2025 TKGS P1&2 SOLUTIONS

1 Solution to Sec 4 Physics Prelim 2025 Paper 1 1 C 6 D 11 B 16 C 21 B 26 D 31 B 36 B 2 C 7 C 12 D 17 D 22 B 27 B 32 C 37 D 3 A 8 B 13 A 18 B 23 D 28 A 33 C 38 C 4 C 9 D 14 C 19 C 24 D 29 C 34 C 39 C 5 B 10 A 15 A 20 B 25 C 30 A 35 B 40 B Qn Solution Ans 1 Derived: Energy, Force, Speed, Weight, Acceleration Base: Length, Mass, Time C 2 12 kHz is within the human hearing range (20 Hz–20 kHz) C 3 Weight, velocity and displacement are vectors; area, volume, distance, pressure, and temperature are scalars. A 4 Graph with decreasing slope (gradient) = decreasing velocity. C 5 Highest point reached is at 20 hr Area under v-t graph (triangle) = 0.5 × 20 × 5000 = 50,000 km. B 6 Constant speed means net force is 0 → friction = 22 N. D 7 Between 7.0 N and 4.0 N, the net force is 3.0 N in the direction of the 7.0 N. (Fnet)2 = (3.02 + 4.02) Fnet = 5.0 N C 8 Free fall – Acceleration due to gravity – 10 ms-2 B 3.0 N 4.0 N

2 At t, Acceleration is a constant a = (v- u)/ (t) u = 0 v = at L = ½ vt = ½ (at)t = ½ at2 At 0.5t, Acceleration is a constant a = (v- u)/ (½ t) u = 0 v(½ t) = ½ at L(½ t) = ½ v(½ t) (½ t) = ½ ( ½ at) (½ t) = ½ at2( ¼ ) = ¼ L 9 Acceleration due to gravity is always 10 m/s² downward. D 10 Inertia causes the ball to swing backward (towards the left) when the car moves right. A 11 The toy is balanced. CG is directly below the point of contact with the table. B 12 Use moments about one point: Tensions are 95.2 N and 74.8 N. Taking moments about Y, CW = ACW (5.00) x Tx = (2.50 x 140) + (4.20 x 30) Tx = (350 + 126) / 5.00 Tx = 95.2 N Upward forces = Downward forces Tx + Ty = 140 + 30 Ty = 74.8 N D 13 Pascal’s principle Force = 60 × (30/2) = 900 N; Conservation of energy, distance = (60×15)/900 = 1 cm. A 14 Same level, same liquid, same pressure C

3 PP= PQ 40 + Patm= PQ Patm= 760 mm PQ = 760 + 40 = 800 mm. 15 Equal mass; density inversely proportional to volume: P = ρQ / 2. A 16 Using energy conservation: speed when hitting target = 12 m/s. C 17 Work done = F × d is same → same KE. D 18 Temperature increase → internal energy increase. B 19 Power out = 100×20 = 2000 W. Efficiency = 2000 / 3000 = 0.67. C 20 Compression → more collisions per unit time with walls. B 21 Brownian motion shows air molecules are in constant random motion. B 22 X is cooler because a dull surface is a good emitter. Explanation: 1. Surface Texture and Thermal Radiation: • A dull (matte) surface is both a good absorber and a good emitter of thermal radiation. B

4 • A polished (shiny) surface is a poor absorber and a poor emitter of thermal radiation. 2. Cooling Process: • Both cans X (dull surface) and Y (polished surface) are filled with water at the same initial temperature (boiling point) and left in the room. • The cans cool down primarily by emitting thermal radiation to the