Cambridge A Level Physics 9702 — 2019 Feb/March Paper 5 · Variant 2

1 A student is investigating the behaviour of a capacitor-resistor circuit as shown in Fig. 1.1. R C input output Fig. 1.1 A neon lamp flashes on and off when it is connected across the capacitor with a potential difference VF across the lamp of approximately 90 V. The student has a number of unmarked resistors. It is suggested that the period T of the flashes of the lamp is related to the resistance R of the resistor by the expression T = RCK where C is the capacitance of the capacitor and K is a constant. The constant K is given by Vi – VL K = ln  Vi – VF where Vi is the potential difference across the input, VF is the potential difference required to make the lamp flash and VL is a constant. Design a laboratory experiment to test the relationship between T and R. Explain how your results could be used to determine a value for K and VL. You should draw a diagram, on page 3, showing the arrangement of your equipment. In your account you should pay particular attention to: • the procedure to be followed • the measurements to be taken • the control of variables • the analysis of the data • any safety precautions to be taken. 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Mark scheme: 1 Defining the problem R is the independent variable and T is the dependent variable, or vary R and measure T 1 keep C constant 1 Methods of data collection labelled diagram or correct symbols of workable circuit including: • (d.c.) power supply correctly positioned • (neon) lamp correctly positioned do not accept ohmmeter in circuit 1 circuit diagram to determine resistance of resistors e.g. using ammeter and voltmeter OR ohmmeter 1 method to determine period or T, e.g. use a stopwatch / timer / oscilloscope do not accept counting the flashes in a specified time 1 circuit diagram showing voltmeter(s) or oscilloscope(s) to determine Vi and VF 1 Method of Analysis plots a graph of T against R 1 gradient K C = 1 ( ) ( ) gradient K C L i i F i i F V V V V e V V V e = − − = − − 1 Question Answer Marks 1 Additional detail including safety considerations Max 6 switch off (high voltage) circuit (before changing the resistor) / wear insulating gloves to prevent electrocution / shock D1 resistance of resistors linked to diagram is V / I for ammeter / voltmeter method or gradient of appropriate graph or resistance from ohmmeter D2 input voltage or Vi is constant D3 repeat experiment for each value of R and average T D4 90 V (or larger) power supply do not accept a.c. or signal generator D5 for stopwatch method: time 10 or more flashes and divide by number of flashes for oscilloscope method: length of wave × timebase D6 record value of capacitance from the capacitor or method to determine capacitance D7 appropriate circuit to enable capacitance to be determined D8 relationship valid if a straight line passing through the origin is produced D9 method to obtain a measurable time period e.g. do a preliminary experiment to choose appropriate resistors, use large values of R or C D10

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2 A student is investigating the motion of a small steel ball in cooking oil. A measure of the oil’s resistance to the ball’s motion is called viscosity. Viscosity has the units pascal second (Pa s). The student drops a ball into a cylinder of oil as shown in Fig. 2.1. ball cooking oil heat Fig. 2.1 The velocity of the ball is measured when it becomes constant and then the viscosity of the oil is determined. The experiment is repeated for different temperatures of oil. It is suggested that the viscosity η and the Celsius temperature θ are related by the equation η = p θ q where p and q are constants. (a) A graph is plotted of lg η on the y-axis against lg θ on the x-axis. Determine expressions for the gradient and the y-intercept. gradient = ............................................................... y-intercept = ............................................................... [1] (b) Values of θ and η are given in Fig. 2.2. θ/ °C η/ 10–3 Pa s lg (θ/ °C) lg (η/ 10–3 Pa s) 38 41 ± 1 46 32 ± 1 55 25 ± 1 64 20 ± 1 72 17 ± 1 79 14 ± 1 Fig. 2.2 Calculate and record values of lg (θ/ °C) and lg (η/ 10–3 Pa s) in Fig. 2.2. Include the absolute uncertainties in lg (η/ 10–3 Pa s). [2] (c) (i) Plot a graph of lg (η/ 10–3 Pa s) against lg (θ/ °C). Include error bars for lg (η/ 10–3 Pa s). [2] (ii) Draw the straight line of best fit and a worst acceptable straight line on your graph. Both lines should be clearly labelled. [2] (iii) Determine the gradient of the line of best fit. Include the absolute uncertainty in your answer. gradient = ......................................................... [2]

Mark scheme: 2(a) gradient = q y-intercept = lg p 1 2(b) 1.58 or 1.580 1.61 or 1.613 1.66 or 1.663 1.51 or 1.505 1.74 or 1.740 1.40 or 1.398 1.81 or 1.806 1.30 or 1.301 1.86 or 1.857 1.23 or 1.230 1.90 or 1.898 1.15 or 1.146 1 absolute uncertainties in lg η: ± 0.01 to ± 0.03 1 2(c)(i) six points plotted correctly must be accurate to the nearest half small square diameter of points must be less than half a small square 1 error bars in lg η plotted correctly all error bars to be plotted total length of bar must be accurate to less than half a small square and symmetrical 1 2(c)(ii) line of best fit drawn points must be balanced do not allow line from top plot to bottom plot if points are plotted correctly then lower end of line should pass between (1.820, 1.275) and (1.835, 1.275) and upper end of line should pass between (1.640, 1.525) and (1.650, 1.525) 1 worst acceptable line drawn steepest or shallowest possible line mark scored only if all error bars are plotted 1 Question Answer Marks 2(c)(iii) gradient determined with clear substitution of data points into ∆y / ∆x; distance between data points must be at least half the length of the drawn line must be negative 1 uncertainty = (gradient of line of best fit – gradient of worst acceptable line) or uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) 1 2(c)(iv) y-intercept determined by substitution of correct point into y = mx + c 1 y-intercept of worst acceptable line determined by substitution into y = mx + c uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line, or uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept) no ECF from false origin method 1 2(d) -intercept 10y p = and given to 2 or 3 sf 1 q = gradient and q and p have correct power of ten from (c)(iii) and (c)(iv) 1 absolute uncertainty in p = intercept of WAL 10y p − − absolute uncertainty in q = uncertainty in gradient correct substitution of numbers must be seen 1 2(e) 100 p q θ = or ( ) lg 100 lg 2 intercept lg gradient p y q θ − − − = = correct substitution of numbers must be seen 1

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