Physics 9702/51 May/June 2022: question paper and mark scheme

Mark scheme

1 Defining the problem d is the independent variable and V is the dependent variable or vary d and measure V 1 keep A or area (of overlap) of plates constant 1 Methods of data collection labelled diagram of workable experiment including: 1 circuit diagram with voltmeter connected in parallel with the capacitor capacitor and voltmeter connected to the metal plates with no power supply in discharge part of the circuit correct symbols for capacitor and voltmeter method to charge parallel plates, e.g. separate circuit diagram showing plates connected to a d.c. power supply or combined 1 circuit with switches and d.c. power supply use calipers to measure d 1 or use micrometer/calipers to measure thickness of spacers use rule(r) to measure lengths to determine A and A = length × breadth 1 Method of analysis 1 1 1 plot a graph of against d or equivalent (e.g. d against ) V V (Do not accept log graphs.) y -intercept × C C 1 K= or K = gradient × A gradient × AW 1 y -intercept × C (for d against : K = − ) V A 1 1 1 W = y -intercept 1 gradient gradient × C (for d against : W = − or W = ) V y -intercept AK Additional detail including safety considerations 6 D1 use gloves to prevent electric shock or do not touch metal plates to avoid shocks D2 keep the initial p.d. across plates or initial charge constant D3 method to determine the value of C, e.g. description of an experiment to measure p.d. or current against time during discharge through a resistor D4 method of operation of circuit(s) using switch(es) D5 description of method to fully discharge capacitor, e.g. between experiments, short-circuit the capacitor or use of switch in parallel with capacitor D6 repeat measurements of d at different points across plates and average D7 repeat measurements of V for same d and average V D8 bottom plate resting on insulating material or top plate supported by strings D9 use high voltage power supply to increase charge on plates or use a very small value of capacitance to increase voltmeter reading D10 relationship valid if a straight line is produced (not passing through the origin)

Mark scheme

2(a) gradient = n 1 y-intercept = lgSZ 2(b) 1 lg (M / 1030 kg) lg (L / 1028 W) 0.68 or 0.681 ± 0.03 or 0.04 0.15 or 0.146 0.81 or 0.806 ± 0.02 or 0.03 0.49 or 0.491 1.08 or 1.079 ± 0.07 or 0.08 1.51 or 1.505 1.36 or 1.362 ± 0.04 2.54 or 2.544 1.63 or 1.633 ± 0.04 3.56 or 3.556 1.96 or 1.959 ± 0.02 4.82 or 4.820 Values of lg (M / 1030 kg) and lg (L / 1028 W) correct as shown above. Absolute uncertainties in lg (M / 1030 kg) correct as shown above. 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 lg (M / 1030 kg) plotted correctly. 1 All error bars must 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 Points must be balanced. Do not accept line from top point to bottom point. Line must pass between (0.92, 1.0) and (0.96, 1.0) and between (1.86, 4.5) and (1.90, 4.5) 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 Δy / Δx. 1 Distance between data points must be greater than half the length of the drawn line. Gradient of worst acceptable line determined. 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 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 allow methods using a false origin. 2(d) n = gradient = (c)(iii) and n and Z both given to two or three significant figures. 1 Value of Z determined using y-intercept. Correct method must be seen. 1 10y -intercept × 1028 10(c)(iv) × 1028 Z= = S 3.85 × 1026 or Z = 10 y -intercept −lg S × 1028 or 26 Z = 10(c)(iv)−lg 3.85×10 × 1028 Absolute uncertainty in n = absolute uncertainty in gradient 1 and ΔZ = (10 y -intercept ) − 10WAL y -intercept × 1028 S Correct substitution of numbers must be seen. 2(e) L determined from (d) or (c)(iii) and (c)(iv) with correct substitution and correct power of ten(s). 1 Do not accept incorrect POT for n or Z. L = 3.85 × 1026 × (d) × 3.0(c)(iii) or lg L = (c)(iii) × lg3.0 + y -intercept