2021 JPJC Prelim H2 Physics P4 Revised Mark Scheme (v2)

1 2021/JPJC/Prelim/9749/04 Suggested Answers to 2021 JC2 Preliminary Examination Paper 4 (H2 Physics) No. Solution Remark 1(a)(i) 1 1 length = 1.000 m 16.2 Ω m 28.4 Ω m A B R R     [1] unit and 3 s.f. [1] range (15RA 17) (27RB 33) 1(a)(ii) 0.24 0.22 0.23 0.23 mm3d  [1] - repeat at least twice - 2 d.p. in mm (0.15  d  0.25) 1(a)(iii)     2 23 7 2 (1.000)(16.2)(1.000) 0.23 10 2 6.7 10 m A AA LRL d              [1] - ans (10-7 m) - unit - 2 or 3 s.f. (e.c.f.) 1(b) 50.0 50.0 50.0 cm2 60.0 60.0 60.0 cm2 0.2151 A L x   I [1] L and x, 3 d.p. in m (49cm  L  51cm) [1] I 1 d.p. in mA, or 4 d.p. in A 1(c) x/m I /A 11 /mx  0.600 0.2151 1.67 0.700 0.2079 1.43 0.800 0.2021 1.25 0.850 0.1998 1.18 0.900 0.1975 1.11 0.950 0.1956 1.05 Accept 111 1 1/ and / B A BR x R L R x       [1] - heading, units - min range of x at least 30 cm above 50.0cm mark - 6 sets of data [1] - raw data’s d.p. [1]- processed data correctly calculated and in 3 s.f. (no mark if table only has x and I)

2 2021/JPJC/Prelim/9749/04 W

3 2021/JPJC/Prelim/9749/04 1(d) Plot I against 1/x Gradient = 0.1966 0.2160 1.080 1.690   =0.03180 0.03180 0.03180 28.4 0.90312 0.90 V B E R E E     Alternate method Use (1.080,0.1966) to find intercept 0.1966 (0.03180)(1.080) 0.1966 (0.03180)(1.080) (16.2)(0.500) 1.31 V A E RL E E    Graph: [1] axis, units, scales [1] plotted points [1] best fit, correct trend, correct linearized equation Calculation: [1] gradient substitution with big gradient triangle [1] substitution to find E [1] calculated E 1(e) New equation is AB EE R x R LI Since ABRR , W’s gradient is larger, y-intercept is smaller [1] Penalise no-repeat once for this question. 2(a) l = 17.8 17.8 2  =17.8 cm [1] - repeat, -1 d.p. in cm (16.0l19.0cm) 2(b) H = 28.2 28.4 2  = 28.3 cm [1] - repeat, -1 d.p. in cm (25.0 H35.0cm) 2(c)(i) () (17.8)(28.3 17.8) 13.7 cm 0.137 m bH    ll [1] substitution [1] ans e.c.f. 2(c)(ii) 22 22 ()bH bH Hb    ll ll ll Plot l2 against lH. b is calculated using -b2=vertical intercept b = (-vertical intercept)1/2 [1] graph [1] method to find b

4 2021/JPJC/Prelim/9749/04 For each variable of C,  and t, penalise once for – no repeat – wrong / no unit (for k too) No. Solution Remark 3(a)(ii) 40 40 402 oo o  [1] - no d.p. in degree - repeat - 39 o to 41o 3(a)(iii) 2percentage uncertainty in 100% 5%40 o o    [1] - 2 to 4    - ans in 1 or 2 sf 3(a)(iv) 40sin 0.342 o   [1] - correct ans - 2 s.f. 3(b)(ii) 24.5 24.4 24.5 cm2C  0.2percentage uncertainty in C 100% 0.8%24.5   [1] - 1 d.p. in cm - repeat C [1] 20 cm  C  30 cm [1] - 0.2 to 0