Cambridge A Level Physics 9702 — 2023 May/June Paper 5 · Variant 2

1 Two coils, P and Q, are placed close to each other, as shown in Fig. 1.1. R coil P coil Q Fig. 1.1 A resistor of resistance R is connected in series with coil P. A changing magnetic flux of frequency f in coil P causes an electromotive force (e.m.f.) E to be induced across the terminals of coil Q. It is suggested that E is related to R by the relationship V E = 2πf M ( R + k) where V is the potential difference across the resistor and coil P, and k and M are constants. Plan a laboratory experiment to test the relationship between E and R. Draw a diagram showing the arrangement of your equipment. Explain how the results could be used to determine values for k and M. In your plan you should include: ● 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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[15]

Mark scheme: 1 Defining the problem R is the independent variable and E is the dependent variable or vary R and measure E 1 keep V constant 1 Methods of data collection labelled diagram of workable experiment including:  coil P placed close to coil Q  separate workable circuit for coil Q  (a.c.) voltmeter or oscilloscope connected across coil Q (Do not accept a power supply connected to coil Q.) 1 a.c. power supply/signal generator connected to resistor and coil P in series 1 workable circuit with power supply and (a.c.) voltmeter/oscilloscope in parallel with resistor and coil P or across terminals of power supply/signal generator 1 method to determine R, e.g. measure current in R and p.d. across R and use R = VR/I or measure R using an ohmmeter 1 Question Answer Marks 1 Method of Analysis plot a graph of 1 E against R or equivalent (e.g. R against 1 E ) Do not accept logarithms. 1 1 2 gradient M fV    (for R against 1 E : gradient 2 M fV   ) 1 2 -intercept k fVM y   or -intercept gradient y k  (for R against 1 E : k =  y-intercept) 1 Question Answer Marks 1 Additional detail including safety considerations 6 D1 precaution linked to hot coil (P) / hot resistor, e.g. use of (heat-proof) gloves, wait until circuit cools down or precaution linked to shocks from high voltages e.g. use of (insulating) gloves or switch off supply before touching the circuit (to change R) D2 keep the number of turns on (both) coils constant D3 keep f constant D4 keep distance between the coils constant D5 method to keep distance between the coils constant, e.g. fix/clamp coils to bench D6 method to measure f, e.g. read from signal generator or use of oscilloscope D7 method to determine f from oscilloscope, e.g. period from oscilloscope T = time-base  horizontal distance and f = 1/T D8 method to determine V or E from oscilloscope, e.g. V = y-gain  vertical distance D9 method to increase E e.g. use iron core/more turns on coil Q/high frequency/high p.d. (across R and coil P) D10 relationship valid if a straight line is produced (not passing through the origin) Do not accept straight line passing through the origin.

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2 A student investigates how the volume of a gas varies with its temperature. Air is trapped in a transparent cylinder of diameter d with a movable piston as shown in Fig. 2.1. d cylinder movable piston trapped air h Fig. 2.1 The distance between the base of the cylinder and the bottom of the piston is h. The trapped air is heated by placing the cylinder in water of temperature θ. The increase in temperature of the trapped air causes the piston to move. When the piston stops moving, the value of h is measured. For each value of h, the volume V of the trapped air is calculated. The experiment is repeated for different values of θ. It is suggested that V and θ are related by the equation pV = Yk (θ + Z ) where k is the Boltzmann constant, p is the atmospheric pressure, and Y and Z are constants. (a) A graph is plotted of V on the y-axis against θ on the x-axis. Determine expressions for the gradient and y-intercept. gradient = ............................................................... y-intercept = ............................................................... [1] (b) Values of θ and h are given in Table 2.1. Table 2.1 θ / °C h / mm V / 10–5 m3 23 62.4 ± 0.1 35 65.2 ± 0.1 48 68.1 ± 0.1 62 70.9 ± 0.1 73 73.3 ± 0.1 88 76.1 ± 0.1 The value of d is (27.9 ± 0.1) mm. The volume V is calculated using the relationship πd 2h V = . 4 Calculate and record values of V / 10–5 m3 in Table 2.1. Include the absolute uncertainties in V. [2] (c) (i) Plot a graph of V / 10–5 m3 against θ / °C. Include error bars for V. [2] (ii) Draw the straight line of best fit and a worst acceptable straight line on your graph. Label both lines. [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 = Yk p y-intercept = YkZ p 2(b) V / 10–5 m3 absolute uncertainty 3.81 or 3.815  0.03 3.99 or 3.986  0.03 4.16 or 4.163  0.04 4.33 or 4.335  0.04 4.48 or 4.481  0.04 4.65 or 4.652  0.04 Values of V correct as shown above. 1 Absolute uncertainties in V correct as shown above. 1 2(c)(i) Six points from (b) plotted correctly. Must be within half a small square. Diameter of points must be less than half a small square. 1 Error bars in V 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 Question Answer Marks 2(c)(ii) Straight line of best fit drawn. Do not accept line from top point to bottom point. Points must be balanced. Line must pass between (27.5, 3.90) and (29.5, 3.90) and between (82.0, 4.60) and (84.0, 4.60). 1 Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). All error bars must be plotted. 1 2(c)(iii) Gradient determined with clear substitution of data points into y / x. Distance between data points must be greater than half the length of the drawn line. 1 Gradient of worst acceptable line determined with clear substitution of data points into y / x. 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 with consistent power of ten in m and y 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) Do not accept ECF from false origin method. 1 Question Answer Marks 2(d)(i) Y determined using gradient and Y and Z given to 2 or 3 significant figures. 27 gradient 7.3188 10 gradient p Y k      1 Z determined using y-intercept and Y and Z given with SI units. -intercept p y Z Yk   or -intercept gradient y Z  Units: Y: no unit Z: °C 1 2(d)(ii) Percentage uncertainty in Y with method shown. gradient percentage uncertainty 100 gradient p p            or Correct substitution for max/min methods. 1 Question Answer Marks 2(e)  determined to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution and correct powers of ten.   2 5 0.0279 0.0600 3.67 10 4 V       and pV Z Yk   or gradient V Z   or -intercept gradient V y    or using h directly: 2 4 p d h Z Yk     or 2 4 gradient d h Z      or 2 -intercept 4 gradient d h y     1

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