Physics 9702/33 February/March 2025: question paper and mark scheme

Mark scheme

1(a) Value of x in range 0.440 to 0.460 m. 1 Value of I in range 0.0300 to 0.0600 A 1 1(b) Six sets of readings of x and I (different x) with correct trend and without help scores 4 marks, five sets scores 3 marks etc. 4 Correct trend is I increases as x increases. Range: 1 xmin ⩽ 0.200 m and xmax ⩾ 0.800 m Column headings: 1 Each column heading must contain a quantity and a unit where appropriate. The presentation of quantity and unit must conform to accepted scientific convention e.g. 1 / I / A–1. Consistency: 1 All values of x must be given to the nearest mm. Significant figures: 1 All values of 1 / I given to the same s.f. as (or one more than) the s.f. in raw I Calculation: 1 Correct calculation of 1 / I 1(c)(i) Axes: 1 Axes must be labelled with the correct quantities. Scales must be chosen so that the plotted points occupy at least half the graph grid in both the x and y directions. Scale markings are no more than 2 cm (one large square) apart. Sensible scales must be used, no awkward scales (e.g. 3:10 or fractions). Plotting of points: 1 All observations in the table must be plotted on the grid. Diameter of plotted points must be ⩽ half a small square. Points must be plotted to an accuracy of half a small square in both x and y directions. Quality: 1 Trend of points must be negative. All points in the table (at least 5) must be plotted on the grid for this mark to be awarded. It must be possible to draw a straight line that is within ±2.0 cm (to scale) on the x axis (normally x-axis) of all plotted points. 1(c)(ii) Line of best fit: 1 ‘Best fit’ is judged by the balance of all points on the grid (at least 5 points) about the candidate’s line. There must be an even distribution of points either side of the line along the full length. Lines must not be kinked or thicker than half a square. Some candidates may choose to identify an anomalous point. If they identify one point as anomalous (e.g. by circling or labelling) then this point is to be disregarded when judging the line of best fit. There must be at least 5 points left after the anomalous point is disregarded. 1(c)(iii) Gradient: gradient sign on answer line consistent with graph drawn. 1 The hypotenuse of the triangle used should be greater than half the length of the drawn line. Both read-offs must be accurate to half a small square in both the x and y directions. Method of calculation must be correct, not Δx / Δy. y-intercept: 1 Either Intercept read directly from the graph, with read-off at x = 0, accurate to half a small square in y direction. Or Correct read-off from a point on the line is substituted into y = mx + c or an equivalent expression. Read-off accurate to half a small square in both x and y directions. 1(d) a equal to candidate’s gradient value and b equal to candidate’s intercept value. Values must not be written as fractions or 1 to only one significant figure. Units for a and b correct and consistent with readings 1 (e.g. m–1A–1 for a and A–1 for b) 1(e) Correct calculation of P 1

Mark scheme

2(a) Raw L to nearest mm and final value for L in range 11.0 to 13.0 cm, with unit. 1 Value for raw T 0 to the nearest degree 1 2(b)(i) Values for raw s and raw d to the nearest mm, with units. 1 2(b)(ii) Value of raw H2 to the nearest mm and less than H1, with units 1 2(b)(ii) Value of T greater than T 0 1 2(b)(iii) Absolute uncertainty in H1–H2 in range 2 to 4 mm 1 Correct method of calculation to find percentage uncertainty e.g. absolute uncertainty/value from (b)(iii)) × 100. If repeated readings have been taken, then the uncertainty can be half the range if the working is clearly shown, but not zero if values are equal. 2(b)(iv) Correct calculation of ΔL, with unit 1 2(c)(i) Second value of L 1 2(c)(ii) Second values of H1, H2 and T 1 Quality: Second L greater than first L, 1 And both H1–H2 values positive and second H1-H2 greater than first H1–H2 2(d) Two values of k calculated correctly. 1 Values not written as fractions or given to only one significant figure. 2(e) Calculation of percentage difference between candidate’s two k values. Comparison of percentage difference with 30% 1 leading to a consistent conclusion. 2(f)(i) A Two readings are not enough to draw a (valid) conclusion , e.g. reference to a relationship 4 B Difficult to measure s and / or d with reason, e.g. parallax error / difficult to judge centre of nail / bolt C Difficult to see mark inside measuring cylinder when adding water, with reason. D Problem with measurement of temperature explained: E.g. Water temperature may vary with position / room temperature changes affect T–T 0 in 2nd set of values / temperature varies with time, so measurement of H 2 at known temperature is difficult E Pipe above mark heats up: E.g. Conduction or steam may heat parts of pipe above mark F (H1 – H2) has large percentage uncertainty G Difficult to measure L with reason: E.g. because end of pipe is uneven / not square / pipe is curved H Difficult to measure H with reason: e.g. rule / knocks/moves/displaces wooden strip 1 mark for each point up to a maximum of 4. 2(f)(ii) A Take more readings and plot a graph / calculate more k values and compare. 4 B Measure s and / or d before setting up apparatus C Use mark on outside of measuring cylinder D Use a stirrer / (thermostatically controlled) water bath E Record / film / video with thermometer, metre rule and wooden strip in view. F Use shorter measuring cylinder G Use longer L value / shorter s and/or longer d / travelling microscope for (H1–H2) directly H Sand end of pipe I Use pipe-cutter (to ensure square cut) J Clamp rule 1 mark for each point up to a maximum of 4.