Prelim SAJC (H1) P1 2024 Ans

1 SAJC 2024 Prelim / 8867 [Turn Over] SAJC JC2 H1 Physics Prelims 2024 Paper 1 Solutions Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 C A C D C B D B D C Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 D A C A A D B C B B Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 B B C A B C B C D D 1 Ans: C Wire diameter  0.1 cm (can estimate by using your ruler thus estimate in cm) Area  r2 = (0.05/100)2 = 7.8 x 10-7 m2 2 Ans: A Charge. Current not charge is the SI base quality for electricity quantities. 3 Ans: C Since the plane is moving at constant speed, a = 0 Fnet = 0 Along the path, W sin + R = T Perpendicular to path, W cos = L 4 Ans: D Average value = 48.00 mm is far from the true diameter of 42.03 mm thus it is NOT ACCURATE The set of repeated readings are close to each other thus PRECISE. 5 Ans: C s = ut + ½ at2 s = ½ (9.81)TE 2 --- (1) s = ½ (3.71)TM 2 --- (2) (2) / (1): TM / TE = ( 9.81/3.71)1/2 = 1.63

2 SAJC 2024 Prelim / 8867 [Turn Over] 6 Ans: B acc = gradient of v-t graph Initially, gradient is positive and decrease to zero. Then gradient is negative, much steeper than the positive gradient and then become zero. C, D – incorrect as initial acceleration should be positive A – incorrect as the negative acceleration is not bigger than the positive acceleration 7 Ans D The force the Earth acts on sky-diver is equal to the force the sky-diver acts on Earth as they are action-reaction pair but they will not cancel each other as they are not acting on the same body. 8 Ans: B X: Air resistance increases with velocity till a constant value Y: Fnet = weight – air resistance decreases till zero as weight = constant 9 Ans: D For m, KEf = ¼ KEi ½ m vm2 = ¼ ( ½ mu2) vm = u / 2 By COM, mu = m (- u /2) + 4mv4m 3u / 8 = v4m 10 Ans: C Since p i is not zero as the momenta of the spheres are not equal, p2kg = 4x 2  p3kg = 6 x 3 thus by COM, pf  0 as pf = pi A – Impulse Ft = p is equal and opp when COM can be applied as Fnet = 0 (N3L) B, D – momentum not mass or speed is considered in COM 11 Ans: D Torque of a couple = 15sin 65 (0.4) = 5.4 Nm

3 SAJC 2024 Prelim / 8867 [Turn Over] 12 Ans: A For 0.50 kg, 0.50(9.81) – T = 0.50 (2.0) T = 3.9 For 0.20 kg, T – f = 0.20(2.0) f = 3.9 – 0.4 = 3.5 N 13 Ans: C Horizontally, F = p/t =(200/1000) [ 7 – (-14)]/0.6 = 7.0 N 14 Ans: A Initially, F = kx for each wire 1/3 W = kx (Weight of lamp W is shared equally by the 3 wires) W = 3kx Finally, ½ W = kx’ (W is now shared equally by the 2 wires) ½ (3 kx) = kx’ x ‘ = 3/2 x difference between h = x’ – x = 3/2 x – x = 0.5 x = 0.5 (0.4 cm) = 0.20 cm 15 Ans: A Since the car is travelling at constant speed, friction f =driving force F P = Fv = fv Pt = fvt E = fvt Since fv2 and t = d/v Ev2 E60km/ E = (60)2/(70)2 E60km = 0.73 E 16 Ans: D

4 SAJC 2024 Prelim / 8867 [Turn Over] W = Fd = P + K 4W = 4Fd = 4 (P + K) = 4 P + 4 K 17 Ans: B Forces acting on the ball is weight and tension. Do not draw Fnet which is the centripetal force in a FBD 18 Ans: C A minute hand takes 1 hr to cover 2  = 2 / T = 2 / 3600 = 1.7 x 10 -3 rad s-1 19 Ans: B a = r2 = r (2/T)2 (a/r)1/2 = 2/T T = 2 (r/a)1/2 12T = 24(r/a)1/2 20 Ans: B V = WD / Q 21 Ans: B Px = I2R = 9R 1.5 Px = 13.5 R = I’2R I’ = (13.5)1/2 = 3.7 A 22 Ans: B If current in Y is I, then current in F is 2I, so current in C is 3I Let the resistance of one resistor be R, RAB = (1/1+1/2)-1 R = 2/3 R, so RC + AB + D = 1 + 2/3 + 1 = 8/3 R Since current in RC + AB + D is 3I, then current in E is 8I so current in X = 3I + 8I = 11I

5 SAJC 2024 Prelim / 8867 [Turn Over] 23 Ans: C E = Ir + V V = E – Ir V = 3.0 – 0.50I Y -intercept = 3.0 Gradient = 0.50 = 3.0 / x- intercept x -intercept = 3.0 / 0.50 = 6.0 24 Ans: A Current increases means the overall resistance of the circuit decreases RT decreases temperature increases VT = 12 – 40(0.10) = 8 V VT‘ = 12 – 40(0.12) = 7.2 V R = V/I = 7.2 / 0.12 = 60  or I =  / (RT + 40) 0.12 = 12/ (RT + 40) RT = 60  25 Ans: B When  = 0, I is perpendicular to B thus F = BIL = max As  increases, F = BIL cos A cosine graph starting with max F at  = 0

6 SAJC 2024 Prelim / 8867 [Turn Over] 26 Ans: C When a current flows in the wire from A to B, using RHGR, the B-field due to the current is vertically downwards at position of the compass. Since the B-field due to the current is much larger than B-field due to Earth, a possible resultant B-field is vertically downwards. 27 Ans: B tan = Fp / Wp tan  = FQ / WQ Since >  tan > tan Fp / Wp > FQ / WQ (Since FP = FQ) Wp < WQ mp < mQ 28 Ans: C Energy released in 1 reaction = m c2 = (2.014102 + 3.016049 – 4.002602 – 1.008665) x1.66 x 10-27 x (3 x 108)2 = 2.821 x 10-12 J Energy used per month = 2000 kWh = (2000 x 103 x 3600) J No of reaction = Energy used per month/ Energy released in 1 reaction = 2.6 x 1021 29 Ans: D A random process is defined as a process in which the exact time of decay of a nucleus cannot be predicted. Instead, the nucleus has a constant probability, ie. the same chance, of decaying in a given time. Therefore, with large numbers of nuclei, it is possible to statistically predict the behaviour of the entire group. One cannot know which nucleus will decay and when it will decay because it is down to chance. 30 Ans: D

7 SAJC 2024 Prelim / 8867 [Turn Over] For paper, beta to be used as all alpha would be stopped and all gamma would penetrate For steel, gamma to be used as all alpha and beta would be stopped.