2025 DHS H2 Prelim P1 Ans_v1

2025 DHS H2 Physics Prelim Paper 1 Suggested Solutions 1 2 3 4 5 6 7 8 9 10 B D C B C C A D B D 11 12 13 14 15 16 17 18 19 20 B A D B B A D B D A 21 22 23 24 25 26 27 28 29 30 D B B C A D A C A C Worked Solutions & Explanations: 1 B < ρ > = (1002 + 998 + 997 + 1001 + 999)/5 = 999.4 kg m-3, closer to 1000 kg m-3 difference from true value = 1000 – 999.4 = 0.6 kg m-3 ∆ ρ = ρ max − ρ avg = 1002 – 997 = 5 kg m-3 2 D Maximum velocity when R = mg 0.60 v = 3.0(9.81) v = 49 m s-1 At 12 m s-1, Fnet = mg – R = ma 𝑎 = 3.0(9.81)−0.6(12) 3.0 = 7.4 m s−1 3 C The total area under the velocity-time graph is the change in displacement. 4 B For an object to float in equilibrium, U = mice g Since the pressure of the water increases with depth, the re is a pressure difference between the top and bottom surface of the ice which results in an upward force on the ice that results in upthrust

5 C = =+ −= ' ' Clockwise moments = Anticlockwise moment s about the hinge (10000 )(2.0) (50 9.81)(2.5) --- Eqn (1) (10000x )(2.0) (40 9.81)(5.0) (50 9.81)(2.5) --- Eqn (2) Eqn (2) - Eqn (1), x 0.0981 m xx xx x 6 C The net force on the sledge is zero when it slides down at constant velocity. The vector triangle of forces on the sledge forms a closed polygon. 7 A Work done by the pump = Fd = PAd = (5000)(600 x 10-4)(0.40) = 120 J 8 D Einput = GPE = mgh P = 1.3×109×9.81×2.0 24×60×60 = 300 kW 9 B Minute hand makes one complete revolution in 1 hour. 𝑣 = 𝑟𝜔 = 3.00 ( 2𝜋 3600) = 5.24 × 10−3 m s-1 10 D Such a motion is possible. (Option A is incorrect) In a binary star system, the centre of mass of two stars of unequal mass will not be at the midway point. It will be closer to star with mass M. Hence the radius of orbit of M is smaller than 2M. (Option C is incorrect) For an orbit centred around its common centre of mass, position of centre of mass is fixed in space. For this condition to be satisfied, the two stars have to constantly be on opposite sides of the common centre of mass. Hence, they will have the same peri od of rotation and hence same angular velocity. (Option D is the answer) Since they have the same angular velocity and different radius, the speed at which each star orbits will also be different from v = r𝜔. (Option B is incorrect)

11 B At point P, 0netF = 22 () EM E M GM m GM m r R r Mr R r M = − =− Rearranging and making r the subject, 1E M Rr M M = + 12 A In moving from Q to P, 1 ( 800)2 400 kJ kg 1.0[ 800 ( 400)] 400 kJ Q P PQ Q r r r r GPE m     − = =− =−  =  = − − − =− 13 D From the equation: 𝜃 (K) = 𝜃 (°C) + 273.15 Gradient = 1 y-intercept = 273.15 14 B When the basketball is pumped, work is done on the air in the ball to cause an increase in pressure. Work done on the gas is positive. Since this process is a fast process, little to no heat exchange happens and hence Q = 0 By 1 st law of thermodynamics, the internal energy increases which