DHS 2024 Prelim Phy P4 Soln

Prepared by CKW2024 [For Internal Use Only] [Turn over Answers to 2024 JC2 Preliminary Examination Paper 4 (H2 Physics) Suggested Solutions No. Solution Remark 1(b) x = 15.0 15.0 15.0 cm2 + = Period T = 20.8 20.7 1.04 s2 20 + = [1] - for correct measurements with units - 1 d.p in cm - repeat - X should be within range of 14.0 cm to 16.0 cm [1] - Recorded no of oscillations, n (can be deduced from working) - 1 or 2 d.p in timing - repeat - T in 3 s.f. or 4 s.f. depending on the d.p. of t1 and t2 - t  10.0 s 1(c) x/cm N Time for N oscillation Period T/s T4/s4 t1/s t2/s 11.0 25 24.2 24.1 0.966 0.871 13.0 25 25.1 25.2 1.01 1.04 15.0 20 20.8 20.7 1.04 1.17 17.0 20 21.4 21.6 1.08 1.36 19.0 20 22.1 22.2 1.11 1.52 21.0 20 22.9 22.8 1.14 1.69 [1] - headings and units - 6 sets of data Note: 20T is not accepted as time for 20 oscillations. [1] - d.p. of raw data - t  10.0 s with repeated measurement [1] - s.f. of processed data - correct calculation, allow 1 slip [1] Range of x to be at least 10.0 cm Don’t accept x = 0 cm and x > 26 cm. These will not be considered one set of data. Note: Marks may be deducted for poor presentation: • No table (including border must be drawn) • Table lines not drawn using ruler • Poor presentation of data in table (e.g. Show working in table) 1(d) Refer to attached graph. [1] axes: units, scale (1:1, 1:2 or 1:5 only) [1] plotted points accurate to half of smallest division [1] best fit line MC: For scale, students should use 1:1, 1:2 or 1:5 scale. In most cases, the data obtained would be able to accommodate the use of these scales which are easy to plot to half the smallest division. • Students are allowed to plot their graph in landscape should they find that portrait does not allow for these scales. In general, the use of these

2 scales would also save students some marks for the plotted points mark and the gradient calculations. 1(d) Given T4 = Px + Q Graph of T4 vs x is plotted, where P is the gradient and Q is the y-intercept. ( ) ( )( ) 41 4 1.67 0.89Gradient= 0.0830 20.8 11.4 = 0.0830 s cm Substitute 20.8,1.67 into the equation, 1.67 0.0830 20.8 0.0564 s P Q Q − − =− =+ =− [1] - linearization (statement) [1] - Big triangle - substitution of gradient coordinates accurate to half of smallest division. - P calculated correctly with units [1] Q calculated correctly with units

3 Prepared by CKW2024 [For Internal Use Only] [Turn over

4 No. Solution Remarks 2(a)(i) w = 1.9 +1.9 1.9 mm2 = t = 1.9 +1.9 1.9 mm2 = L0 = 7.5 +7.5 7.5 cm2 = [1] - Repeated w, t & L0 - Recorded w and t to the nearest 0.01 mm with unit - Value of w and t within the range 1.50 to 2.50 mm - Recorded L0 to the nearest mm with unit - Value of L0 within the range 7.0 to 9.0 cm 2(a)(ii) Volume V = (1.910−3)(1.910-3)(27.510−2) = 5.4  10−7 m3 [1] - Correct calculation of V within range of 3.1 x 10-7 m3 to 1.12 x 10-6 m3 - Recorded to correct no. of s.f. and units 2(b)(i) L = 8.8 8.8 8.8 cm2 + = Extension e = 8.8 – 7.5 = 1.3 cm = 0.013 m Force F = 10010−3  9.81 = 0.981 N [1] - Correct calculation of e - Recorded e to correct no. of d.p. - F = 0.981 N (2 or 3 s.f.) - repeat measurement for L 2(b)(ii) m/kg L/m e/m F/N 0.000 0.075 0.000 0.000 0.100 0.088 0.013 0.981 0.200 0.116 0.041 1.96 0.300 0.165 0.090 2.94 0.400 0.212 0.137 3.92 0.500 0.261 0.186 4.91 [1] - headings and units - at least 5 sets of data (excluding 0) Do not accept masses such as 150g, 250g, 350g as no 50g mass were provided. [1] - d.p. of raw data - m in 3 d.p [1] - correct calculation, allow 1 slip - s.f of processed data MC: Candidates tend to forget th at “0” is also considered a data point/set when included in the table. Hence it should be recorded to the appropriate precision. 2(b)(iii) Refer to attached graph. [1] - plotted points accurate to half of smallest division - best fit curve / line - No marks awarded if best fit curve / line does not pass through origin

5 Prepared by CKW2024 [For Internal Use Only] [Turn over 2(b)(iv) When the extended length is 2L0, the extension e is L0 = 0.075 m and force F is 2.7 N. Energy stored = area under the graph 1 (0.981 1.96)(0.028)(0.013)(0.981)22 +=+ (1.96 2.7)(0.034) 2 ++ = 0.127 J [1] correct calculation OR estimate via counting squares MC: Inaccurate read-offs from the graph were a common error. Another common error was taking the area under the graph from L = 0 m to L = 2L0 Some candidates who managed to obtain a curve estimated the area under the graph using the area of a triangle which tends to be an underestimate and hence were penalized. 2(b)(v) Energy stored per unit volume = − 7 0.127 5.42 10 = 2.34 105 J m−3 [1] correct calculation

6 No. Solution Remarks 3(a) DY = 4.5 4.5 4.5 cm2 + = Diameter dY = 0.30 0.30 0.30 mm2 + = [1] - correct measurement for DY. Accept 4.0 cm to 5.5 cm - repeat [1] - correct measurement for dY. Accept 0.22 mm to 0.35 mm - repeat 3(b)(i) There are 13 turns on cardboard tube Y. LY = 13  DY   + 2(5) = 13  4.5   + 2(5) = 194 cm [1] - Estimated Ly using circumference  no of turns [1] - Added length of tail ends and value of Ly within the range 180 to 220 cm recorded to correct s.f. 3(b)(ii) =+Y Y tail13 2L D L  =  + Y Y tail13 2L D L  +  = Y tailY YY 13 2 DLL LL L L  +==Y Y 13 (0.2) 2(0.2) x 100% x 100% 4.4 %194 [1] for correct percentage uncertainty (1 or 2 s.f.) 3(c) R = 15  I = 140.610−3 A [1] - Record R as 15  - Record I to 0.1 x 10-3 or 0.0001 A - Value of I within the range of 0.1350 to 0.1550 A 3(d) R /  I / A IR / V 15 0.1406 2.1 18 0.1329 2.4 22 0.1227 2.7 27 0.1148 3.1 33 0.1128 3.7 [1] - headings and units - 5 sets of data [1] - d.p, units of raw data [1] - correct calculation - s.f of processed data

7 Prepared by CKW2024 [For Internal Use Only] [Turn over 3(e) ( ) ( )( ) Y 3.4 2.2Gradient= 0.085730 16 0.0857 A Substitute 30,3.4 into the equation, 3.4 0.0857 30 0.829 V 0.829 9.67 0.0857 G H H HX G − =− = =+ = = = =  [1] - points plotted correctly - best fit line drawn [1] - value of G calculated correctly (with or without unit) [1] value of XY calculated correctly in 2 or 3 sig. fig 3(f)(i) DZ = 4.5 4.5 4.5 cm2 + = Diameter dZ = 0.20 0.20 0.20 mm2 + = (0.14 to 0.25 mm) = = =Y Z 3 3(194) 146 cm44 LL [1] - correct measurement for DZ - Accept 4.0 cm to 5.0 cm - (repeat) - correct measurement for dZ - Accept 0.15 mm to 0.25 mm - (repeat) - correct calculation for Lz 3(f)(ii) R/ I/A IR/V 15 0.1258 1.9 18 0.1165 2.1 22 0.1079 2.4 27 0.1001 2.7 33 0.0972 3.2 ( ) ( )( ) Z 3.05 2.0Gradient= 0.070031.5 16.5 0.0700 A Substitute 31.5,3.05 into the equation, 3.05 0.0700 31.5 0.845 V 0.845 12.1 0.0700 G H H HX G − =− = =+ = = = =  [1] - headings and units - 5 sets of data - d.p, units of raw data - s.f of processed data [1] value for XZ calculated correctly

8 Y Z

9 Prepared by CKW2024 [For Internal Use Only] [Turn over 3(f)(iii) Difference: The calculated value is 12.1  and the measured value is 21.6 . Reason: Contact resistance of crocodile clip is not accounted for in the calculated value . Since R is proportional to length for a given resistivity and cross - sectional area, measured R is greater than the calculated value OR The coil heats up due to the current passing through it and the resistance of the coil increase s with increase in the temperature of the coil. [1] - Measured value is higher than the calculated val