Pure Physics Nov 2014 TYS ANSWERS

Nov 2014_Suggested ANSWERS_July 2023 Page 1 of 8 – Nov 2014 Catholic High School | O-Level Physics NOT IN SYLLABUS: 5059 Nov 2014 P1: - Suggested Answers P2: - Paper 1 [40 marks] 1 C 11 B 21 D 31 C 2 B 12 D 22 C 32 B 3 B 13 B 23 D 33 B 4 C 14 C 24 B 34 B 5 A 15 B 25 C 35 D 6 C 16 B 26 A 36 A 7 A 17 A 27 C 37 D 8 B 18 B 28 D 38 D 9 A 19 B 29 A 39 C 10 D 20 A 30 B 40 C *Q. 14: (C) This question required the application of a standard fact to an unfamiliar situation. Some found this difficult. [Note: The pressure of a gas is inversely proportional to its volume. When the tube is held upright: Pair = Patm + PHg. When the tube is held upside down: Pair + PHg = Patm  Pair = Patm - PHg. The reduction in pressure causes the volume of the air to increase, assuming temperature is constant. (B is incorrect: Atmospheric pressure acts equally in all directions.) (D is incorrect: Pressure is exerted equally in all directions.) *Q. 15: (B) This question tested factual knowledge. A significant minority opted for A which incorrectly suggests that a gas has a fixed volume. (A is incorrect.) *Q. 16: (B) [Note: The better insulating materials take a longer time to heat up.] (D is incorrect.) *Q. 20: (A) This question required the application of a standard fact to an unfamiliar situation. Some found this difficult. [Note: Evaporation causes cooling. There is no rise in temperature.] (D is incorrect.) Pair Pair PHg PHg Patm Patm

Nov 2014_Suggested ANSWERS_July 2023 Page 2 of 8 – Nov 2014 *Q. 22: (C) Some found this question challenging. [Note: Determine the angle between the original incident ray and the reflected ray.] (A is incorrect.) *Q. 27: (C) [Note: The current flowing through a solenoid produces a magnetic field, and not an electric field.] (A is incorrect.) *Q. 28 (D) [Note: The two materials must be made of insulators and not conductors. so that they will not lose their excess charges when earthed.] (A is incorrect.) *Q. 32 (B) This question asked about a poorly wired mains plug and the consequence of an error shown in the diagram. Many found this a challenging question. [Note: The live and the neutral wires are both correctly plugged into the mains. The metal case is wrongly connected to the mains , through the live terminal.] (A and D are incorrect.) appliance metal case L N E fuse switch usual, correct situation before plug appliance metal case L N E fuse switch this question before plug

Nov 2014_Suggested ANSWERS_July 2023 Page 3 of 8 – Nov 2014 Paper 2 [80 marks] 1 ai Elastic potential energy 1 aii Gravitational potential energy 1 bi K.E. = ½ mv2 1900 = ½(45)(v2) v = 9.19 m s-1 (3 s.f.) 1 1 bii Not all the elastic potential energy in the spring is transferred to the gymnast as K .E. Some of it is lost as thermal or sound energy in the stretched spring. [Note: Some of it is also transferred to her as G.P.E., as B is higher than A). 1 2 ai The precision of a ruler is 0.1 cm . It is thus not possible to accurately measure the i nternal diameter of the washer, 0.80 cm , with the correct precision. [Or: It is difficult to place the ruler accurately across the diameter of the washer.] 1 aii Vernier calipers 1 bi Volume = r2h = External volume - Internal volume = [( 2.5 2 )2 − ( 0.80 2 )2](0.15) = 0.661 cm3 (3 s.f.) [Note: There is no need to convert the answer to m3.] 1 1 bii Density = mass volume = 5.2 ÷ 0.661 = 7.87 g cm-3 (3 s.f.) [Note: There is no need to convert the answer to kg m-3.] 1 1 3 ai Mass is the amount of substance in an object. Weight is the amount of gravitatonal force acting on an object. 1 aii W = mg [Note: g = gravitational field strength, N / kg] m = W g = 150 000 ÷ 10 = 15 000 kg (3 s.f.) 1 bi Taking moments about point X: Sum of clockwise moments = sum of anti-clockwise moments (Liftfront)(7.0 + 5.0) = (150 000)(5.0) Liftfront = (150 000) (5.0) 12.0 = 62 500 N (3 s.f.) 1 1 bii Sum of upward forces = sum of downward forces Liftfront + Liftback = 150 000 Liftback = 150 000 - 62 500 = 87 500 N (3 s.f.) [Or: Take moments about C.G. or at liftfront] 1 1 c * Reduce lift force at front rotor and increase lift force at back rotor. [Note: It is insufficient to just decrease the lift force at front rotor; or just increase the lift force at back rotor. This is because the question has stated that the C.G. of the helicopter stays at the same height.] 1

Nov 2014_Suggested ANSWERS_July 2023 Page 4 of 8 – Nov 2014 4 a 1 b 1. The air molecules are in constant, random motion. 2. The air molecules collide unevenly with the smoke particles. 1 1 c When the air molecules move faster, the frequency and force of collisions with the walls of the glass box increases. Thus the force acting per unit area increases, and results in an increase in pressure of the air. 1 1 5 a As a liquid solidifies, the molecules have lower potential energy as intermolecular forces of attraction between the molecules increase. The particles lose the ability to move freely and now vibrate about their fixed positions. 1 b Total energy removed = mc + mlf = (400)(4.2)(20 - 0) + (400)(330) = 33 600 + 132 000 = 166 000 J (3 s.f.) 1 1 1 c The air around the freezing compartment contracts as it cools, becomes denser and sinks. Warmer, less dense air at the bottom will rise to the top, and be cooled by the freezing compartment. This forms convection currents. [Note: It is wrong to state that ‘the cooler, denser air is warmed at the bottom and expands’.] 1 1 1 6 a Critical angle is the angle of incidence in the optically denser medium , for which the angle of refraction in the optically less dense medium is 90°. 1 b 2 ci Within the circular area, the angle of incidence is less than the critical angle and light rays will be refracted into the air. Outside the circular area, the angle of incidence is greater than the critical angle and light rays will undergo total internal reflection , reflecting totally back into the pool. As light travels in all directio ns, the refracted light rays will thus appear to form a circular area. 1 1

Nov 2014_Suggested ANSWERS_July 2023 Page 5 of 8 – Nov 2014 cii tan 49° = r 2 r = 2 × tan 49° = 2.30 m (3 s.f.) 1 1 7 a Light-dependent resistor (LDR) 1 b Rtotal = ( 1 R1 + 1 R2 )-1 = ( 1 60 + 1 30)-1 + ( 1 80 + 1 80)-1 = 20 + 40 = 60 Ω 2 1 ci Current is the rate of flow of electric charges passing a point in the circuit. 1 cii When the light intensity increases, the resistance of the LDR decreases. This reduces the overall resistance of the circui t (I = V R), and hence the current in the battery increases. 1 8 ai Efficiency is the ratio of useful power output to total power input. [Or: - Ratio of useful energy output to total energy input. - Ratio of useful work done to total work done.] 1 aii Filament lamp: Output power = 9 100 × 40 = 3.60 W Fluorescent lamp: Output power = 40 100 × 10 = 4.00 W Fluorescent lamp produces greater useful output power. 1 aiii Energy savings = (No. of kW)(No. of h) = ( 40 1000)(600) - ( 10 1000)(600) = 18 kWh Cost savings = 18  $0.25 = $4.50 1 1 1 bi Infra-red radiation 1 bii According to Fig. 8.2, at point P where most energy is emitted , it is in the infra-red region and is radiated as thermal energy. 1 biii * Using Fig. 8.1: With 96% of the radiation transmitted, the efficiency o f the filament lamp is only 9%. Even if 100% of the radiation is transmitted, the efficiency of the filament lamp is not likely to increase much and is still much lower than the 40% efficiency of the flouroscent lamp. [Note: 96% of radiation transmitted → 9% efficiency 100% of radiation transmitted → 9% × 100 96  9.4%