Cambridge A Level Physics 9702 — 2024 Oct/Nov Paper 5 · Variant 2

1 Fig. 1.1 shows a coil made from resistance wire. Fig. 1.1 The coil is placed in cooking oil of mass m. The total length of the resistance wire in the oil is L. A potential difference V is applied to the coil. The temperature of the oil increases by Δθ in time t. It is suggested that Δθ is related to L by the relationship AtV 2 = mKΔθ + Z L where A is the cross-sectional area of the wire, and K and Z are constants. Plan a laboratory experiment to test the relationship between Δθ and L. Draw a diagram showing the arrangement of your equipment. Explain how the results could be used to determine values for K and Z. 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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Mark scheme: Question Answer Marks 1 Defining the problem L is the independent variable and or temperature change/increase is the dependent variable 1 or vary L and measure or temperature change/increase keep t constant 1 Methods of data collection labelled diagram of workable experiment including: 1 • oil in a beaker/container (on a bench) • coil fully submerged in oil • (bulb of) thermometer in the oil • at least three labels from thermometer, coil or resistance wire, oil, beaker/container, clamp/stand, bench Do not accept other heating sources. method to determine V – diagram of workable circuit including: 1 • power supply connected to wire • voltmeter positioned to measure V across the coil measure the initial and final temperature and find the difference  1 method to determine t, e.g. use stopwatch/timer 1 and method to determine L e.g. use a rule(r) to measure L / length of wire or e.g. using number of turns and measure the diameter of the coil with rule(r) / calipers 1 Method of Analysis 1 1 1 plot a graph of against or equivalent, e.g. against  L L Do not accept logarithms. 1 1 1 for against for against  L L AtV 2 AtV 2  gradient K = K = m  gradient m 1 1 1 for against for against  L L Z = −mK  y -intercept Z = AtV 2  y -intercept or AtV 2  y -intercept Z = − gradient 1 Additional detail including safety considerations 6 D1 precaution linked to hot oil / beaker / wire, e.g. use of gloves to prevent burns from oil or precaution linked to spillage of oil, e.g. perform experiment in a tray D2 keep A and m and V constant d 2 D3 use a micrometer to measure the diameter (d) of the wire and A = 4 D4 repeat measurements of d along the wire and average D5 method to reduce heat loss e.g. add insulation around the container / add a lid to the container D6 method to keep V constant, e.g. adjust / change a variable resistor / power supply to keep V or voltmeter reading constant D7 use a balance to determine the mass of the oil and mass of oil = mass of (beaker + oil) − mass of beaker or place beaker on balance and zero balance, then add oil and read balance D8 stir the oil for uniform temperature or keep the initial temperature (of oil) constant D9 repeat the experiment for the same value of L and average  / average temperature change  Z  D10 relationship valid if a straight line is produced (passing through − )    mK  Do not accept line passing through the origin. D11 method to determine L accurately, e.g. measure length of unwound coil

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2 A student investigates the refraction of white light entering a transparent rectangular block. A narrow beam of light enters the block at the midpoint of one of the shorter sides. The angle of incidence θ is measured, as shown in Fig. 2.1. block beam of light θ d Fig. 2.1 (not to scale) The distance d between the corner of the block and the point where the beam of light touches the boundary of the block is measured. The experiment is repeated for different values of θ. It is suggested that d and θ are related by the equation B sin2 θ = B + d 2 n2 where B and n are constants. 1 (a) A graph is plotted of d 2 on the y-axis against on the x-axis. sin2 θ Determine expressions for the gradient and y-intercept. gradient = ............................................................... y-intercept = ............................................................... [1] 1 (b) Values of θ, and d are given in Table 2.1. sin2 θ Table 2.1 1 θ / ° d / cm d 2 / cm2 sin2 θ 28.5 4.39 24.8 ± 0.2 33.5 3.28 21.4 ± 0.2 42.5 2.19 17.1 ± 0.2 50.0 1.70 14.7 ± 0.2 57.5 1.41 12.9 ± 0.2 63.5 1.25 11.8 ± 0.2 Calculate and record values of d 2 / cm2 in Table 2.1. Include the absolute uncertainties in d 2. [2] 1(c) (i) Plot a graph of d 2 / cm2 against sin2 θ. Include error bars for d 2. [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 = Bn2 1 y-intercept = − B 2(b) 1 d2 / cm2 615 or 615.0 458 or 458.0 292 or 292.4 216 or 216.1 166 or 166.4 139 or 139.2 Values correct as shown above. Uncertainties in d2 decreasing from 10 to 4 or 5. 1 2(c)(i) Six points from (b) plotted correctly. 1 Must be within half a small square. Diameter of points must be less than half a small square. Error bars in d2 plotted correctly. 1 All error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical. 2(c)(ii) Straight line of best fit drawn. 1 Do not accept line from top point to bottom point. Points must be balanced. Line must pass between (1.90, 250) and (2.00, 250) and between (3.55, 500) and (3.65, 500). Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1 All error bars must be plotted. 2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1 Distance between data points must be greater than half the length of the drawn line. Gradient determined of worst acceptable line with clear substitution of data points into y / x. 1 uncertainty = (gradient of line of best fit – gradient of worst acceptable line) or uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) 2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and x into y = mx + c. 1 y-intercept of worst acceptable line determined by substitution into y = mx + c. 1 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. 2(d)(i) B determined using y-intercept (B = – y-intercept) and B and n given to 2 or 3 or 4 significant figures. 1 n determined using gradient 1 and B and n given with SI units with correct powers of ten. gradient gradient n = or n = B − y -intercept Unit for B: cm2 No unit for n. 2(d)(ii) Percentage uncertainty in n determined with method shown. 1 1  y -intercept gradient  percentage uncertainty =  +   100 2  y -intercept gradient  or correct substitution for max/min methods. 2(e) determined to a minimum of two significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution and correct 1 power of ten. −1 gradient  = sin − y -intercept + 900 or −1 n 2 B  = sin B + 900

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