2023 AHS Physics Prelim Paper 1 solutions

AHS 2023 S4 Physics Prelim Exam Paper 1 solutions Q Ans Explanation 1 C diameter of Earth = 10 x 106 m = 1.0 x 107 m = 1.0 x 104 x 103 m = 1.0 x 104 km diameter of atom = 10 x 10-9 m = 1.0 x 10-1 x 10-9 m = 1.0 x 10-1 m length of football field = 100 m = 10 x 101 m 2 B From P to Q to R → ¾ of a complete oscillation period of oscillation = (1.5 / 3) x 4 = 2.0 s number of oscillations in one minute = 60 / 2.0 = 30 3 A 4 D Since velocity is positive throughout, there is no change in direction. Option B and C can be eliminated. Before opening his parachute, the velocity should not decrease for the parachutist. 5 B Distance travelled from t = 20 s to t = 50 s = area under the graph = (½ x 20 x (12.0 + 16.0)) + (½ x 10 x 12.0) = 340 m. Average speed = total distance / total time = 340 / 30 = 11.333…. = 11 m / s (2 s.f.) 6 A There are no action-reaction pair as all the forces are acting on the same ladder. 7 D The block of wood is moving to the left. Hence forces acting towards the left is positive. First 5.0 s: Using “F = ma”, since a = 0 m / s2, net force = 0 N Thus, there must be friction of 20 N acting on the wood. After 5.0 s: Net force = F = F1 – friction – F2 = 20 N – 20 N – 10N = - 10 N (this means the net force is acting to the right, opposite to the motion of the wood) From “F = ma”, ‘a’ is negative when net force F is negative. Negative ‘a’ means deceleration. 30 N South 30 N West Combined effect of 2 horizontal forces = 30 N West Together with the 30 N South force, the resultant force is acting approximately in the South-West direction. Hence to keep the mass in equilibrium, an additional force with the same magnitude as the resultant force acting in the opposite direction to the resultant force is required. Resultant force South-West Additional force North-East

8 C Using “ ρ = m / V ” V of each ball bearing = m / ρ = 18.0 / 9.0 = 2.0 cm3 V of eight steel balls = 2.0 x 8 = 16.0 cm3 Reading on measuring cylinder = 25.0 + 16.0 = 41.0 cm3 9 D d4 is perpendicular to the line of action of the force F. 10 D Diagram Taking moments about K, sum of clockwise moments = sum of anti-clockwise moments (11.0 x 3.0) + (11.0 x 1.2) = (12.0 x d) + (10.0 x 1.5) d = 31.2 / 12.0 = 2.6 m 11 C Pressure in the mercury column increases with depth. Atmospheric pressure is acting on the surface of the mercury. So the point is 10 cm above the surface of the mercury. 12 B Liquid pressure is only affected by height of the liquid column above it, not on volume, cross-sectional area or shape of column. See textbook page 128. 13 B liquid pressure difference in mercury column = Δh ρ g = (6.0 / 100) x 13600 x 10 = 8160 Pa Since there is no change in atmospheric pressure, the difference in pressure for the two sides of the manometer must be the same. Using pressure difference = Δh ρ g 816 0 = h x 1000 x 10 h = 0.82 m Or: 6.0 cm x 13.6 = 0.816 ≈ 0.82 m (because mercury is 13.6 times denser than water 14 B Total work done = work done against friction + work done against gravity + EK gained F x d = (f x d) + (mgh) + EK gained 15 x 12 = (4 x 12) + (2.0 x 10 x 5.0) + EK gained EK gained = 32 ½ m vQ2 - ½ m vP2 = 32 2.0 x vQ2 – 2.0 x 2.02 = 64 vQ2 = 36 ➔ vQ = 6.0 m/s H J K not drawn to scale 12 N 1.2 m 3.0 m 1.5 m x m 11 N 11 N 10 N

OR : EK at the start + Total work done = EP at the end + EK at the end + work done against friction (½ m vP2) + (F x d) = (mgh) + (½ m vQ2) + (f x d) (½ x 2.0 x 2.02) + (15 x 12) = (2.0 x 10 x 5.0) + (½ x 2.0 x vQ2) + (4 x 12) 4 + 180 = 100 + vQ2 + 48 vQ2 = 36 ➔ vQ = 6.0 m/s 15 A Power = Work Done / time (where work done is measured in J and time in s, hence J / s is equivalent to the unit of watt W. kW h is a measure of energy, not power. 16 C Chemical potential energy, Elastic potential energy, Gravitational potential energy 17 A As they are made of the same material, they have the same density in the shape of a solid block. But as they have different masses, they do not have the same inertia nor heat capacity. Though they have the same temperature, only the average kinetic energy of the particles in both are the same. The total KE and total PE of the particles are not the same because Y has more particles (more mass). 18 D This is similar to Brownian motion as the dust particles are light enough to be affected by the invisible air particles but large enough to be seen under bright light. 19 C At a lower temperature, the average kinetic energy of the particles is lower. Hence, to maintain the same gas pressure, the frequency of collision have to be greater than before. 20 B εα∆θ + 6.00 mV → change of 100 °C (temperature difference 100 °C) + 8.40 mV → change of (100/6.00) x 8.40 = 140 °C (temperature difference 140 °C) The refence point (fixed point) is the hot junction (steam point, not ice point), Hence boiling point of propane = 100 – 140 = - 40 °C 21 B 22 D angle of incidence i = 90o – 15o = 75o using Snell’s Law, n = sin i / sin r 1.5 = sin 75o / sin r angle of refraction r = 40o x = 90 – r = 50o x

23 C Critical angle is the angle of incidence in the optically denser medium for which the angle of refraction in the optically less dens medium is 90o. 24 D Rays A and B are wrong because they diverge away. Ray C is wrong because it did not bend. 25 C Remember the EM spectrum song? 26 A higher-pitched → higher frequency (lower period) same loudness → same amplitude. 27 A v = 2d / t (because echo travels to and fro), so d = vt / 2 depth of seabed, d1 = 1500 x 1.0 / 2 = 750 m distance of whale from the surface, d2 = 1500 x 0.60 / 2 = 450 m distance of whale above the seabed = 750 – 450 = 300 m (Note: The thickness of a whale is only a few metres, which is only a small percentage of the answer. Hence it can be ignored.) 28 C There is an electric field due to the charge in the plastic ball. There is Earth’s gravitational field as well. 29 A The ball is neutral, which is represented by no charges on the ball. 30 D p.d. = work done / charge p.d. across a resistor = 12 / 4.0 = 3.0 V Total p.d.across the 3 resistors = 3.0 x 3 = 9.0 V. Hence e.m.f. of battery = 9.0 V. 31 D Let cross-sectional areas of wire 1 and wire 2 be A1 and A2 respectively. A = πr2 = πd2 / 4. So when the diameter of wire 2 is 50% of wire 1, the cross-sectional area of wire 2 is 25% of wire 1 because of the square factor. Hence A2 = 0.25 A1 Proof: A1 = πd2 / 4 = π(0.30 x 10-3)2 / 4 = 7.0714 x 10-8 m2 A2 = π(0.15 x 10-3)2 / 4 = 1.7679 x 10-8 m2 = 0.25 A1 So A2 = 0.25 A1. Moreover, length of wire 2 is 3 times the length of wire 1. So L2 = 3L1. Using R = ρl / A R1 = ρL1 / A1 = R R2 = ρL2 / A2 = ρ(3L1 ) / (0.25A1 ) =12 ρL1 / A1 = 12 R

32 B R = V / I When V = 1.0 V, R = 1.0 / 0 = very large When V = 2.0 V, R = 2.0 / 11 x 10-3 = 182 Ω When V = 3.0 V, R = 3.0 / 30 x 10-3 = 100 Ω Hence the value of R decreases for values of V above 1.4 V. Although the line is straight after V = 1.4V, it does not pass through the origin. Hence the resistance is not constant. Moreover, the diode is a non-Ohmic conductor. 33 B When the current is 0.50 A, the effective resistance R = V / I = 12 / 0.50 = 24 Ω. When X = 18 Ω and Y = 12 Ω Effective resistance = (1/12 + 1/12)-1 + 18 = 24 Ω 34 A When the variable resistor is set at 0 Ω, t