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 te