Cambridge A Level Physics 9702 — 2005 May/June Paper 4 · Variant 1

This document consists of 16 printed pages. SP (SLM/AR) S74754/4 © UCLES 2005 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level PHYSICS 9702/04 Paper 4 May/June 2005 1 hour Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen in the spaces provided on the Question Paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. The number of marks is given in brackets [ ] at the end of each question or part question. You may lose marks if you do not show your working or if you do not use appropriate units. Centre Number Candidate Number Name If you have been given a label, look at the details. If any details are incorrect or missing, please fill in your correct details in the space given at the top of this page. Stick your personal label here, if provided. For Examiner’s Use 1 2 3 4 5 6 7 8 Total

2 9702/04/M/J/05 © UCLES 2005 Data speed of light in free space, c = 3.00 × 108 m s–1 permeability of free space,
0 = 4 × 10–7 H m–1 permittivity of free space, 0 = 8.85 × 10–12 F m–1 elementary charge, e = 1.60 × 10–19 C the Planck constant, h = 6.63 × 10–34 J s unified atomic mass constant, u = 1.66 × 10–27 kg rest mass of electron, me = 9.11 × 10–31 kg rest mass of proton, mp = 1.67 × 10–27 kg molar gas constant, R = 8.31 J K–1 mol–1 the Avogadro constant, NA = 6.02 × 1023 mol–1 the Boltzmann constant, k = 1.38 × 10–23 J K–1 gravitational constant, G = 6.67 × 10–11 N m2 kg–2 acceleration of free fall, g = 9.81 m s–2

3 9702/04/M/J/05 © UCLES 2005 [Turn over Formulae uniformly accelerated motion, s = ut +
 at 2 v2 = u2 + 2as work done on/by a gas, W = pV gravitational potential, φ = – simple harmonic motion, a = – ω2x velocity of particle in s.h.m., v = v0 cos ωt v = ± ω √(x0 2 – x2) resistors in series, R = R1 + R2 + . . . resistors in parallel, 1/R = 1/R1 + 1/R2 + . . . electric potential, V = capacitors in series, 1/C = 1/C1 + 1/C2 + . . . capacitors in parallel, C = C1 + C2 + . . . energy of charged capacitor, W =
 QV alternating current/voltage, x = x0 sin ωt hydrostatic pressure, p = ρgh pressure of an ideal gas, p =
 <c2> radioactive decay, x = x0 exp(– λt) decay constant, λ = critical density of matter in the Universe, ρ0 = equation of continuity, Av = constant Bernoulli equation (simplified), p1 +
 ρv2 1 = p2 +
 ρv2 2 Stokes’ law, F = Arv Reynolds’ number, Re = drag force in turbulent flow, F = Br2ρv2 ρvr  3H0 2 8G 0.693 t
 Nm V Q 40r Gm r

4 9702/04/M/J/05 Answer all the questions in the spaces provided. 1 The orbit of the Earth, mass 6.0 × 1024 kg, may be assumed to be a circle of radius 1.5 × 1011m with the Sun at its centre, as illustrated in Fig. 1.1. Fig. 1.1 The time taken for one orbit is 3.2 × 107s. (a) Calculate (i) the magnitude of the angular velocity of the Earth about the Sun, angular velocity = … rad s–1 [2] (ii) the magnitude of the centripetal force acting on the Earth. force = … N [2] Earth, mass 6.0 x 1024 kg Sun 1.5 x 1011 m For Examiner’s Use © UCLES 2005

5 9702/04/M/J/05 [Turn over (b) (i) State the origin of the centripetal force calculated in (a)(ii). … …[1] (ii) Determine the mass of the Sun. mass = … kg [3] For Examiner’s Use © UCLES 2005

6 9702/04/M/J/05 2 (a) State what is meant by an ideal gas. … … …[2] (b) The product of pressure p and volume V of an ideal gas of density ρ at temperature T is given by the expressions and where N is the number of molecules and k is the Boltzmann constant. (i) State the meaning of the symbol <c2>. …[1] (ii) Deduce that the mean kinetic energy EK of the molecules of an ideal gas is given by the expression EK =  kT. [2] (c) In order for an atom to escape completely from the Earth’s gravitational field, it must have a speed of approximately 1.1 × 104m s–1 at the top of the Earth’s atmosphere. (i) Estimate the temperature at the top of the atmosphere such that helium, assumed to be an ideal gas, could escape from the Earth. The mass of a helium atom is 6.6 × 10–27kg. temperature = … K [2] (ii) Suggest why some helium atoms will escape at temperatures below that calculated in (i). … …[1] p =
 ρ<c2> pV = NkT, For Examiner’s Use © UCLES 2005

7 9702/04/M/J/05 [Turn over 3 (a) Define specific latent heat of fusion. … … …[2] (b) A mass of 24 g of ice at –15 °C is taken from a freezer and placed in a beaker containing 200 g of water at 28 °C. Data for ice and for water are given in Fig. 3.1. Fig. 3.1 (i) Calculate the quantity of thermal energy required to convert the ice at –15 °C to water at 0 °C. energy = … J [3] (ii) Assuming that the beaker has negligible mass, calculate the final temperature of the water in the beaker. temperature = … °C [3] For Examiner’s Use © UCLES 2005 specific heat capacity specific latent heat of fusion / J kg–1K–1 / J kg–1 ice 2.1 × 103 3.3 × 105 water 4.2 × 103 –

8 9702/04/M/J/05 4 A tube, closed at one end, has a constant area of cross-section A. Some lead shot is placed in the tube so that the tube floats vertically in a liquid of density ρ, as shown in Fig. 4.1. Fig. 4.1 The total mass of the tube and its contents is M. When the tube is given a small vertical displacement and then released, the vertical acceleration a of the tube is related to its vertical displacement y by the expression a = – y , where g is the acceleration of free fall. (a) Define simple harmonic motion. … … …[2] (b) Show that the tube is performing simple harmonic motion with a frequency f given by f =
. [3] Aρg M 1 2π Aρg M tube, area of cross-section A lead shot liquid, density For Examiner’s Use © UCLES 2005

9 9702/04/M/J/05 [Turn over (c) Fig. 4.2 shows the variation with time t of the vertical displacement y of the tube in another liquid. Fig. 4.2 (i) The tube has an external diameter of 2.4 cm and is floating in a liquid of density 950 kg m–3. Assuming the equation in (b), calculate the mass of the tube and its contents. mass = … kg [3] (ii) State what feature of Fig. 4.2 indicates that the oscillations are damped. … …[1] –3 –2 –1 0 1 2 3 0 0.4 0.2 0.6 0.8 1.0 1.2 1.4 y / cm t / s For Examiner’s Use © UCLES 2005

10 9702/04/M/J/05 5 An isolated conducting sphere of radius r is given a charge +Q. This charge may be assumed to act as a point charge situated at the centre of the sphere, as shown in Fig. 5.1. Fig. 5.1 Fig. 5.2. shows the variation with distance x from the centre of the sphere of the potential V due to the charge +Q. Fig. 5.2 (a) State the relation between electric field and potential. …[1] V 00 r 2r 3r 4r x +Q r For Examiner’s Use © UCLES 2005

11 9702/04/M/J/05 [Turn over (b) Using the relation in (a), on Fig. 5.3 sketch a graph to show the variation with distance x of the electric field E due to the charge +Q. [3] Fig. 5.3 E x 0 0 r 2r 3r 4r For Examiner’s Use © UCLES 2005

12 9702/04/M/J/05 6 An ideal iron-cored transformer is illustrated in Fig. 6.1. Fig. 6.1 (a) Explain why (i) the supply to the primary coil must be alternating current, not direct current, … … …[2] (ii) for constant input power, the output current must decrease if the output voltage increases. … … …[2] core output input primary coil secondary coil For Examiner’s Use © UCLES 2005

13 9702/04/M/J/05 [Turn over (b) Fig. 6.2 shows the variation with time t of the current Ip in the primary coil. There is no current in the secondary coil. Fig. 6.2 Fig. 6.3 Fig. 6.4 (i) Complete Fig. 6.3 to show the variation with time t of the magnetic flux Φ in the core. [1] (ii) Complete Fig. 6.4 to show the variation with time t of the e.m.f. E induced in the secondary coil. [2] (iii) Hence state the phase difference between the current Ip in the primary coil and the e.m.f. E induced in the secondary coil. phase difference = … [1] 0 0 t E 0 0 t 0 0 Ip t For Examiner’s Use © UCLES 2005

14 9702/04/M/J/05 For Examiner’s Use © UCLES 2005 7 The isotope Manganese-56 decays and undergoes β-particle emission to form the stable isotope Iron-56. The half-life for this decay is 2.6 hours. Initially, at time t = 0, a sample of Manganese-56 has a mass of 1.4 µg and there is no Iron-56. (a) Complete Fig. 7.1 to show the variation with time t of the mass of Iron-56 in the sample for time t = 0 to time t = 11 hours. [2] Fig. 7.1 (b) For the sample of Manganese-56, determine (i) the initial number of Manganese-56 atoms in the sample, number = …[2] (ii) the initial activity. activity = … Bq [3] 0 2 4 6 8 10 12 mass of Iron-56 t / hours

15 9702/04/M/J/05 [Turn over (c) Determine the time at which the ratio is equal to 9.0. time = … hours [2] mass of Iron-56 mass of Manganese-56 For Examiner’s Use © UCLES 2005

16 9702/04/M/J/05 8 (a) Define capacitance. … …[1] (b) (i) One use of a capacitor is for the storage of electrical energy. Briefly explain how a capacitor stores energy. … … …[2] (ii) Calculate the change in the energy stored in a capacitor of capacitance 1200 µF when the potential difference across the capacitor changes from 50 V to 15 V. energy change = … J [3] For Examiner’s Use © UCLES 2005 Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the June 2005 question paper 9702 PHYSICS 9702/04 Paper 4 (Core), maximum raw mark 60 This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. This shows the basis on which Examiners were initially instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the Examination. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the Report on the Examination. • CIE will not enter into discussion or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the June 2005 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Grade thresholds for Syllabus 9702 (Physics) in the June 2005 examination. minimum mark required for grade: maximum mark available A B E Component 4 60 41 35 19 The thresholds (minimum marks) for Grades C and D are normally set by dividing the mark range between the B and the E thresholds into three. For example, if the difference between the B and the E threshold is 24 marks, the C threshold is set 8 marks below the B threshold and the D threshold is set another 8 marks down. If dividing the interval by three results in a fraction of a mark, then the threshold is normally rounded down.

June 2005 GCE A LEVEL MARK SCHEME MAXIMUM MARK: 60 SYLLABUS/COMPONENT: 9702/04 PHYSICS Paper 4 (Core)

Page 1 Mark Scheme Syllabus Paper A LEVEL - JUNE 2005 9702 4 © University of Cambridge International Examinations 2005 1 (a) (i) angular speed = 2π/T C1 = 2π/(3.2 × 107) = 1.96 × 10-7 rad s-1 A1 [2] (ii) force = mrω2 or force = mv2/r and v = rω C1 = 6.0 × 1024 × 1.5 × 1011 × (1.96 × 10-7)2 = 3.46 × 1022 N A1 [2] (b) (i) gravitation/gravity/gravitational field (strength) B1 [1] (ii) F = GMm/x2 or GM = r3ω2 C1 3.46 × 1022 = (6.67 × 10-11 × M × 6.0 × 1024)/(1.5 × 1011)2 C1 M = 1.95 × 1030 kg A1 [3] 2 (a) obeys the law pV/T = constant or any two named gas laws M1 at all values of p, V and T A1 [2] or two correct assumptions of kinetic theory of ideal gas (B1) third correct assumption (B1) (b) (i) mean square speed B1 [1] (ii) mean kinetic energy = ½m<c2> M1 ρ = Nm/V and algebra leading to [do not allow if takes N = 1] M1 ½m<c2> = 3/2 kT A0 [2] (c) (i) ½ × 6.6 × 10-27 × (1.1 ×104)2 = 3/2 × 1.38 × 10-23 ×T C1 T = 1.9 × 104 K A1 [2] (ii) Not all atoms have same speed/kinetic energy B1 [1] 3 (a) (thermal) energy/heat required to convert unit mass/1 kg of solid to liquid M1 with no change in temperature/at melting point A1 [2] (b) (i) energy required to warm ice = 24 × 10-3 × 2.1 × 103 × 15 (= 756 J) C1 energy required to melt ice at 0 °C = 24 × 10-3 × 330 × 103 (= 7920 J) C1 total energy = 8700 J A1 [3] (ii) energy lost by warm water = 200 × 10-3 × 4.2 × 103 × (28 - T) C1 200 × 4.2 × (28 - T) = 24 × 4.2 × T + 8676 C1 T = 16 °C A1 [3] [allow 2 marks if ∆T calculated] [allow 2 marks if (24 x 4.2 x T) omitted] [allow 1 mark for 224 x 4.2 x (28 - T) = 8676, T - 19 °C]

Page 2 Mark Scheme Syllabus Paper A LEVEL - JUNE 2005 9702 4 © University of Cambridge International Examinations 2005 4 (a) acceleration proportional to displacement (from a fixed point) M1 or a = - ω2x with a, ω and x explained and directed towards a fixed point A1 [2] or negative sign explained (b) for s.h.m., a = (-)ω2x B1 identifies ω2 as Aρg/M and therefore s.h.m. (may be implied) B1 2πf = ω B1 hence f = π 2 1 √M Apg A0 [3] (c) (i) T = 0.60 s or f = 1.7 Hz C1 0.60 = (2π√M)/√(π × {1.2 × 10-2}2 × 950 × 9.81) C1 M = 0.0384 kg A1 [3] (ii) decreasing peak height/amplitude B1 [1] 5 (a) field strength = potential gradient [- sign not required] B1 [1] [allow E = ∆V/∆x but not E = V/d] (b) No field for x < r B1 for x > r, curve in correct direction, not going to zero B1 discontinuity at x = r (vertical line required) B1 [3] 6 (a) (i) flux/field in core must be changing M1 so that an e.m.f./current is induced in the secondary A1 [2] (ii) power = VI M1 output power is constant so if VS increases, IS decreases A1 [2] (b) (i) same shape and phase as IP graph B1 [1] (ii) same frequency M1 correct phase w.r.t. Fig. 6.3 A1 [2] (iii) ½π rad or 90° B1 [1] 7 (a) curve levelling out (at 1.4 µg) M1 correct shape judged by masses at nT½ A1 [2] [for second mark, values must be marked on y-axis) (b) (i) N0 = (1.4 × 10-6 × 6.02 × 1023)/56 C1 = 1.5 × 1016 A1 [2] (ii) A = λN C1 λ = ln2/(2.6 × 3600) (= 7.4 × 10-5 s-1) C1 A = 1.11 × 1012 Bq A1 [3] (c) 1/10 of original mass of Manganese remains C1 0.10 = exp(-ln2 × t/2.6) t = 8.63 hours A1 [2] [use of 1/9, giving answer 8.24 hrs scores 1 mark]

Page 3 Mark Scheme Syllabus Paper A LEVEL - JUNE 2005 9702 4 © University of Cambridge International Examinations 2005 8 (a) Q/V, with symbols explained [do not allow in terms of units] B1 [1] (b) (i) on a capacitor, there is charge separation/there are + and - charges M1 either to separate charges, work must be done or energy released when charges ‘come together’ A1 [2] (ii) either energy = ½CV2 or energy = ½QV and C = Q/V C1 change = ½ × 1200 × 10-6 (502 - 152) C1 change = 1.4 J (1.37) A1 [3] [allow 2 marks for ½C(∆V)2, giving energy = 0.74 J)