BBSS 4E O Prelim 2024 Physics 6091 P1 MS

Q Answer Key Suggested ExplanaƟon 1 B Displacement is a vector, while charge is a scalar . 2 B Distance travelled = area under graph 3 = ½ (5.0) (Ɵme) Time = 1.20 s 3 D Velocity decreases, hence gradient of displacement Ɵme graph decreases. 4 A For angle between the forces = 0 °, Resultant force = 3.0 + 4.0 N = 7.0 N For angle between the forces = 90 °, resultant force = 5.0 N (Pythagoras theorem). Hence, we expect the resultant force to be between 5.0 N and 7.0 N. 5 C By newton’s 3rd law, force = 4200 N on the propeller due to the air upwards. 6 B Total mass = 450 + 240 = 690 g Total volume = 450/9 + 240/8 = 80 cm3 EffecƟve density = 690 / 80 = 8.63 g/cm3 7 A Taking moments about P , (50)(0.5) = (S)(1.0) S = 25 N 8 C Now, the C.G. is at Y . Later on, when more sand is added, the C.G. goes up verƟcally from Y towards X. 9 C P = W / t = mgh / t = (2.0)(10)(50) / 4.0 = 25 W 10 A Constant volume  same average distance apart Cooled  temperature decreases  average speed decreases 11 B ConvecƟon current cannot reach the bread as heated less dense air will not sink down. 12 B E = P t = I V t = (33)(200)(4.0 x 3600) = 95 MJ Useful energy = 0.90 x 95 MJ = 86 MJ 13 B Pressure at Y = Pressure at X 250/40 = F/10 F = 62.5 N 14 D Factual recall (from X to the vacuum on top, 80 cm). 15 B Factual recall (higher temperature, larger average KE of air molecules, more frequent and more forceful collisions with smoke parƟcles by air molecules). 16 C Factual recall. 17 D HeaƟng at A or cooling at C will result in the downward convecƟon current at point P . 18 D KE = ½ m v2 = ½ (4.0) (1.52) = 4.50 J Total work done = KE converted + work done against fricƟon = 4.5 + (1.0)(1.0) = 5.5 J W = Fd = 5.5 = (F)(1.0) F = 5.5 N 19 A PlasƟc is a poor conductor of heat. Vacuum stops conducƟon and convecƟon. 20 D The white inner surface (wrapped around athlete) reflects IR from athlete back to athlete so that athlete cools down more slowly. [past year O-level paper, 2022] 21 B Internal energy increases as the average kineƟc energy of molecules increases. 22 D 4 wavelengths between P and Q  distance PQ is 8.0 m Speed of wave = d/t = 8.0 / 10 = 0.80 m/s Time to travel 30 m = 30 / 0.80 = 37.5 s 23 B Same frequency due to same source G is moving downwards from its max (peak) posiƟon so its speed is increasing H is going to move upwards 24 C Cannot see cannot see 2 and 3 Can see 1, 5, 4

Q Answer Key Suggested ExplanaƟon 25 A Angle of incidence = 90 – 33 = 57 ° sin (57) / sin (r) = 1.5 r = 34 ° n = 1 / sin C = 1.5  C = 42 ° 26 C Factual recall. 27 D R is infrared radiaƟon. It is used in television remote controller . 28 B Glass rod loses electrons to silk and hence becomes posiƟvely charged. There are more protons than electrons on glass rod. 29 B p.d. across 1 resistor = 9 / 3 = 3.0 V W = V Q = (3) (2) = 6.0 J 30 B For the same p.d. across, R = V / I Half the current means twice the resistance. Length / Area = 8.0 / 25 = 0.32. New length / area = 0.32 x 2 = 0.64 31 C Highest output when R is high and resistance of LDR is low (bright) 32 D PQ: 12/7 =1.71 Ω QR: 33/14 = 2.36 Ω RS: 20/7 = 2.86 Ω SP: 45/14 = 3.21 Ω 33 C 1A flows though the 8 Ω that is in parallel. V = I R = 1.0 x 8 = 8.0 V 34 D Factual recall. Earth wire is now connected to the live terminal, so the metal case is live. 35 B Factual recall 36 A Iron is a soŌ magneƟc material which gains and loses magneƟsm easily. 37 C Apply Fleming’s LeŌ Hand Rule (on posiƟve charges to the leŌ) 38 A p.d. at secondary coil = 4/1 x 2.5 = 10 V lamp Q will experience a 10 V across it. It will be much brighter since the p.d. is higher than rated. 39 B Factual recall. A is not correct as the decay is not a linear process (count-rate versus Ɵme is not linear) 40 C 81 protons, 124 neutrons There are no electrons in the nucleus. Electrons orbit around the nucleus. There are 43 more neutrons than protons in the nucleus.