2022 VS Phy Prelims P1&2 Ans

Physics Prelim 2022 Answers 1 2 3 4 5 6 7 8 9 10 B D B C C C A B C B 11 12 13 14 15 16 17 18 19 20 D D C D D A A C A B 21 22 23 24 25 26 27 28 29 30 D B D D A A D D C D 31 32 33 34 35 36 37 38 39 40 B A C A C A A D B C Suggested Solution 1 (a)(i) Density is defined as mass per unit volume. (a)(ii) Mass of a 1 m3 = density x volume cube of limestone = 2.71 x (50)3 g = 340 kg (correct to 2 s.f.) (shown) (b)(i) . (b)(ii) Take y to be the number of identical limestone blocks that can be lifted. Taking moments about the front wheel, Sum of clockwise moments = Sum of anti-clockwise moments (1000) x (1.0) x (10) = (340) x (0.9) x (10) x (y) y = 3.27 blocks Therefore the maximum number of blocks is 3. Suggested Solution 2 (a) Increase in gravitational = mg∆h potential energy of ball = (0.60)(10)(5) = 30 J (b) Kinetic energy at A = Potential energy at B + Kinetic energy at B = 30 J + 8.5 J = 38.5 J Force on rear tyre by ground. Force on front tyre by ground. weight of forklift. force on forklift by limestone

2 © VICTORIA SCHOOL 22/4P/6091/02 (c) The student is incorrect as weight is the only force acting on the basketball after it leaves the student’s hand. And weight is the resultant force that causes the (downwards) acceleration of the basketball rather than an upwards acceleration that explains the motion of the ball upwards. Suggested Solution 3 (a) Pressure at X = (0.76)(13600)(10) = 1.03 x 105 Pa (b)(i) (ii) Pressure of trapped gas= Atmospheric Pressure + Pressure due to 10 cm Hg = 76 cm Hg + 10 cm Hg = (86 x 10-2)(13 600)(10) = 1.17 x 103 Pa Both levels are correctly drawn at 55 cm (c) The pressure at Y is still 76 cm Hg as the additional pressure is due to the reaction force acting downward on mercury column by barometer. Suggested Solution 4 (a) (b) The water particles are in constant random motion and slide over one another. 1. As the liquid is heated, its temperature increases that cause the average speed of the particles to increase. 2. This leads to a higher force per collision and higher rate of collision with the walls of the container. Force exerted per unit area increases causing a rise in pressure. (c) Number of divisions between ice point and steam point = 10 divisions Number of divisions from ice point = 6 divisions Temperature of unknown liquid = 6 10 × 100 °C = 60 °C Or any other relevant mathematical calculations Suggested solution 5 (a) It means there is 2 complete rotations/waves produced per second. (b) f = 2 × 8 = 16 Hz v = f λ 330 = 16 × λ λ = 20.6 m (c) New waveform has half the original amplitude (0.5 d). It will have a period that is doubled. So will show 1.75 waves.

3 © VICTORIA SCHOOL 22/4P/6091/02 Suggested solution 6 (a) Electric field is a region in which an electric charge experiences an electrostatic force. (b) (i) the beam of uncharged particles deflect s upwards and is attracted to the positively charged plate. (ii) the beam of electromagnetic wave continues undeflected in a straight line. (c) There is electrostatic induction and the electrons in the uncharged particles are attracted to the side closer to the positively charged plate. Since the negative side is closer to the plate than the positive side of the uncharged particles, the particles are attracted to the plate. The gamma wave is uncharged and will not be attracted to the plate hence it continues moving along a straight line. Suggested solution 7 (a) Voltmeter is connected in parallel to the fixed resistor P. (b) (i) Rp = 𝑉 𝐼 = 1.2 𝑉 0.12 𝐴 = 10 Ω (b) (ii) R = 𝜌𝐿 𝐴  = 𝐴𝑅 𝐿 = (6.0 × 10−3 × 10−4)×(10) 4.0 × 10−2 = 1.5 × 10-4 Ωm (c) The ammeter reading increases. When the fixed resistor Q is connected in parallel with fixed resistor P, the effective resistance of the circuit decreases. Suggested solution 8 (a) (i) 2 V across 0.9 MΩ 4 V across the voltmeter will be twice the resistance according to potential divider principle, total resistance across points X and Y = 0.9 × 2 = 1.8 MΩ (a) (ii) Resistance across the thermistor, R 1 1.8 = 1 𝑅 + 1 5.2 R = 2.75 MΩ From Fig. 8.1, the temperature of the thermistor = 22.0 °C (b) Disadvantage of using the thermistor is its resistance does not vary linearly with temperature.

4 © VICTORIA SCHOOL 22/4P/6091/02 Suggested solution 9 (a) P = 𝑉2 𝑅 𝑅 = 2302 50 = 1060 Ω (3 s.f.) (b) Energy used in a year = 0.050 kW × 3.6 h × 365 = 65.7 kWh Cost in a year = 65.7 × 0.29 × 3 = $57.16 Suggested solution 10 (a) 1. The direction of the magnetic field in M and N are opposite. 2. The strength of the magnetic field in M and N are different. Magnetic field strength of M is stronger than N. (b) (i) The solenoid is an electromagnet that induces magnetism in the paper clip. The opposite pole of the paper clip is attracted to the magnetic pole of the solenoid. (b) (ii) Since the solenoid is made of steel wire, it retains its magnetism after the switch is opened. So it remains attracted to the solenoid. Suggested solution 11 (a) Hot water particles at higher temperature have higher kinetic energy . It collides with the powder and transfers some of its kinetic energy. The average kinetic energy of powder increases, its temperature rises as it gains heat. (b) (i) Heat transfer away by convection as the air above the hot drink heats up by conduction, it expands and rises due to lower density . Cooler air of higher density would move in to be heated and the process repeats. (b) (ii) Cover the cup. This minimizes heat loss by convection of air as the warm air is trapped within the cup. (c) (i) Specific latent heat of fusion of ice is the amount of thermal energy required to change unit mass of ice from solid state to liquid state, without a change of temperature. (c) (ii) Given the assumption of no other heat loss, Heat loss by milo = Heat gain by ice melting + Heat gain by melted ice (0.4)(4200)(90-Ө) = (0.05)(3.36 x 105) + (0.05)(4200)(Ө-0) Ө = 71.1 OC

5 © VICTORIA SCHOOL 22/4P/6091/02 Suggested solution 12 (a)(i) Visible light has a lower frequency than ultraviolet rays. (a)(ii) Lead to premature aging of skin tissues and even skin cancer. (b)(i) V = wavelength / period = 2.0 / 0.02 = 100 m s-1 (b)(ii) Correct wave drawn. (shifted by half a wavelength to the right) (b)(iii) The speed will remain constant. (c)(i) (ii) (i) Correct construction of 2 rays to locate image of point A Correct drawing of image IA’. (ii) Both rays from AB and OC must converge on A’. Note: Minus 1m if no arrow head drawn for rays. (iii) Image becomes smaller and nearer to the lens. Suggested Solution 13 (a) It means covering equal distance in equal interval of time. (b)(i) Distance travelled = Area bounded by graph and the time axis = 1 2(17+23)(4.0-2.5) = 30.0 m (2 or 3 s.f. accepted) (ii) Acceleration of = gradient of velocity-time graph car = 23−17 4.0−2.5 = 4.00 m s-2 Resultant force = ma = (1500)(4.00) = 6.00 kN (2 or 3 s.f. accepted)

6 © VICTORIA SCHOOL 22/4P/6091/02 (c) Correct graph shape. Correct values for both key positions (d)