Cambridge A Level Physics 9702 — 2011 May/June Paper 4 · Variant 2

1 (a) State what is meant by a field of force. .......................................................................................................................................... .................................................................................................................................... [1] (b) Gravitational fields and electric fields are two examples of fields of force. State one similarity and one difference between these two fields of force. similarity: .......................................................................................................................... .......................................................................................................................................... difference: ........................................................................................................................ .......................................................................................................................................... .......................................................................................................................................... [3] (c) Two protons are isolated in space. Their centres are separated by a distance R. Each proton may be considered to be a point mass with point charge. Determine the magnitude of the ratio force between protons due to electric field . force between protons due to gravitational field ratio = ............................................... [3]

Mark scheme: 1 (a) region (of space) where a particle / body experiences a force B1 [1] (b) similarity: e.g. force ∝ 1 / r 2 potential ∝ 1 / r B1 [1] difference: e.g. gravitation force (always) attractive B1 electric force attractive or repulsive B1 [2] (c) either ratio is Q1Q2 / 4πε0m1m2G C1 = (1.6 × 10–19)2 / 4π × 8.85 × 10–12 × (1.67 × 10–27)2 × 6.67 × 10–11 C1 = 1.2 × 1036 A1 [3] or FE = 2.30 × 10–28 × R –2 (C1) FG = 1.86 × 10–64 × R –2 (C1) FE / FG = 1.2 × 1036 (A1)

More questions on Gravitational field

2 (a) State what is meant by a mole. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... .................................................................................................................................... [2] (b) Two containers A and B are joined by a tube of negligible volume, as illustrated in Fig. 2.1. container A container B 3.1 × 103 cm3 4.6 × 103 cm3 17 °C 30 °C Fig. 2.1 The containers are filled with an ideal gas at a pressure of 2.3 × 105 Pa. The gas in container A has volume 3.1 × 103 cm3 and is at a temperature of 17 °C. The gas in container B has volume 4.6 × 103 cm3 and is at a temperature of 30 °C. Calculate the total amount of gas, in mol, in the containers. amount = ........................................ mol [4]

Mark scheme: 2 (a) amount of substance M1 containing same number of particles as in 0.012 kg of carbon-12 A1 [2] (b) pV = nRT C1 amount = (2.3 × 105 × 3.1 × 10–3) / (8.31 × 290) + (2.3 × 105 × 4.6 × 10–3) / (8.31 × 303) C1 = 0.296 + 0.420 C1 = 0.716 mol A1 [4] (give full credit for starting equation pV = NkT and N = nNA)

More questions on Equation of state

3 A capacitor consists of two metal plates separated by an insulator, as shown in Fig. 3.1. For Examiner’s Use metal plate insulator metal plate Fig. 3.1 The potential difference between the plates is V. The variation with V of the magnitude of the charge Q on one plate is shown in Fig. 3.2. 20 15 Q / mC 10 5 00 5 10 15 V / V Fig. 3.2 (a) Explain why the capacitor stores energy but not charge. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .................................................................................................................................... [3] (b) Use Fig. 3.2 to determine For Examiner’s (i) the capacitance of the capacitor, Use capacitance = .......................................... μF [2] (ii) the loss in energy stored in the capacitor when the potential difference V is reduced from 10.0 V to 7.5 V. energy = ......................................... mJ [2] (c) Three capacitors X, Y and Z, each of capacitance 10 μF, are connected as shown in For Fig. 3.3. Examiner’s Use Y X A B Z Fig. 3.3 Initially, the capacitors are uncharged. A potential difference of 12 V is applied between points A and B. Determine the magnitude of the charge on one plate of capacitor X. charge = ......................................... μC [3]

Mark scheme: 3 (a) charges on plates are equal and opposite M1 so no resultant charge A1 energy stored because there is charge separation B1 [3] (b) (i) capacitance = Q / V C1 = (18 × 10–3) / 10 = 1800 µF A1 [2] (ii) use of area under graph or energy = ½CV2 C1 energy = 2.5 × 15.7 × 10–3 or energy = ½ × 1800 × 10–6 × (102 – 7.52) = 39 mJ A1 [2] (c) combined capacitance of Y & Z = 20 µF or total capacitance = 6.67 µF C1 p.d. across capacitor X = 8 V or p.d. across combination = 12 V C1 charge = 10 × 10–6 × 8 or 6.67 × 10–6 × 12 = 80 µC A1 [3] GCE AS/A LEVEL – May/June 2011 9702 42

More questions on Capacitors and capacitance

4 (a) The first law of thermodynamics may be expressed in the form For Examiner’s ΔU = q + w. Use Explain the symbols in this expression. + ΔU ................................................................................................................................. + q .................................................................................................................................... + w ................................................................................................................................... [3] (b) (i) State what is meant by specific latent heat. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [3] (ii) Use the first law of thermodynamics to explain why the specific latent heat of vaporisation is greater than the specific latent heat of fusion for a particular substance. .................................................................................................................................. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [3]

Mark scheme: 4 (a) +∆U: increase in internal energy B1 +q: thermal energy / heat supplied to the system B1 +w: work done on the system B1 [3] (b) (i) (thermal) energy required to change the state of a substance M1 per unit mass A1 without any change of temperature A1 [3] (ii) when evaporating greater change in separation of atoms/molecules M1 greater change in volume M1 identifies each difference correctly with ∆U and w A1 [3]

More questions on Specific heat capacity and specific latent heat

5 A bar magnet is suspended vertically from the free end of a helical spring, as shown in For Fig. 5.1. Examiner’s Use helical spring magnet coil V Fig. 5.1 One pole of the magnet is situated in a coil. The coil is connected in series with a high-resistance voltmeter. The magnet is displaced vertically and then released. The variation with time t of the reading V of the voltmeter is shown in Fig. 5.2. V 0 0 0.5 1.0 1.5 2.0 2.5 t / s Fig. 5.2 (a) (i) State Faraday’s law of electromagnetic induction. .................................................................................................................................. .................................................................................................................................. ............................................................................................................................ [2] (ii) Use Faraday’s law to explain why For Examiner’s 1. there is a reading on the voltmeter, Use .................................................................................................................................. ............................................................................................................................ [1] 2. this reading varies in magnitude, .................................................................................................................................. ............................................................................................................................ [1] 3. the reading has both positive and negative values. .................................................................................................................................. ............................................................................................................................ [1] (b) Use Fig. 5.2 to determine the frequency f0 of the oscillations of the magnet. f0 = .......................................... Hz [2] (c) The magnet is now brought to rest and the voltmeter is replaced by a variable frequency alternating current supply that produces a constant r.m.s. current in the coil. The frequency of the supply is gradually increased from 0.7 f0 to 1.3 f0, where f0 is the frequency calculated in (b). On the axes of Fig. 5.3, sketch a graph to show the variation with frequency f of the amplitude A of the new oscillations of the bar magnet. A 0 0.7 f0 f0 1.3 f0 f [2] Fig. 5.3 (d) (i) Name the phenomenon illustrated on your completed graph of Fig. 5.3. For Examiner’s ............................................................................................................................ [1] Use (ii) State one situation where the phenomenon named in (i) is useful. .................................................................................................................................. ............................................................................................................................ [1]

Mark scheme: 5 (a) (i) (induced) e.m.f. proportional to M1 rate of change of (magnetic) flux (linkage) / rate of flux cutting A1 [2] (ii) 1. moving magnet causes change of flux linkage B1 [1] 2. speed of magnet varies so varying rate of change of flux B1 [1] 3. magnet changes direction of motion (so current changes direction) B1 [1] (b) period = 0.75 s C1 frequency = 1.33 Hz A1 [2] (c) graph: smooth correctly shaped curve with peak at f0 M1 A never zero A1 [2] (d) (i) resonance B1 [1] (ii) e.g. quartz crystal for timing / production of ultrasound A1 [1]

More questions on Electromagnetic induction

6 An alternating current supply is connected in series with a resistor R, as shown in Fig. 6.1. For Examiner’s Use R Fig. 6.1 The variation with time t (measured in seconds) of the current I (measured in amps) in the resistor is given by the expression I = 9.9 sin(380t). (a) For the current in the resistor R, determine (i) the frequency, frequency = .......................................... Hz [2] (ii) the r.m.s. current. r.m.s. current = ............................................ A [2] (b) To prevent over-heating, the mean power dissipated in resistor R must not exceed For 400 W. Examiner’s Calculate the minimum resistance of R. Use resistance = ........................................... Ω [2]

Mark scheme: 6 (a) (i) 2πf = 380 C1 frequency = 60 Hz A1 [2] (ii) IRMS × √ 2 = I0 C1 IRMS = 9.9 / √ 2 = 7.0 A A1 [2] (b) power = I2R C1 R = 400 / 7.02 = 8.2 Ω A1 [2] GCE AS/A LEVEL – May/June 2011 9702 42

More questions on Characteristics of alternating currents

7 (a) State what is meant by the de Broglie wavelength. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... .................................................................................................................................... [2] (b) An electron is accelerated in a vacuum from rest through a potential difference of 850 V. (i) Show that the final momentum of the electron is 1.6 × 10–23 N s. [2] (ii) Calculate the de Broglie wavelength of this electron. wavelength = ........................................... m [2] (c) Describe an experiment to demonstrate the wave nature of electrons. For You may draw a diagram if you wish. Examiner’s Use .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .................................................................................................................................... [5]

Mark scheme: 7 (a) wavelength of wave associated with a particle M1 that is moving A1 [2] (b) (i) energy of electron = 850 × 1.6 × 10–19 M1 = 1.36 × 10–16 J energy = p2 / 2m or p = mv and EK = ½mv 2 momentum = √ (1.36 × 10–16 × 2 × 9.11 × 10–31) M1 = 1.6 × 10–23 N s A0 [2] (ii) λ = h / p C1 wavelength = (6.63 × 10–34) / (1.6 × 10–23) = 4.1 × 10–11 m A1 [2] (c) diagram or description showing: electron beam in a vacuum B1 incident on thin metal target / carbon film B1 fluorescent screen B1 pattern of concentric rings observed M1 pattern similar to diffraction pattern observed with visible light A1 [5]

More questions on Wave-particle duality

8 (a) State what is meant by the binding energy of a nucleus. For Examiner’s .......................................................................................................................................... Use .......................................................................................................................................... ..................................................................................................................................... [2] (b) Show that the energy equivalence of 1.0 u is 930 MeV. [3] (c) Data for the masses of some particles and nuclei are given in Fig. 8.1. mass / u proton 1.0073 neutron 1.0087 deuterium (21H) 2.0141 zirconium (9740Zr) 97.0980 Fig. 8.1 Use data from Fig. 8.1 and information from (b) to determine, in MeV, (i) the binding energy of deuterium, binding energy = ....................................... MeV [2] (ii) the binding energy per nucleon of zirconium. For Examiner’s Use binding energy per nucleon = ....................................... MeV [3]

Mark scheme: 8 (a) energy required to separate nucleons in a nucleus M1 to infinity A1 [2] (b) 1u = 1.66 × 10–27 kg E = mc2 C1 = 1.66 × 10–27 × (3.0 × 108)2 M1 = 1.49 × 10–10 J = (1.49 × 10–10) / (1.6 × 10–13) M1 = 930 MeV A0 [3] (c) (i) ∆m = 2.0141u – (1.0073 + 1.0087)u = –1.9 × 10–3 u C1 binding energy = 1.9 × 10–3 × 930 =1.8 MeV A1 [2] (ii) ∆m = (57 × 1.0087u) + (40 × 1.0073u) – 97.0980u C1 = (–)0.69 u binding energy per nucleon = (0.69 × 930) / 97 C1 = 6.61 MeV A1 [3] GCE AS/A LEVEL – May/June 2011 9702 42 Section B

More questions on Mass defect and nuclear binding energy

9 (a) Describe the structure of a metal wire strain gauge. You may draw a diagram if you wish. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .................................................................................................................................... [3] (b) A strain gauge S is connected into the circuit of Fig. 9.1. +4.5 V RF strain gauge S +9 V R – R + –9 V V1 VOUT 1.0 kΩ V2 RF Fig. 9.1 The operational amplifier (op-amp) is ideal. The output potential VOUT of the circuit is given by the expression RF VOUT = × (V2 – V1). R RF (i) State the name given to the ratio R . ............................................................................................................................ [1] (ii) The strain gauge S has resistance 125 Ω when not under strain. For Examiner’s Calculate the magnitude of V1 such that, when the strain gauge S is not strained, the output VOUT is zero. Use V1 = ........................................... V [3] (iii) In a particular test, the resistance of S increases to 128 Ω. V1 is unchanged. RF The ratio is 12. R Calculate the magnitude of VOUT . VOUT = ........................................... V [2]

Mark scheme: 9 (a) thin / fine metal wire B1 lay-out shown as a grid B1 encased in plastic B1 [3] (b) (i) gain (of amplifier) B1 [1] (ii) for VOUT = 0, then V + = V – or V1 = V2 C1 V1 = (1000/1125) × 4.5 C1 V1 = 4.0 V A1 [3] (iii) V2 = (1000 / 1128) × 4.5 = 3.99 V C1 VOUT = 12 × (3.99 – 4.00) = (–) 0.12 V A1 [2]

More questions on Potential difference and power

10 Explain briefly the main principles of the use of magnetic resonance to obtain diagnostic For information about internal body structures. Examiner’s Use ................................................................................................................................................. ................................................................................................................................................. ................................................................................................................................................. ................................................................................................................................................. ................................................................................................................................................. ................................................................................................................................................. ................................................................................................................................................. ................................................................................................................................................. ................................................................................................................................................. ................................................................................................................................................. ........................................................................................................................................... [8]

Mark scheme: 10 strong / large (uniform) magnetic field B1 nuclei precess / rotate about field direction (1) radio frequency pulse B1 at Larmor frequency (1) causes resonance / nuclei absorb energy B1 on relaxation / de-excitation, nuclei emit r.f. pulse B1 pulse detected and processed (1) non-uniform field superposed on uniform field B1 allows position of resonating nuclei to be determined B1 allows for location of detection to be changed (1) (six points, 1 each plus any two extra – max 8) [8]

More questions on Concept of a magnetic field

11 The use of ionospheric reflection of radio waves for long-distance communication has, to a For great extent, been replaced by satellite communication. Examiner’s Use (a) State and explain two reasons why this change has occurred. 1. ...................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... 2. ...................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... [4] (b) The radio link between a geostationary satellite and Earth may be attenuated by as much as 190 dB. Suggest why, as a result of this attenuation, the uplink and downlink frequencies must be different. .......................................................................................................................................... .......................................................................................................................................... .......................................................................................................................................... .................................................................................................................................... [2]

Mark scheme: 11 (a) e.g. unreliable communication (M1) because ion layers vary in height / density (A1) e.g. cannot carry all information required (M1) bandwidth too narrow (A1) e.g. coverage limited (M1) reception poor in hilly areas (A1) (any two sensible suggestions, M1 & A1 for each, max 4) [4] (b) signal must be amplified (greatly) before transmission back to Earth B1 uplink signal would be swamped by downlink signal B1 [2] GCE AS/A LEVEL – May/June 2011 9702 42

More questions on Electromagnetic spectrum

12 (a) The signal-to-noise ratio in an optic fibre must not fall below 24 dB. The average noise For power in the fibre is 5.6 × 10–19 W. Examiner’s Use (i) Calculate the minimum effective signal power in the optic fibre. power = ........................................... W [3] (ii) The fibre has an attenuation per unit length of 1.9 dB km–1. Calculate the maximum uninterrupted length of fibre for an input signal of power 3.5 mW. length = ......................................... km [3] (b) Suggest why infra-red radiation, rather than ultraviolet radiation, is used for long-distance communication using optic fibres. .......................................................................................................................................... .................................................................................................................................... [1]

Mark scheme: 12 (a) (i) ratio / dB = 10 lg(P1 / P2) C1 24 = 10 lg(P1 / {5.6 × 10–19}) C1 P1 = 1.4 × 10–16 W A1 [3] (ii) attenuation per unit length = 1 / L × 10 lg(P1 / P2) C1 1.9 = 1 / L × 10 lg({3.5 × 10–3}/{1.4 × 10–16}) C1 L = 1 km A1 [3] or attenuation = 10 lg({3.5 × 10–3}/{5.6 × 10–19}) (C1) = 158 dB attenuation along fibre = (158 – 24) (C1) L = (158 – 24) / 1.9 = 71 km (A1) (b) less attenuation (per unit length) / longer uninterrupted length of fibre B1 [1]

More questions on Physical quantities