2022 FHSS Phy Prelims Ans

Paper 1 1 2 3 4 5 6 7 8 9 10 C A C B C D A B D D 11 12 13 14 15 16 17 18 19 20 A C B D D B C A A C 21 22 23 24 25 26 27 28 29 30 D D A D A A D B A A 31 32 33 34 35 36 37 38 39 40 D B D B C C B C D B

Section A 1 (a) [1] (a) W = mg = (0.900 kg)(10 N/kg) = 9.00 N (3sf) or 9.0 N (2sf) [1] (b) 1m – appropriate scale (1.0 cm to 2.0 N) 1m – correct vector diagram (right angle triangle, direction of arrows) 1m – correct magnitude of friction force Allow for e.c.f. from (a) [1] [1] [1] (c) Newton’s first law. Forces are balanced. Resultant force of friction, weight and normal contact force = 0 N. (Any one) [1] 2 (a) moment by brake pad = moment by master piston (220)(32) = (F)(8) F = 880 N [1] [1] Weight of box Normal Contact Force Friction

(b) Pmaster piston = Pslave piston F1/A1 = F2/A2 880 / 1.5 = F2/5.0 F2 = 2930 N or 2900 N (allow for e.c.f.) [1] [1] 3 (a) The product of force applied and distance moved in the direction of the force [1] (b) KE = ½ x m x v2 = 0.5 x 0.20 x 202 = 40 J [1] [1] (c) G.P.E at max height = m x g x h = 0.200 x 10 x 12 = 24 J (2sf) [1] [1] (d) At max height, KE remaining = 40 J – 24 J = 16 J (allow for e.c.f.) 0.5 x 0.200 x v2 = 16 v = 12.6 m / s or 13 m / s [1] [1] 4 (a) The focal length is the distance along the principal axis, between the principal focus and the optical centre of the lens. OR Distance between the optical centre of the lens and the principal focus (focal point). [1] (b) Accept either position of principal focus of (iii) Rays and arrowheads should be in solid lines [1] [1] [1] (c)(i) 60.0 cm [1] (ii) 2f = 60.0 cm f = 30.0 cm [1] (iii) 1. Virtual 2. Upright 3. Magnified [1] 5 (a) Negatively charged [1] (b) The negative charges will be transferred from rod Y to sphere B. [1] I

The negative charges are attracted by positively charged rod X as unlike charges attract. [1] Hence, there is more negative charges than positive charges in sphere B upon removal of the rods. (Sphere B has a net negative charge) [1] (c) Sphere A will become positively charged (and B is negatively charged). [1] They will attract each other. [1] Since unlike charges attract. [1] 6 (a)(i) I = V/R = 4.8 / 32 = 0.15 A [1] [1] (ii) Q = I(t) = (0.15)(25) = 3.75 C or 3.8 C (allow for e.c.f.) [1] [1] (iii) Pd across parallel branch = 6.0 – 4.8 V = 1.2 V R = V/I = 1.2 / 0.15 = 8.0 Ω Alternative method: Potential Divider method [32 / (32 + R)] x 6.0 = 4.8 (32 + R) / 32 = 6.0 / 4.8 R = 8.0 Ω [1] [1] (iv) Method 1 (1/R1 + 1/R2 = 1/Rtotal) 1/R + 1/24 = 1/8 R = 12 Ω allow for e.c.f. Method 2 Current flowing through Y = 1.2 / 24 = 0.050 A Current flowing through X = 0.15 – 0.050 = 0.10 A R = V/I = 1.2 / 0.10 = 12 Ω [1] [1] [1] [1] b Overall resistance of circuit decreases, so overall current (I) increases. Voltmeter reading will increase since pd across Z increases (or V = RI). Alternative Answer Overall resistance of parallel branch decreases. Pd across Z will increase since Z will receive a larger proportion of the e.m.f and voltmeter reading increases. (Potential Divider) [1] [1] [1] [1] 7 (a) Either AB – Downward arrow Or CB – Upward arrow [1] (b) Current flows from A to B. By Fleming’s Left Hand rule, the induced force which is perpendicular to the magnetic field and current will be downwards. For side, CD, current flows from C to D and the force will be upwards. The coil rotates in an anticlockwise direction. [1] [1] [1] (c) The coil is horizontal. [1]

That is when the perpendicular distance from the centre of rotation to the line of action of the force is maximum. [1] 8 (a) Iron is easily magnetised and demagnetised (or soft magnetic material) whereas steel does not magnetise or demagnetise easily (or hard magnetic material). This ensures better magnetic flux linkage between the 2 coils if iron is used instead of steel. Any other plausible answer. [1] [1] (b) Reduce energy loss during transmission Since heat loss is P = I2R, the lower the current, the lower the energy loss during transmission. [1] [1] Section B 9 (a) Water is used as a coolant because of its very high specific heat capacity. It can take in a large amount of thermal energy with only a small rise in its temperature. [1] [1] (b)(i) Thermal energy required to be removed as claimed = (0.8 x 5.0 X 107 ) x 80% = 3.2 x 107 J [1] (ii) Actual amount of thermal energy removed Q = mc∆ = (0.22 x 4 x 60) (4200)(80-30) = 1.1088 x 107 J = 1.1 x 107 J [1] (iii) Some thermal energy is lost to the surroundings, apart from it being absorbed by the cooling water. [1] (c) Metal pipes are used as they are good conductors of heat and allows heat to be conducted faster away from the hot water to the external wall of the pipe. The metal pipes being coloured in black are good emitters of radiation and therefore radiates heat to the surrounding air at a higher rate. Using narrow pipes increase the surface area to facilitate a higher rate of emission of heat to the surrounding air. [1] [1] [1] (d) Energy absorbed by air = 1.1088 x 107 J (allow for e.c.f.) (1.25 x 4 x 60)(760)(  - 20) = 1.1088 x 107 J  = 68.6 oC = 69 oC (68.6 oC also accepted) [1] [1] 10 (a)(i) As the coil moves away, there is a changing magnetic field experienced by it. Or [1]

There is a changing magnetic flux linkage between the magnet and the solenoid. According to Faraday’s Law, there is an induced emf in a closed circuit, hence there is a flow of an induced current. [1] (ii) North-pole [1] (iii) Deflect right; According to Lenz’s Law, the direction of the induced e.m.f. opposes the change producing it. Hence, the induced current flows in opposite direction as compared with the original motion. [1] [1] (b)(i) 4.0 V ; T = 0.04 s f = 1 / 0.04 = 25 Hz [1] [1] [1] (ii) Vertical line across 4 divisions [1] (iii) 1 division above and 1 division below the x-axis 8 divisions along the x-axis [1] 11E (a) (i) microwaves (satellite communication) (ii) radio waves (television broadcast) (iii) visible light (optic fibre communication) [1] (b) n = 𝑐 𝑣 1.5 = 3.0 ×108 𝑚/𝑠 𝑣 v = 2.0 x 108 m/s [1] [1]

(c) Signal with least time = (ii) radio communication time = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑠𝑝𝑒𝑒𝑑 Time taken for (i) satellite communication = 2 × 35 000 000 𝑚 3.0 × 108 𝑚/𝑠 = 0.23 s (2sf) Time taken for (ii) radio communication = 3 000 000 𝑚 3.0 × 108 𝑚/𝑠 = 0.010 s (2sf) Time taken for (iii) optic fibre = 3 000 000 𝑚 2.0 × 108 𝑚/𝑠 = 0.015 s (2sf) [1] [1] (d) Signal / wave / light experiences total internal reflection. Angle of incidence is greater than critical angle. Light is traveling from optically denser medium (glass) to optically less dense medium (air). [1] [1] [1] (e) (i) Sound waves cannot be transmitted in vacuum. Sound waves require a medium for propagation. (ii) speed of sound in air = 330 m/s. This much slower speed of sound would mean a very long time between transmitting the signal and receiving the signal. (time = 3 000 000 𝑚 330 𝑚/𝑠 = 9090 s = 2.5 hours) [1] [1] 11 O (a)(i) The imaginary line drawn by joining all adjacent points of a wave that are on the same phase. [1] (a)(i) T = 2 x 0.80 s = 1.60 s f = 1/T = 1/1.60 = 0.625 Hz [1] [1] (a) (iii) wavelength = 2.0 m speed = 2.0 x 0.625 = 1.25 m / s or 1.3 m / s (allow for e.c.f.) [1] [1] (a) (iv) Arrows correct