ASRJC 2025 Prelim H2P4 Physics MS

ANDERSON SERANGOON JUNIOR COLLEGE PHYSICS 9749 1 2025 JC2 Prelim Paper 4 Solution & Mark Scheme Q1 Answer Mark (a)(i) There is no zero error. d1 = 0.25 mm, d2 = 0.24 mm +==0.25 0.24 0.25 mm2d Value of d to 0.01 mm. 0.21 mm ≤ d ≤ 0.27 mm, repeated reading Examiner’s Comments: Many students did not take repeated measurements. Some students did not record the measurement to the correct precision and some had their values out of range. 1 (a)(ii)  −= =  2 80 25 4 9 104 1000 .A. m2 - (b)(iv) l = 0.500 m V = 0.505 V l measured to the nearest mm with unit. 0.45 m ≤ l ≤ 0.55 m V measured to the nearest 0.001 V with unit. 0.1 V < V < 2.0 V Examiner’s Comments: Some students did not record the measurements to the correct precision. The setting on the DMM may be wrong for students who recorded the voltmeter reading to 2 d.p. Marks are only awarded if all measurements recorded in b(iv), (c) and (d) are correct. 1 1 (b)(v) I = 53.1 x 10–3 A I measured to the nearest 0.0001 A. Examiner’s Comments: A handful of students did not record the measurements to the correct precision. Some students did not realise the measurements taken ar ein mA. Students are expected to convert from mA to the given unit A and present the answer to the correct precision. It is not acceptable to change given the unit to mA. 1 (c)(iii) l = 0.650 m V = 0.442 V Examiner’s Comments: Any errors in unit and precision penalised in (b)(iv) -

ANDERSON SERANGOON JUNIOR COLLEGE PHYSICS 9749 2 Q1 Answer Mark (d) Successfully collected 6 sets of values for l and V with no intervention or assistance provided. Range of l : Δl ≥ 60 cm Each column heading must contain a quantity and unit. Examiner’s Comments: Any errors in unit and precision of V and l penalised in (b)(iv). Most students who had assistance rendered was due to the connection of the rheostat. Majority of the students did not meet the criteria of a wide range for l. Students are expected to use a wide range for independent variable when no range is stipulated in the question. For a straight line graph, 6 points are needed. Students are remined to record their data using pen and draw the table with clear headings. l / m V / V l 1 / m–1 l V / Vm–1 0.100 0.781 10.0 7.81 0.200 0.699 5.00 3.50 0.350 0.595 2.86 1.70 0.500 0.505 2.00 1.01 0.650 0.442 1.54 0.680 0.800 0.363 1.25 0.454 2 1 1 (e) ll =−VM N Plot a graph of l V against l 1 where gradient = M and y-intercept = –N (alternative: Plot a graph of V against l where gradient = –N and y-intercept = M) − = − 9.10 1.20Gradient = 0.823 11.8 2.2 =+ =+ =− y mx c c c 9.10 0.823(11.8) 0.611 M = 0.823 V N = 0.611 Vm–1 Axes: Sensible scales must be used, no awkward scales (e.g. 3:10). Scales must be chosen so that the plotted points occupy at least half the graph in both x and y directions. Scales must be labelled with the quantity which is being plotted. Plotting of points: All observations must be plotted on the grid. Work to an accuracy of half a small square. 1 1

ANDERSON SERANGOON JUNIOR COLLEGE PHYSICS 9749 3 Q1 Answer Mark Line of best fit There must be a fair scatter of points on either side of the line. (accept: 1 vs 1, 2 vs 1 but not 4 vs 2). and in each half of the line there should be roughly the same number of points above as below. Gradient • The hypotenuse of the Δ must be greater than half the length of the drawn line. • Read-offs must be accurate to half a small square • Check for Δy/Δx (i.e. do not allow Δx/Δy). • No unit for gradient. Rounded to 3 s.f. y-intercept • Correct method used to find y-intercept: E.g: y-intercept read correctly from graph if the x-axis starts from zero. Or sub a point on the straight line into the eqn to find y-intercept. Using a pt on the table to find the y-intercept is acceptable provided the pt lies on the straight line. • Read-offs must be accurate to half a small square if y-intercept is read directly from graph (same as gradient’s coordinate readings) M and N: Recorded to 3 s.f with units. M is equated to gradient. Unit for M is V. N is equated to negative of y-intercept of graph. Unit for N is V m−1. Examiner’s Comments: Most students were able to linearise and plot the correct graph. Students are reminded to use sharp 2B pencil to draw the graph. Some students still use odd scale and had points plotted wrong. Points should be plotted accurately on the graph grid and all points recorded in the table must be plotted. A handful of students did not fulfil the 2nd criteria of the best fit line. Students should adopt the good practice of calculating gradient and y- intercept separately before relating to the constants, Most students did not recorded to final answer of M and N to the correct s.f and unit. 1 1 1 1 (f) I − − − = = =   NA . . . . 8 7 3 0 611 4 9 10 5 6 10 Ω m 53 1 10 Correct substitution of values consistent with units 2.0 x 10−7 Ωm ≤ ρ ≤ 9.9 x 10−7 Ωm Examiner’s Comments: Most students did not gain credit to this part as the answers are usually out of range which may due to POT errors or inaccurate data collected. 1 (g) Since ρ α N, when ρ increases N increases, hence Graph Z will have the same gradient but more negative y-intercept (i.e. below the original graph) (alternative: for graph of V against l, Z will have a greater gradient and same y-intercept) 1

ANDERSON SERANGOON JUNIOR COLLEGE PHYSICS 9749 4 Q1 Answer Mark Examiner’s Comments: Majority of the students did not gain credit as they did not realise is ρ proportional to N. Those who realized failed to keep M constant.

ANDERSON SERANGOON JUNIOR COLLEGE PHYSICS 9749 Turn Over l V / Vm–1 1.00 2.00 3.00 9.00 4.00 8.00 5.00 6.00 7.00 0 2.0 4.0 6.0 8.0 10.0 12.0 x x x x x x (11.8, 9.10) (2.2, 1.20) l -11/m Z

ANDERSON SERANGOON JUNIOR COLLEGE PHYSICS 9749 Turn Over Q2 Answer Mark (b) No. of oscillations = 10 −+== ..k. 26 5 7 0 6 8 m s2 Successfully collected at least 2 sets of (α and t) without assistance or intervention. Correct trend, Δα ≥ 10 º α with unit and recorded to nearest 1º t > 10 s and measured to nearest 0.1 s. Evidence of repeated reading T calculated to correct s.f, with correct unit 1.0 s ≤ T ≤ 2.0 s k calculated correctly to 2 s.f. Examiner’s Comments: Students are reminded that the question required the results to be presented clearly and it is a good practice to tabulate the results as many who recorded their data failed to take care of the units, precision of the raw data. Many also did not take care of the s.f of the calculated data. For oscillations, raw timing, t, for N oscillations must be >10.0s, repeated and recorded. Many students just recorded final period which will lead to no credit for raw data as well as period T. Students are also penalize for not stating N, no of oscillations as T cannot be determined without N. Some students were careless in the calculation of k and did not record k to the correct sf. α / ⁰ t1 / s t2 / s T / s k / m s−2 45 14.8 14.6 1.47 6.5 60 12.7 12.8 1.28 7.0 1 1 1 1 1 (c) It is difficult to determine the centre of the pendulum bob as there is a lack of reference. This will affect the accuracy of L. Or It is difficult to keep the protractor still when measuring angle α as the hands are not steady. This will affect the accuracy of α. Error must relate to this experiment, and the measurement of t and θ . Ex