JPJC 2024 JC2 H2 Prelim P4_Solution

2024/JPJC/Prelim/9749/04 [Turn over Suggested Solutions No. Solution Remark 1(a) R = 68  I = 121.1 mA y = 41.2 cm [1] - for correct d.p and unit in I and y 1(b) R/ y/cm I/mA R y /  m−1 IR/V 56 42.5 121.1 130 6.8 68 41.2 121.1 170 8.2 82 40.0 121.0 210 9.9 100 38.5 121.0 260 12.1 120 37.7 121.2 318 14.5 [1] - headings and units - 5 sets of data [1] - d.p. of raw data - s.f. of processed data [1] correct calculation, allow 1 slip 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 R Q R QyF=− I Graph of R y vs IR is plotted, where Q F is the gradient and −Q is the y-intercept. ( ) ( )( ) ( ) 305 137.5Gradient= 23.9 14.0 7.0 = 23.9 Substitute 14.0,305 into the equation, 305 23.9 14.0 29.6 Q F Q Q − =−  = + − = Since Q F =23.9  29.6 F = 23.9 F = 1.23 F = 1.23 V Q = 29.6  m−1 [1] - Big triangle - substitution of gradient coordinates [1] Q F calculated correctly [1] Q calculated correctly [1] F calculated correctly [1] Q and F with correct units

2 2024/JPJC/Prelim/9749/04

3 2024/JPJC/Prelim/9749/04 [Turn over 1(d) d1 = 0.19 mm d2 = 0.19 mm Diameter of wire, d = 0.19 0.19 0.19 mm2 + = Given 2 4Q d  = 2 4 Qd = ( ) 2329.6 0.19 10 4  −   = = 8.39  10−7  m [1] - correct measurements with unit and d.p - repeated measurement [1]  calculated correctly

4 2024/JPJC/Prelim/9749/04 No. Solution Remarks 2(a)(i) 1 = 60 2 = 60  = 60 + 60 602  = [1] - correct measurements - repeat measurement - nearest degree 2(a)(ii) n1= 11 n2 = 11 n = 11 + 11 112 = [1] correct measurements [1] repeat measurement 2(b) Given n = k tan k = 11 6.4tan tan60 n  ==  [1] k calculated correctly [1] answer for k in 2 s.f 2(c) / n1 n2 n k = tan n  60 11 11 11 6.4 70 14 14 14 5.1 80 26 26 26 4.6 [1] - headings and units - 3 sets of data - no d.p for raw data - correct s.f for processed data [1] correct calculation 2(d) For  = 90, the two pendulums will make contact with the wooden rod at the same length. Hence both pendulums will always oscillate in phase. [1]

5 2024/JPJC/Prelim/9749/04 [Turn over No. Solution Remarks 3(a)(i) L = 26.2 26.2 26.2 cm2 + = h = 71.8 71.8 71.8 cm2 + = [1] - correct unit and measurement for h and L. - repeat measurement 3(a)(ii) 71.8 2.7426.2 h L == [1] - correct calculation - answer in 3 s.f 3(a)(iii) Let Y = h L Y h L Y h L   =+ Percentage uncertainty in 0.2 0.2 100%71.8 26.2 h L = +  = 1 % [1] for correct percentage uncertainty (1 or 2 s.f.) [1] Accept h & L = 0.2 or 0.3 cm 3(b) m/g h/cm L/cm h L 100 71.8 26.2 2