2023 CHS Phy Prelims Ans

2023 SEC 4 PRELIMINARY EXAMINATION Paper 1 1 B 2 B 3 C 4 D 5 C 6 A 7 D 8 D 9 D 10 A 11 B 12 A 13 C 14 A 15 D 16 B 17 A 18 C 19 C 20 A 21 B 22 A 23 A 24 D 25 D 26 C 27 B 28 B 29 B 30 C 31 C 32 D 33 A 34 D 35 A 36 C 37 A 38 C 39 A 40 C Paper 2 Section A S/N Mark Scheme Remarks 1ai displacement = area under graph [formula] = ½ (2.50) (4.0) [B1] = 5.0 m [B1] 1aii Acceleration due to gravity = gradient of graph = (0.0 – 4.0) / (2.5 – 0.0) [B1] = -1.6 m / s2 Deceleration = 1.6 m / s2 [B1] 1bi Decreasing + -ve gradient; v = 0.0 m / s before t = 2.5 s [B1] 1bii Air resistance = W = mg [formula] = 7.5 (1.6) = 12 N [B1] 2a Fnet = ma [formula] F – W = ma F – mg = ma F – (2.8 × 106) (10) = (2.8 × 106) (1.4) [B2] F = 3.2 × 107 N or 3.19 × 107 N [B1] W = mg & F = ma calculations are one mark each 2b Mass of rocket decreases (as fuel is used up) [B1] Value of g on rocket decreases (as rocket rises) [B1] 3ai Point where (all) the weight (of the object) appears to act [B1] (regardless of the orientation of the object). 3aii No, because centre of gravity of XY is not at the (geometric) centre (or 2.0 m mark) of rod. [B1] 3b Taking moments about (pivot) P, Sum of clockwise moments = sum of anticlockwise moments M = F x d [formula – all above] (3.0 – 2.5) (2.90) + 44 (0.60) = F (1.10) [B1] F = 25.3 N [B1] 3c Resultant moment clockwise (when sphere is not in water) OR Larger anticlockwise moment due to F (needed to balance larger clockwise moment). [B1] Shifting P to the right increases perpendicular distance from (line of action of) F to P/pivot. [B1] Accept if discuss about smaller clockwise moment due to weight of sphere and / or rod

2 4a Same volume for two different temperatures. [B1] Volume does not vary continuously with temperature. [B1] 4bi Kinetic energy of the gas particles increases (with temperature). [B1] Force and frequency of collisions of gas particles on walls of container increases. [B1] Force per unit area (and hence pressure) of gas increases. [B1] 4bii Conduction [B1] Flask particles vibrate and collide against gas particles (causing them to vibrate and collide more often). [B1] 5a Change in mass = 120 – 80 = 40 g [B1] Lv = IVt2 = m (lv) [formula] 2.0 (12) (3750) [B1] = 40 (lv) [B1] lv = 2250 J / g (or 2.25 x 106 J / kg) [B1] 5b Large amount of energy needed to overcome (large) intermolecular forces (of attraction) between liquid molecules in order to separate them. OR Energy is needed for the steam to do work against the surrounding air. [B1] 6a f = 1 / T [formula] = 1 / 0.8 [B1] = 1.3 Hz or 1.25 Hz [B1] 6b Correct shape [B1] 6c The first trough on Fig. 6.2 [B1] 6d Shorter wavelength (compared to Fig. 6.2) [B1] Speed of wave decreases at shallower depth (while frequency remains constant). [B1] If mention frequency changing, - 1 m 7a (from L to R) microwave / infra-red / ultra-violet/X rays/gamma rays Any 2 correct [B1] All correct [B2] 7bi The process where one or more electrons are (added/)removed from a neutral atom by the action of radiation. [B1] infra-red radiation OR microwaves [B1] Must be from answer in (a) 7bii Heating/raising temperature of the cells [B1] 8a Electric field lines directed away from sphere (or from positive to negative). [M1] Hence, it is positively charged. [A1] 8b Tangent to curve in correct position and direction [B1] 8c (Drawing) negative charges drawn on the inside of the metal box [B1] As the box is earthed, electrons from earth are attracted to the inside of the box as unlike charges attract. [B1] (Alternative: Electrons are attracted to the inside of the metal box as unlike charges attract. As the metal box is earthed, electrons from earth flow to the metal box to neutrali se the positive charges on the outside) 9ai Coil / wire cuts magnetic field (of magnet) OR changing magnetic field

3 (through coil) OR changing magnetic flux linkage between coil and magnet [B1] electromotive force induced by Faraday’s Law (of electromagnetic induction) [B1] 9aii (plane of coil) Horizontal OR in position shown in diagram [B1] Coil cutting magnetic field the fastest / Greatest (rate of) change in magnetic flux linkage (between coil and magnet) [B1] 9b (Induced) c urrent in magnetic field (of magnet) experiences a force [B1] Direction of force o pposes the change causing it (ie force by student to rotate generator) by Lenz’ Law [B1] Paper 2 Section B 10ai Thermistor with correct circuit symbol. [B1] Resistance increases (non-linearly OR at decreasing rate) as temperature decreases (or resistance decreases as temperature increases). [B1] 10aii Temperature of the bulb increases and hence, the resistance of the bulb increases. [B1] 10b Solenoid is an electromagnet when current flows through solenoid. [B1] It attracts the soft iron connected to the compressor (closing the gap or forming a closed circuit from the live wire to the compressor). [B1] 10ci 8 A to 13 A 10cii The current is large when it is just switched on (or calculate). [B1] Causes overloading of circuit if other appliances are connected (in parallel) to the same electrical socket. [B1] 10ciii Energy used = Pt [formula] = 0.180 (24) = 4.32 kWh [B1] Cost = $0.32 (4.32) = $1.38 [B1] 11ai E = mgh [formula] = (0.20) (10) (0.60) [B1] = 1.2 J 11aii By conservation of energy, E = ½ mv2 [formula] 1.2 = ½ (0.20) (v2) v = 3.5 m / s [B1] 11aiii Will not change [B0] Speed of block is independent of mass. [B1] Can show by calculation 11b Velocity is a vector and speed is a scalar. [B1] Direction of the block changes. [B1] 11ci Kinetic energy of block to gravitational potential energy of block and thermal energy of slope (and block) [B1] 11cii Work done against friction = Energy at E – Energy at F F (d) = 1.2 – mgh [formula] F (0.35) = 1.2 – 0.2 (10) (0.25) [B1] F = 2.0 N [B1]

4 EITHER 12ai All three rays are horizontal [B1] 12aii Double-headed arrow drawn between point of divergence of the 3 rays of light and the optical centre of lens. [B1] 12bi Two correct principal rays [B1] Correct position and size of real image (same size and distance from lens as object) [B1] Deduct one mark for no arrows or wrong direction 12biii Larger [B1] 12ci Bend away from normal after exiting prism [B1] 12cii Greater because violet light bends more towards normal / has a smaller angle of refraction / travels more slowly [B1] 12di i = 90° – 25° = 65° n = sin i / sin r 1.54 = sin 65° / sin r r = 36° (or 36.1°) [B1] 12dii Largest possible angle of incidence at the inner wall is the critical angle c (and hence cannot exceed c). [B1] OR 12ai Three concentric circles of increasing distance apart [B1] Anticlockwise d