2023 JPJC Prelim H2 Phy P4 Solutions

2023/JPJC/Prelim/9749/04 [Turn over Answers to 2023 JC2 Preliminary Examination Paper 4 (H2 Physics) Suggested Solutions No. Solution Remark 1(a) 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 [1] - 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 ≥ 20.0 s 1(b) 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 [1] - d.p. of raw data - t ≥ 20.0 s [1] s.f. of processed data [1] correct calculation, allow 1 slip Don’t accept x = 0 cm. This will not be considered one set of data. 1(c) Refer to attached graph. [1] axes: units, scale [1] plotted points accurate to half of smallest division [1] best fit line 1(c) Given T4 = Px + Q Graph of T4 vs x is plotted, where P is the gradient and Q is the y-intercept. [1] - Big triangle - substitution of gradient coordinates - linearisation

2 2023/JPJC/Prelim/9749/04 ( ) ( )( ) 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] P calculated correctly with units [1] Q calculated correctly with units

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4 2023/JPJC/Prelim/9749/04 No. Solution Remarks 2(a)(i) L0 = 7.5 +7.5 7.5 cm2 = [1] correct measurements with unit and d.p 2(a)(ii) Volume V = (1.9×10−3)(1.9×10-3)(2×7.5×10−2) = 5.42×10−7 m3 [1] ans 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] - both e and F calculated correctly - repeat measurement for L - answer for F in 2 or 3 sig. fig 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 - 6 sets of data (award full credit if m = 0.000 kg not included in the table) [1] - d.p. of raw data - m in 3 d.p - s.f of processed data [1] correct calculation, allow 1 slip 2(b)(iii) Refer to attached graph. [1] - plotted points accurate to half of smallest division - best fit curve / line 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 No marks awarded if best fit curve / line does not pass through origin 2(b)(v) Energy stored per unit volume = 7 0.127 5.41 10 −× = 2.35 ×105 J m−3 [1] correct calculation

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6 2023/JPJC/Prelim/9749/04 No. Solution Remarks 3(a) DY = 4.5 4.5 4.5 cm2 + = Diameter d Y = 0.30 0.30 0.30 mm2 + = [1] - correct measurement for DY. Accept 4.0 cm to 5.0 cm - repeat [1] - correct measurement for dY. Accept 0.25 mm to 0.35 mm - repeat 3(b)(i) There are 13 turns on cardboard tube Y. LY = 13 × 2 × π × r = 13 × DY × π = 13 × 4.5 × π = 184 cm [1] sub [1] ans 3(b)(ii) YY 13LD π= YY YY LD LD ∆∆ = YY YY 100% 100%LD LD ∆∆ ×= × Hence Y 0.2percentage uncertainty 1in 4.5 00% 4.4%L == × [1] for correct percentage uncertainty (1 or 2 s.f.) 3(c) R = 15 Ω I = 140.6×10−3 A [1] - correct R - correct I with d.p 3(d) R/Ω 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 - s.f of processed data [1] correct calculation

7 2023/JPJC/Prelim/9749/04 [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 − =− = = + = = = = Ω [1] - points plotted correctly - best fit line drawn [1] - value of G calculated correctly (with or without unit) [1] value of X Y calculated correctly in 2 or 3 sig. fig 3(f)(i) DZ = 4.5 4.5 4.5 cm2 + = Diameter d Z = 0.20 0.20 0.20 mm2 + = (0.15 to 0.25 mm) Y Z 3 3(184) 138 cm44 LL = = = [1] - correct measurement for DZ - Accept 4.0 cm to 5.0 cm - (repeat) - correct measurement for d Z - Accept 0.15 mm to 0.25 mm - (repeat) - correct calculation 3(f)(ii) R/Ω 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 [1] - headings and units - 5 sets of data - d.p, units of raw data - s.f of processed data

8 2023/JPJC/Prelim/9749/04 ( ) ( )( ) 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 − =− = = + = = = = Ω [1] value for X Z calculated correctly

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10 2023/JPJC/Prelim/9749/04 3(f)(iii) Difference: The calculated value is 12.1 Ω and the measured value is 21.6 Ω. Measured value is higher than the calculated value. Reason: Due to the contact resistance of crocodile clip. [1] for difference and reason 3(g)(i) Given 2 kLX d = ⇒ 2Xdk L= First value of k (for wire Y) 32 Y 9.67(0.30 10 ) 1.84k −×= = 4.73×10−7 Ω m Second value of k (for wire Z) 32 Z 12.1 (0.20 10 ) 1.38k −×= = 3.51×10−7 Ω m [1] correct calculation of value of both k. 3(g)(ii) 7 7 (4.73 3.51) 10Percentage difference 100% 35% 3.51 10 − − −×= ×= × The percentage difference of 35% between the two k values are higher than the percentage uncertainties of LY at 4.4%. As such, the result of my experiment does not support the suggested relationship. [1] valid evaluation based on comparing percentage difference with percentage uncertainties of LY 3(h)(i) From the equation B = CnI, magnetic flux density B at each end of the tube depends on n, number of turns of wire per unit length. Since tube Y has greater number of turns of wire per unit length than tube Z, tube Y has greater magnetic flux density at its ends. [1] explanation 3(h)(ii) Fig. 1 [1] Diagram to show - battery connected directly across the coil - compass placed beside on end of the tube