2023 JPJC Prelim H2 Phy P2 Solutions

2023/JPJC/Prelim/9749/02 [Turn over Answers to 2023 JC2 Preliminary Examination Paper 2 (H2 Physics) Suggested Solutions: No. Solution Remarks 1(a)(i) 6 4 50 10 0.125 T4.00 10B A φ − − ×= = =× 5 0.01 0.102550 4.00 BA BA φ φ ∆∆∆=+=+ = ∆B = (0.125 T)(0.1025) = 0.01 T (1 s.f.)  B = (0.13 ± 0.01) T [1] value of B B ∆ [1] value of ∆B to 1 s.f. [1] answer 1(a)(ii) The percentage (or fractional) uncertainty of φ is more significant than that of A. [1] 1(b)(i) The perpendicular distance from A to line of action of T is given by d = 1.2 sin 50° Taking moments about A, (36)(0.45 cos 60°) = T (1.2 sin 50°) (36)(0.45cos60 ) 1.2sin50 8.8115 8.8 N (shown) T = = =   OR Anti-clockwise moment = Ty (1.2 cos 60°) + Tx (1.2 sin 60°) = (T cos 20°)(1.2 cos 60°) + (T sin 20°)(1.2 sin 60°) [1] moment equation [1] numerical answer before 8.8 N T d A 50°

2 2023/JPJC/Prelim/9749/02 No. Solution Remarks 1(b)(ii) Let Fx and Fy be the horizontal and vertical components of F respectively. Consider vertical forces: 36 N = T cos 20° + Fy  Fy = 36 − 8.81 cos 20° = 27.72 N Consider horizontal forces: Fx = Tx = T sin 20° = 8.81 sin 20° = 3.01 N 22 2 2 3.01 27.72xyF FF= += + = 27.9 N [1] value of F y [1] value of F x [1] answer 2(a) The phase difference is due to the extra distance x travelled by the wave from the student to the wall and back to him (x = 0.60 m) and also the phase change of π rad due to reflection at the wall. Phase difference 2 2 (0.60) 0.40 4 rad (or 0 or 2 rad) xπφπ λ π π ππ = + = + = [1] value of x [1] substitution [1] answer 2(b)(i) When x1 = 1.10 m, A1 = 1.2 × 10−8 m When x2 = 1.70 m, A2 = ? 821 1 21 12 2 8 1.101.2 10 1.70 0.78 10 m Ax x AAAx x − −  =→= =×   = × Assumption: no energy is absorbed by the wall upon reflection [1] substitution [1] answer [1] assumption 2(b)(ii) Since φ = 4π rad, they are in phase, so Amplitude of resultant wave = (1.2 + 0.78) × 10−8 m = 2.0 × 10−8 m [1] answer 2(c) 22 28 6 1 8 11 1 62 2.0 101.0 10 1.2 10 2.8 10 W m rr r r AA AA Ι ΙΙΙ − − − −−    ×= →= = ×   ×  = × [1] substitution [1] answer 3(a)(i) The electric field strength E at a point is the negative of the potential gradient dV dx− at that point. [1] 3(a)(ii) The equipotential lines are equally spaced with the same potential difference between consecutive lines. [1]

3 2023/JPJC/Prelim/9749/02 [Turn over No. Solution Remarks 3(a)(iii) Since VE d= , we have ( ) 2 600 600 2.5 10 E − −−= × 44.8 10E = × NC−1 [1] for correct substitution [1] for correct answer 3(b) [1] for six uniform field lines between the plates [1] for correct direction of field lines 3(c)(i) Since the gain in electric potential energy of the particle isqV∆ , we have ( ) 16400 500 1.44 10q −×− − = × 191.60 10q −= −× C [1] for correct substitution [1] for correct answer 3(c)(ii) The particle is an electron. [1] No e.c.f. from (c)(i) 3(c)(iii) By conservation of energy, loss in k.e. = gain in e.p.e. ( ) 231 7 2 161 9.11 10 3.50 10 1.44 102 fv−− ×× × × −=×  73.01 10fv ≈× ms−1 [1] for applying conservation of energy [1] for correct answer

4 2023/JPJC/Prelim/9749/02 No. Solution Remarks 4(a)(i) ( )400p.d. across LDR 4. 0 2.0 0.27273 V400 8400 = − − =+ ( )p.d. across LDR 2.0 0.27273 2.0 0.27273 2.0 1.7 V p p p V V V = −− = + = − ≈− [1] for p.d. across LDR [1] for correct answer 4(a)(ii) When a burglar blocks the laser beam, the resistance of the LDR will be high (or increases). Hence the p.d across the LDR and alarm will be high (or increases). This will turn on the alarm. [1] [1] 4(b)(i) There will be no deflection in the galvanometer when the p.d across JY = 1.5 V. Let the balance length (between J and Y) be LJY. Then ( ) ( ) JY JY JY JY 1.6 4.01.00 1.6 0.50 1.61.5 4 .01.00 1.6 0.50 0.4922 0.49 m LV L L =  + =  + = ≈ [1] for correct substitution [1] for correct answer 4(b)(ii) When 1.5 Ω resistor is connected across points A and B , the terminal p.d across cell E is ( )terminal 1.5 1.51.5 0.50 1.125 V V = + = There will be null deflection when JY 1.125 VV′ = ( ) ( ) JY JY JY JY 1.6 4.01.00 1.6 0.50 1.61.125 4. 01.00 1.6 0.50 0.3691 0.37 m LV L L ′ ′ =  + ′ =  + ′ = ≈ [1] for correct terminalV [1] for correct substitution [1] for correct answer 5(a) Gain in kinetic energy = Loss in electric potential energy 21 002 mv eV−= − ( ) ( )( ) 31 2 191 9.11 10 1.60 10 2002 v−−×= × 6 61 8.382 10 8.38 10 m s v − = × ≈× [1] statement [1] correct answer before rounding off

5 2023/JPJC/Prelim/9749/02 [Turn over No. Solution Remarks 5(b)(i) [1] path 5(b)(ii) ( ) ( ) 2 31 6 19 2 3 For circular motion, 9.11 10 8.38 10 1.60 10 2.8 10 1.7 10 T mvBqv r mvB qr B − −− − = = ×× = ×× = × [1] for correct substitution [1] for correct answer 5(c)(i) The component of the velocity parallel to the magnetic field, is unaffected by the magnetic field. The component of the velocity perpendicular to the magnetic field results in a magnetic force acting on the electron perpendicular to its motion. Combining these 2 effects produces a helical path of the electron. [1] for correct effect [1] for correct effect [1] correct path 5(c)(ii) Radius of curvature, ( ) ( ) 31 6 19 3 2 9.11 10 8.38 10 sin30 1.60 10 1.7 10 1.4 10 m mvr Bq ⊥ − −− − = ×× = ×× = ×  [1] for correct substitution [1] for correct answer P1 P2 v V = 200 V magnetic field flux density B

6 2023/JPJC/Prelim/9749/02 No. Solution Remarks 6(a)(i) If all the electron’s energy is absorbed, the electron could reach −5.17 + 3.7 = −1.47 eV level. Since there are no such energy level, the electron can only excite to −1.56 eV or n = 4 level. [1] for transitions to n = 1 [1] for transitions to n = 2 and n = 3 6(a)(ii) [1] for 3 spectral lines to the left and 2 spectral lines to the right [1] for correct spacing for the extreme 2 spectral lines on the right 6(a)(iii) If the electron absorbed the incoming photon, there is no energy levels for it to excite to. As such, there will be no emission lines observed. [1] 6(b)(i) The X-ray tube is evacuated so that the electrons emitted from the filament can move towards the metal target without colliding with air molecules and getting scattered. [1] 6(b)(ii) A photon is a quantum of electromagnetic radiation having a discrete amount of energy E hf= where h is the Planck constant and f the frequency of radiation. [1] 6(b)(iii) From Fig. 6.3, 11 min 0.40 10 mλ −= × , energy / eV n = 5 n = 4 n = 3 n = 1 − 5.17 n = 2 −3.07 −1.98 −1.56 −1.42 n = 2 to n = 1 n = 4 to n = 1 n = 3 to n = 1 n = 4 to n = 2 n = 3 to n = 2 n = 4 to n = 3 increasing frequency

7 2023/JPJC/Prelim/9749/02 [Turn over No. Solution Remarks ( )( ) ( ) min 34 8 11 14 6.63 10 3.00 10 0.40 10 4.97 10 J hcE λ − − − = ×× = × = × [1] correct substitution [1] correct answer 6(b)(iv) ( ) 14 19 14 4.97 10 1.60 10 4.97 10 310.6 310 kV qV V V − −− = × ×= × = ≈ [1] substitution 7(a)(i) The engine takes in air at a rate of 1300 kg s−1 2 in in 2 1 4 1300 1.2 3.04 153.3 150 m s (show