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