1 1(c)(i) n = (4.32 × 105) / (7.15 × 104) or n = 6.04 C1 ρ × (42.5 × 3600)2 = (3π × 6.043) / (6.67 × 10–11) C1 ρ = 1.33 × 103 kg m–3 A1 1(c)(ii) Jupiter likely to be a gas/liquid (at high pressure) [allow other sensible suggestions] B1

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2 (a) State what is meant by specific latent heat. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) A beaker of boiling water is placed on the pan of a balance, as illustrated in Fig. 2.1. A V d.c. supply heater balance pan boiling water Fig. 2.1 The water is maintained at its boiling point by means of a heater. The change M in the balance reading in 300 s is determined for two different input powers to the heater. The results are shown in Fig. 2.2. voltmeter reading ammeter reading M / g / V / A 11.5 5.2 5.0 14.2 6.4 9.1 Fig. 2.2 (i) Energy is supplied continuously by the heater. State where, in this experiment, 1. external work is done, ........................................................................................................................................... ........................................................................................................................................... 2. internal energy increases. Explain your answer. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... [3] (ii) Use data in Fig. 2.2 to determine the specific latent heat of vaporisation of water. specific latent heat = .................................................. J g–1 [3] [Total: 8]

Mark scheme: 2(a) (thermal) energy per (unit) mass (to cause change of state) B1 (energy required to cause/released in) change of state at constant temperature B1 2(b)(i) 1. (work done on/against) the atmosphere B1 2. water as it turns from liquid to vapour M1 as potential energy of molecules increases A1 or surroundings as its temperature rises (M1) as energy is lost/transferred to surroundings (A1) 2(b)(ii) VI – h = M / t × L (where h = power loss) or L = (VIt – Q) / M (where Q = energy loss) C1 (14.2 × 6.4) – (11.5 × 5.2) = (9.1 – 5.0) × L / 300 or L = [(14.2 × 6.4) – (11.5 × 5.2)] × 300 / (9.1 – 5.0) C1 L = 2300 J g–1 A1

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3 (a) (i) Define the radian. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) State, by reference to simple harmonic motion, what is meant by angular frequency. ........................................................................................................................................... .......................................................................................................................................[1] (b) A thin metal strip, clamped horizontally at one end, has a load of mass M attached to its free end, as shown in Fig. 3.1. clamp L x oscillation of load metal strip load mass M Fig. 3.1 The metal strip bends, as shown in Fig. 3.1. When the free end of the strip is displaced vertically and then released, the mass oscillates in a vertical plane. Theory predicts that the variation of the acceleration a of the oscillating load with the displacement x from its equilibrium position is given by c a = – 3 x c ML m where L is the effective length of the metal strip and c is a positive constant. (i) Explain how the expression shows that the load is undergoing simple harmonic motion. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) For a metal strip of length L = 65 cm and a load of mass M = 240 g, the frequency of oscillation is 3.2 Hz. Calculate the constant c. c = ........................................... kg m3 s–2 [3] [Total: 8]

Mark scheme: 3(a)(i) angle (subtended) where arc (length) is equal to radius M1 (angle subtended) at the centre of a circle A1 3(a)(ii) angular frequency = 2π × frequency or 2π / period B1 3(b)(i) c / ML3 is a constant so acceleration is proportional to displacement B1 minus sign shows that acceleration and displacement are in opposite directions B1 3(b)(ii) c / ML3 = (2πf )2 C1 c = 4π2 × 3.22 × 0.24 × 0.653 C1 = 27 kg m3 s–2 A1

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4 (a) Explain the principles behind the generation of ultrasound waves for diagnosis in medicine. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[5] (b) Ultrasound frequencies as high as 10 MHz are used in medical diagnosis. Suggest one advantage of the use of high-frequency ultrasound rather than lower-frequency ultrasound. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[1] [Total: 6]

Mark scheme: 4(a) quartz/piezo-electric and crystal/transducer B1 p.d. across crystal causes it to distort B1 applying alternating p.d. causes oscillations/vibrations B1 when applied frequency is natural frequency, crystal resonates B1 natural frequency of crystal is in ultrasound range B1 4(b) small(er) structures can be resolved/observed/identified B1

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5 The analogue signal from a microphone is to be transmitted in digital form. The variation with time t of part of the signal from the microphone is shown in Fig. 5.1. 16 14 microphone output / mV 12 10 8 6 4 2 0 0 0.2 0.4 0.6 0.8 1.0 1.2 t / ms Fig. 5.1 The microphone output is sampled at a frequency of 5.0 kHz by an analogue-to-digital converter (ADC). The output from the ADC is a series of 4-bit numbers. The smallest bit represents 1.0 mV. The first sample is taken at time t = 0. (a) Use Fig. 5.1 to complete Fig. 5.2. time t / ms microphone output / mV ADC output 0.2 ................................................... ............................ 0.8 ................................................... ............................ Fig. 5.2 [2] (b) After transmission of the digital signal, it is converted back to an analogue signal using a digital-to-analogue converter (DAC). Using data from Fig. 5.1, draw, on the axes of Fig. 5.3, the output level from the DAC for the transmitted signal from time t = 0 to time t = 1.2 ms. 16 14 output level 12 10 8 6 4 2 0 0 0.2 0.4 0.6 0.8 1.0 1.2 t / ms Fig. 5.3 [4] (c) It is usual in modern telecommunication systems for the ADC and the DAC to have more than four bits in each sample. State and explain the effect on the transmitted analogue signal of such an increase. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] [Total: 8]

Mark scheme: 5(a) (0.2 ms) 8.0 (mV) 1000 B1 (0.8 ms) 5.8 (mV) 0101 B1 5(b) series of steps B1 all (step) changes are at 0.2 ms intervals B1 steps with correct levels at correct times (1 mark if five levels correct; 2 marks if all levels correct) level 0 8 10 15 5 8 time / ms 0–0.2 0.2–0.4 0.4–0.6 0.6–0.8 0.8–1.0 1.0–1.2 B2 5(c) smaller step heights (possible) B1 smaller changes (in input signal) can be seen/reproduced/represented or (allows) more accurate reproduction (of the input signal) B1

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6 (a) For any point outside a spherical conductor, the charge on the sphere may be considered to act as a point charge at its centre. By reference to electric field lines, explain this. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) An isolated spherical conductor has charge q, as shown in Fig. 6.1. x sphere, charge q P Fig. 6.1 Point P is a movable point that, at any one time, is a distance x from the centre of the sphere. The variation with distance x of the electric potential V at point P due to the charge on the sphere is shown in Fig. 6.2. 14 12 V / 103 V 10 8 6 4 2 0 0 2 4 6 8 10 12 x / cm Fig. 6.2 Use Fig. 6.2 to determine (i) the electric field strength E at point P where x = 6.0 cm, E = ................................................ N C–1 [3] (ii) the radius R of the sphere. Explain your answer. R = .................................................... cm [2] [Total: 7]

Mark scheme: 6(a) electric field lines are radial/normal to surface (of sphere) B1 electric field lines appear to originate from centre (of sphere) B1 6(b)(i) tangent drawn at x = 6.0 cm and gradient calculation attempted C1 E = 9.0 × 104 N C–1 (1 mark if in range ±1.2; 2 marks if in range ±0.6) A2 or correct pair of values of V and x read from curved part of graph and substituted into V = q / 4πε0x (C1) to give q = 3.6 × 10–8 C (C1) (then E = q / 4πε0x2 and x = 6 cm gives) E = 9.0 × 104 N C–1 (A1) or (E = q / 4πε0x2 and V = q / 4πε0x and so) E = V / x (C1) giving E = 5.4 × 103 / 0.060 (C1) = 9.0 × 104 N C–1 (A1) 6(b)(ii) (R =) 2.5 cm B1 potential inside a conductor is constant or field strength inside a conductor zero (so gradient is zero) B1

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7 (a) Feedback is used frequently in amplifier circuits. State (i) what is meant by feedback, ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) two benefits of negative feedback in an amplifier circuit. 1. ....................................................................................................................................... ........................................................................................................................................... 2. ....................................................................................................................................... ........................................................................................................................................... [2] (b) An amplifier circuit incorporating an ideal operational amplifier (op-amp) is used to amplify the output of a microphone. The circuit is shown in Fig. 7.1. +5.0 V P + – 92.5 kΩ –5.0 V V OUT R Fig. 7.1 When the potential at point P is 48 mV, the output potential difference VOUT is 3.6 V. (i) Determine 1. the gain of the amplifier circuit, gain = .......................................................... [2] 2. the resistance of resistor R. resistance = ...................................................... Ω [2] (ii) State and explain the effect on the amplifier output when the potential at P exceeds 68 mV. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] [Total: 10]

Mark scheme: 7(a)(i) (part of) the output is combined with the input M1 reference to potential/voltage/signal A1 7(a)(ii) • increased (operating) stability • increased bandwidth/range of frequencies over which gain is constant • less distortion (of output) Any 2 points. B2 7(b)(i) 1. gain = 3.6 / (48 × 10–3) C1 = 75 A1 2. gain = 1 + RF / R 75 = 1 + (92.5 × 103) / R C1 R = 1300 Ω A1 7(b)(ii) for 68 mV, gain × VIN = 5.1 (V) or output voltage would be greater than the supply voltage M1 amplifier would saturate (at 5.0 V) or output voltage = 5.0 (V) A1

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8 A thin slice of conducting material is placed normal to a uniform magnetic field, as shown in Fig. 8.1. magnetic field F E S R C D P Q current I Fig. 8.1 The magnetic field is normal to face CDEF and to face PQRS. The current I in the slice is normal to the faces CDQP and FERS. A potential difference, the Hall voltage VH, is developed across the slice. (a) (i) State the faces between which the Hall voltage VH is developed. ................................................................ and .............................................................. [1] (ii) Explain why a constant voltage VH is developed between the faces you have named in (i). ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[4] (b) Two slices have similar dimensions. One slice is made of a metal and the other slice is made of a semiconductor material. For the same values of magnetic flux density and current, state which slice, if either, will give rise to the larger Hall voltage. Explain your reasoning. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] [Total: 7]

Mark scheme: 8(a)(i) DERQ and CFSP B1 8(a)(ii) charge carriers moving normal to (magnetic) field B1 charge carriers experience a force normal to I (and B) B1 charge build-up sets up electric field across the slice or build-up of charges results in a p.d. across the slice B1 charge stops building up/VH becomes constant when FB = FE B1 8(b) VH inversely proportional to n/number density of charge carriers B1 number density of charge carriers (n) lower in semiconductors so VH larger for semiconductor slice B1 or VH proportional to v/drift velocity (B1) (for same current) drift velocity (v) higher in semiconductors so VH larger for semiconductor slice (B1)

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9 (a) State what is meant by a field of force. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) Explain the use of a uniform magnetic field and a uniform electric field for the selection of the velocity of charged particles. You may draw a diagram if you wish. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[4] (c) A beam of charged particles enters a region of uniform magnetic and electric fields, as illustrated in Fig. 9.1. region of uniform magnetic and electric fields path of particle mass m charge +q velocity v magnetic field into plane of paper Fig. 9.1 The direction of the magnetic field is into the plane of the paper. The velocity of the charged particles is normal to the magnetic field as the particles enter the field. A particle in the beam has mass m, charge +q and velocity v. The particle passes undeviated through the region of the two fields. On Fig. 9.1, sketch the path of a particle that has (i) mass m, charge +2q and velocity v (label this path Q), [1] (ii) mass m, charge +q and velocity slightly larger than v (label this path V). [2] [Total: 9]

Mark scheme: 9(a) region (of space) B1 where an object/particle experiences a force B1 9(b) electric and magnetic fields normal to each other B1 velocity of particle normal to both fields B1 forces (on particle) due to fields are in opposite directions B1 forces are equal for particles with a particular speed/for a selected speed/for speed given by v = E(q) / B(q) B1 9(c)(i) path labelled Q shown undeviated B1 9(c)(ii) reasonable curve in field and no ‘kink’ on entering, labelled V B1 deviated ‘upwards’ B1

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10 (a) A metal surface is illuminated with light of a single wavelength λ. On Fig. 10.1, sketch the variation with λ of the maximum kinetic energy EMAX of the electrons emitted from the surface. On your graph mark, with the symbol λ0, the threshold wavelength. EMAX 0 λ Fig. 10.1 [3] (b) A neutron is moving in a straight line with momentum p. The de Broglie wavelength associated with this neutron is λ. On Fig. 10.2, sketch the variation with momentum p of the de Broglie wavelength λ. λ 0 0 p Fig. 10.2 [2] [Total: 5]

Mark scheme: 10(a) B1 graph line with λ always < λ0 B1 negative gradient with correct concave curvature B1 10(b) curve with negative gradient and correct concave curvature M1 not touching either axis A1

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11 The circuit for a full-wave rectifier using four ideal diodes is shown in Fig. 11.1. X A input Y R B Fig. 11.1 A resistor R is connected across the output AB of the rectifier. (a) On Fig. 11.1, (i) draw a circle around any diodes that conduct when the terminal X of the input is positive with respect to terminal Y, [1] (ii) label the positive (+) and the negative (–) terminals of the output AB. [1] (b) The variation with time t of the potential difference V across the input XY is given by the expression V = 5.6 sin 380t where V is measured in volts and t is measured in seconds. The variation with time t of the rectified potential difference across the resistor R is shown in Fig. 11.2. 6 rectified potential difference / V 4 2 0 t1 t2 t Fig. 11.2 Use the expression for the input potential difference V, or otherwise, to determine (i) the root-mean-square (r.m.s.) potential difference Vr.m.s. of the input, Vr.m.s. = ...................................................... V [1] (ii) the number of times per second that the rectified potential difference at the output reaches a peak value. number = .......................................................... [2] (c) A capacitor is now connected between the terminals AB of the output. The capacitor reduces the variation (the ripple) in the output to 1.6 V. (i) On Fig. 11.2, sketch the variation with time t of the smoothed output voltage for time t = t1 to time t = t2. [4] (ii) Suggest and explain the effect, if any, on the mean power dissipation in resistor R when the capacitor is connected between terminals AB. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] [Total: 11]

Mark scheme: 11(a)(i) circles drawn only around the top left and bottom right diodes B1 11(a)(ii) B shown as (+)ve and A shown as (–)ve B1 11(b)(i) Vr.m.s. (= 5.6 / √2) = 4.0 V A1 11(b)(ii) 380 = 2πf or f = 60.5 Hz C1 number (= 2f ) = 120 A1 11(c)(i) peak values (all) unchanged B1 (all) minima shown at 4.0 V B1 three lines from near peak showing concave curves after leaving dotted line not ‘kinked’ and not cutting the peak reaching candidate’s minimum at the point where the decay meets the next dotted line B1 three lines drawn along the dotted lines showing rise in voltage from minima back to peak values B1 11(c)(ii) mean p.d. is higher or r.m.s. p.d. is higher or capacitor supplies energy to resistor M1 so (mean) power increases A1

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12 The isotope iodine-131 (13153I) is radioactive with a decay constant of 8.6 × 10–2 day–1. β– particles are emitted with a maximum energy of 0.61 MeV. (a) State what is meant by (i) radioactive, ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) decay constant. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (b) Explain why the emitted β– particles have a range of energies. ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (c) A sample of blood contains 1.2 × 10–9 g of iodine-131. Determine, for this sample of blood, (i) the activity of the iodine-131, activity = .................................................... Bq [3] (ii) the time for the activity of the iodine-131 to be reduced to 1/50 of the activity calculated in (i). time = ................................................. days [2] [Total: 11]

Mark scheme: 12(a)(i) nucleus emits particles/EM radiation/ionising radiation B1 emission/release from unstable nucleus or emission from nucleus is random and/or spontaneous B1 12(a)(ii) probability of decay (of a nucleus) or fraction of (number of undecayed) nuclei that will decay M1 per unit time A1 12(b) energy is shared with another particle B1 mention of antineutrino B1 12(c)(i) number = [(1.2 × 10–9) / 131] × 6.02 × 1023 or number = (1.2 × 10–3 × 10–9) / (131 × 1.66 × 10–27) ( = 5.51 × 1012) C1 A = λN C1 = [0.086 / (24 × 3600)] × 5.51 × 1012 = 5.5 × 106 Bq A1 12(c)(ii) 1 / 50 = exp(–0.086t) or 1 / 50 = 0.5n C1 t = 45 days A1

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