SCGS 2022 Prelim P1 and P2 ANS

2022 PRELIMINARY EXAM SUGGESTED ANSWER SCHEME PAPER 1 (40 marks) Qn Answer Explanation 1 B Micrometer must be a choice for measuring diameter. Vernier calipers are not suitable for measuring length of wires. 2 C Volume of sphere = /6 (0.481cm)3 ; Density = (0.450 / 0.058 )x 1000 kgm-3 = 7720 kgm-3 3 B Returning back to the starting point will result in zero displacement. Total distance is non - zero 4 C A & B are not possible as the value is less than √42 + 52 kN or 6.4 kN. 8.3 kN is closer to 9.0 kN when the two forces are parallel to each other. 5 B 6 D The gradient, which is the acceleration, is greater at X than at Y. Therefore the resultant force is acting on the raindrop is greater at X. 7 D The graph shows increasing velocity ( increasing gradient) followed by constant velocity (constant gradient) 8 A There is no atmosphere to slow down the feather. It moves with constant acceleration. 9 B Distance travelled = ½(3.0s)(22+8)ms-1 = 45 m 10 C At constant speed, resultant force = 0. Total upward force = total downward force. 11 B Patm - Pchamber = P40 cmH2O = (0.4m)(1000kgm-3)(10Nkg-1) = 4000 Nm-2 12 D P = W/A. Since all 5 blocks have the same mass, P  1/A  an inverse function graph 13 C T x 5.0 cm = 40 x 35 cm  T = 280 N 14 C The CG of the block falls through a height of 0.9 m. GPE = 500 N x 0.9 m = 450 J 15 C Shortest time to reach her classroom is by running. More power is used to gain the same amount of GPE as Power = work done / time 16 C Efficiency = (200 x 6000 )W / (500 x 10 x 300 )W x 100% = 80% 17 A Molecules do not expand when a solid is heated 18 A Evaporation resulted in the lowering of the KE of the remaining molecules in the liquid . Molecules leaving need to attain the necessary KE to overcome the bonds and escape into the atmosphere. Therefore they must possess high energy. 19 C In one second 6000 J = m x 4200 Jkg-1oC x 20oC  m = 0.0714 kg 20 C Molecules do not expand. The KE of molecules do not increase – only PE increases during change of state The solid shaded region are of the same area

Question Answer Explanation 21 D 1.8 = sin R/sin 30o  R = sin-1 (0.9) = 64.15o 22 C Total internal reflection occurs when angle of incidence in the optically denser medium is greater than critical angle and can only occur inthe optically denser medium 23 B The dotted lines locate the image to the left and on the same side as the object. This is the image formed when image is less than the focal length of the lens. 24 B R = V/I 25 D All other choices have an open circuit 26 B Variable resistor is connected in parallel to the fixed resistor. Voltmeter will register an increase when resistance is increased. ( alternative explanation : Increasing the variable resistor increases the total resistance of the two parallel resistors . Using potential divider eqn before and after adjustment, V1 decreases and V2 increases) 27 A Resistance is lowest when temperature for thermistor is high and LDR is placed in bright light. 28 D Overheating is due to high current than allowed by the fuse 29 B Mobile electrons are attracted to the positively-charged sphere. Electrons from earth discharges the positive charges at the bottom of the plate. Removing the earthing wire first results in metal having a net negative charge. 30 B Bringing the positive rod near will induce electrons on top and positive charges at the bottom 31 D Like charges repel each other 32 C Currents flowing in difference direction create a strong field in the centre. 33 A Recall question 34 B Distance = (0.04 s x 300 m/s) = 12 m 35 C The crest/ trough moves forward, thus constituting the wave velocity. 36 A /2 = L; V= f   f = V/ = V/2L 37 C Use right-hand grip rule 38 C Apply Fleming’s Left-hand rule with the centre finger pointing in the direction opposite to the motion of the -ve charge 39 A Ns/Np = Vs/Vp  Vs = 1/20 x 240 = 12 V. Ip = 12 V/ 6.0 Ω = 2.0 A. From IpVp = IsVs  Ip(240V) = 2.0 A (12 V) Thus Ip = 0.1 A 40 D Increasing coil speed increases the frequency as well as the amplitude

PAPER 2 SECTION A ( 50 marks) Qn Part Answer Mark Remarks 1 (a) Measuring tape (b)(i) Average density = 18000 kg / ( 2.0 x 15.0 x 0.25 )m 3 = 2400 kgm-3 (ii) Average pressure = ( 18000 x 10 )N / (0.25 x 15)m 2 = 4.8 x 106 Nm-2 (iii) Doubling the length doubles the volume. Since density is constant, mass/weight is doubled. Since area is also doubled, pressure remains unchanged 2 (a) Speed is rate of change of distance while velocity is the rate of change of displacement. Since displacement is a vector, velocity has a direction and magnitude while speed is a scalar (b)(i) Velocity change = (+25m/s)-(-22m/s) = +47 m/s (ii) Average acceleration = (+47m/s)/0.0013 s = +3.6 x 104 ms-2 (iii) Average force F = ma = 0.16 kg x [+3.6 x 104 ]ms-2 = + 5.8 x 103 N 3 (a) A point where the whole weight of the body appears to act regardless of the body’s orientation (b)(i) Applying principle of moments, W x 0.2m = 80 N x 0.5 m +70 N x 1.3 m = 40 Nm + 91 Nm  W = 655 N (iii) Either : Move the 70 N weight to the right until the CG of the weight is directly above B. Reason : the weight of the stationary student must be the maximum to produce the corresponding anticlockwise moment to keep the plank in equilibrium. OR : Move the pivot to the left . Reason: This creates a greater clockwise moment . To keep the plank in equilibrium, the weight of the student will need to increase. 4 (a) ▪ Density of ice decreases as the temperature increases from – 5oC to 0oC ▪ At 0oC, the density increases (when its volume decreases) from 33 cm3 to 30 cm3. (b) Density of ice at volume 33 cm3 = (30/33) gcm-3 = 0.91 gcm-3 Density of ice at volume 30 cm3 = (30/30) gcm-3 = 1.00 gcm-3 Change in density = (1.00 – 0.91) gcm-3 = 0.09 gcm-3

( c) ▪ Water molecules/particles at the surface of the ice obtain sufficient (vibrational) kinetic energy to overcome the forces of attraction between each other and exist as free molecules. ▪ These water molecules must also have sufficient (translational) kinetic energy to escape into the space above the ice and exists as free molecules. 5 (a)(i) Number of pulses = (6 x 60 s) / 0.00462 s – 1 = 77921 = 78000 (ii) Total energy = 77921 x 0.00012 J = 9.35 J (iii) From Q = mc 9.35 J = 50 g x 4.2 Jg-1oC-1 x    = 0.045 oC (b) ▪ Energy absorbed by the skull resulted in less energy transmitted to the brain ▪ Nature of the fluid in the brain is different from water. Thus the specific heat capacity differs. (c) Input energy = 0.20 W x (60 x 6)s = 72 J Fraction of energy converted into radio waves = 9.35 J/72 J = 0.13 6 (a) Patm = 0.76 m x 13.6 x 10-3 kgm-3 x 10 Nkg-1 = 103360 Nm-2 = 100000 or 105 Nm-2 (b) ▪ Since vertical height below mercury meniscus in tube is less than 760mm, higher atmospheric pressure forces air through the hole and bubbles through the mercury upwards. ▪ Air pressure in the space above meniscus increases and pushes the mercury meniscus down slowly. ▪ When the meniscus move past the hole, the meniscus will go down quickly since the pressure