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 0.5 cmC - ans in 1 or 2 sf 3(b)(iii) 12 2 22.8 22.7 2(4) 5.69 s ttT N    [1] - 1t , 2t 1 d.p. in s - 1t , 2t > 20s - repeated - T in 3 s.f. (T>2s) 3(c) 70 70 702 oo o  70sin 0.572 o   12 2 20.4 20.3 2(2) 10.2 s ttT N    [1] -  no d.p. in o -  > 40 o - repeat - ans in 1 or 2 sf [1] - 1t , 2t 1 d.p. in s - 1t , 2t > 20s - repeated - T in 3 s.f. - T in (c) > (b)

5 2021/JPJC/Prelim/9749/04 3(d)(i) /o sin(/o) 1t /s 2t /s N T/s 40 0.34 22.8 22.7 4 5.69 70 0.57 20.4 20.3 2 10.2   1 1 0.5 1 sin / 2 5.69 (0.245)(0.34) 0.050 305 s m kg T kC m k k        2 1 0.5 2 sin / 2 10.2 (0.245)(0.57) 0.050 327 s m kg T kC m k k      [1] Both values of k correct with units 3(d)(ii) Percentage difference of k = 327 305 100% 7.2%305   percentage uncertainty in and are 5% a nd 0.8%C Since percentage difference of k is larger than the percentage uncertainty in  and C, the experiment does not support the relationship. [1] - calculate % diff of k - conclude by comparing with % uncertainties of and C. 3(e)(i) C = 0.120 m  = 40o sin(40/2) = 0.34 1t /s 2t /s N T/s 22.5 22.4 9 2.49 24.3 24.2 9 2.69   3 1 0.5 3 sin / 2 2.49 (0.120)(0.34) 0.050 273 s m kg T kC m k k        4 1 0.5 4 sin / 2 2.69 (0.120)(0.34) 0.050 295 s m kg T kC m k k      [1] - table heading with units - all data in correct d.p. -  : 39 o to 41o - C < 13 cm [1] k1 correct calculation and units [1] K2 correct calculation and units 3(e)(ii) Percentage difference of k in (e) = 295 273 100% 8.1%273   Percentage difference of k in (d) and (e) are around 7 and 8%. [1] - compare % diff of k in (d) and (e)

6 2021/JPJC/Prelim/9749/04 Magnitudes of k in (d) are 305 and 327, while those in (e) are 273 and 295. The values of k in (d) are larger than those in (e). (is there is no obvious relationship, comment that “there’s no relationship”) [1] - compare values of k in (d) and (e) 3(f)(i) - Use the longer wire and set up the apparatus according to Fig. 3.2. Use n the number of loops in the rubber band as 3. Follow step (b)(iii) to determine period T. - keep C,  , m constant - Repeat the experiment by increasing more loops of the rubber band to obtain 6 different values of n and T. Calculate 1 n . - Plot a graph of T against 1 n . If a straight line graph through the origin is obtain ed, then the relationship is proven. (Accept lg T vs lg n with gradient = -1 ,but cannot pass through origin) - There is a limit to how many times the rubber band can be looped to increase n as the circumference of the band is not big enough. [1] simple s teps to obtain data [1] - constants - repeat to vary n [1] - plot graph -straight line conclusion [1] limited n 3(f)(ii) To enable more number of loops n, use rubber band with a smaller width (cross sectional area). Accept longer rubber band [1]

7 2021/JPJC/Prelim/9749/04 4 Solution Remarks To investigate the relationship between the transmittance  of light and the wavelength λ of light and thickness t of glass, by determining n and m. Equation : nmkt Diagram Fig. 1 – top view Fig. 2 – side view Diagram: [1] Fig 1: Laser source, diffraction grating, measurements to take Diagram: [1] Fig 2: Glass, laser source, intensity meter glass block clamped vertically intensity meter laser source screen  laser source diffraction grating L2 L1

8 2021/JPJC/Prelim/9749/04 Procedure : a) Set up the apparatus as shown in Fig 1 and Fig. 2 b) Determine the wavelength  by passing the laser light through a diffraction grating with slit separation d as shown in Fig 1. Determine  using 1 2 tan L L  where L1 is distance from central maxima to 1st order maxima, L2 is distance from grating to screen. Determine  using sindn  c) Measure the thickness t of glass block using Vernier calipers d) Record the intensity Io of the incident light using intensity meter Record the intensity I of the transmitted light using intensity meter Calculate o  I I To determine n: () mnkt Independent variable: λ Dependent variable: , Controlled variables: t e) Rep lace the laser with a different wavelength and repeat th e experiment to get 6 different sets of Io, I, , L1, L2, , . f) From () mnkt , get lg = n(lg) + lg(ktm ) Plot a graph of lg against lg. n = gradient Procedure: Measurements, calculations for: [1] . [1] t [1]  To find n: [1] variables for experiments to find n. [1] Instructions on how