2025 EJC H2 Prelim P3 MS

©EJC 2025 9749/J2H2PRELIM/2025 EUNOIA JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS 2025 9749 PHYSICS MARK SCHEME Qns Answer Marks 1(a) (incident) wave reflects at end/top of tube (incident) wave and reflected wave superpose waves have same frequency, wavelength and speed B1 B1 B1 1(b)(i) line has maximum value of amplitude at h = 0 and h = 0.60 m only AND line has minimum/zero value of amplitude at h = 0.30 m only modulus cosine graph (Sharp at h = 0.30 m) M1 A1 1(b)(ii) line has maximum value of amplitude at h = 0 and h = 0.60 m only AND line has minimum/zero value of amplitude at h = 0.30 m only AND modulus cosine-squared graph B1 1(c)(i) vertical/along length of tube/along axis of tube B1 1(c)(ii) phase difference = 0 A1 1(c)(iii) phase difference = 180o A1 1(d) v = fλ f = 340 / (2 × 0.60) C1

2 ©EJC 2025 9749/J2H2PRELIM/2025 Qns Answer Marks = 280 Hz A1 1(e)(i) f = 340 / 0.60 = 570 Hz (or 2 x 280 = 560 Hz) A1 1(e)(ii) 0.60 / 4 = 0.15 m A1

3 ©EJC 2025 9749/J2H2PRELIM/2025 Qns Answer Marks 2(a)(i) 2 62 32 (1.1 10 )(120 10 ) 1.1( 10 )2 1.38898... 1.4 (2 s.f.) LLR A r    −− − == =  = =  M1 2(a)(ii) Let potential at Y be 5.0 V RPQ//1.8 Ω = 1.4 1.8 1.4 1.8  + = 0.7875 Ω PD across RPQ 0.7875 5.00.7875 2.5= + = 1.198 V C1 PD across XY = 5.0 – 1.198 2 = 4.4 V C1 A1 2(b)(i) Current in wire A is in magnetic field produced by current in wire B. Wire A experiences a magnetic force towards wire B according to Fleming’s Left Hand Rule. B1 From Newton’s third law, an equal and opposite force acts on wire B. (or apply FLH rule again on wire B) B1 2(b)(ii)1. IA = −3.0 cos (200πt) = −3.0 cos (200π(6.5 x 10-3)) = 1.76 A From graph, IB = 1.2 A C1 For both IA and IB ( ) 0 7 6 2 4 10 (1.2) 4 8 102 0 05 BB d .( . )     − − =  = =  I F per unit length on wire A = BBIAL / L = BBIA = 64 8 10 1 76( . )( . ) − C1 = 68 4 10. − N m-1 A1

4 ©EJC 2025 9749/J2H2PRELIM/2025 2(b)(ii)2. • positions and shape of graph during intervals of non-zero force • positions of intervals of zero force B1 B1 2(b)(ii)3. 2 2 30 (15)2 34 W rms .PR ==  = I A1 force per unit length / N m –1 t / ms

5 ©EJC 2025 9749/J2H2PRELIM/2025 Qns Answer Marks 3 (a)(i) (W =) 2.60 ×105 × (3.80 – 2.30) × 10–3 = 390 J A1 3(a)(ii) no (total) change (in internal energy) gas returns to its original temperature/ state/ no change to pressure and volume B1 B1 3(b) A to B row all correct (1370, – 390, 980) B to C row all correct (0, 550, 550) C to A row: ΔU adds to the other two ΔU values to give zero C to A row: w = 0 and q adds to w to give ΔU value complete correct answer: B1 B1 B1 B1 3(c) when vapourising, greater change