2021 Prelim H2Phy P1 Ans

1 2021 DHS H2 Physics Prelim Paper 1 Suggested Solutions Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 D B D B C C C D D A Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 B D A A C C B A B A Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 A C B C C D D B A A 1 D Average car tyre radius, R = 0.20 m Cross sectional radius r = 0.10 m Volume = (2πR)(πr2) = 0.039 m3 2 B Units on right side = (unit of k)(A2)(m/m)2 = (unit of k)(A2) Units on left side = W unit of k =W/ A2= (W/A)(A) = V/A = Ω 3 D v increases as the skydiver accelerates vertically downwards. Resultant force decreases (force due to air resistance) causes acceleration to decrease. 4 B v2 = u2 + 2as, 0 = u2 + 2(-9.81)(12.7) v = u + at , 0 = √ 2(9.81)(12.7) + (-9.81)t t = 1.61s, total time = 2t = 2(1.61) = 3.22 s 5 C Gradient = acceleration = (12.0 – 3.0)/15.0 = 9.0/15.0 Resultant force = ma = (2.5)(9.0/15.0) = 1.5 N 6 C perfectly elastic collision : vQ – vp = uP – uQ = 1.2 – (-1.8) = 3.0 -- Eqn 1 COM :

2 (0.30)(1.2) – (0.60)(1.8) = 0.30 vP + 0.60 vQ -- Eqn 2 Solve Eqn 1 and 2, vP = 2.8 m s-1, vQ = 0.20 m s-1 7 C Max pressure the probe can take = atmospheric pressure + pressure due to the liquid On Earth: Max pressure = 100 x 103 + (64)(1000)(9.81) = 727840 Pa On Moon of Saturn: 727840 = 35 x 103 + h (740) (3.6) h = 260 m 8 D When T = (D + W2), the net force on the lift is 0. Hence by Newton’s first law, the lift must have a constant velocity. If it was originally at rest, it will be at rest (constant velocity of 0). 9 D Negative work is done by the upthrust on the cylinder. 10 A At the top of circle, mg + N = mv2/r At minimum speed, N = 0 mg = mv2/r v2 = rg = (1.20) (9.81) v = 3.43 m s -1 11 B A is wrong because magnitude of gravitational force GMm/r2 decreases with increasing distance. In an orbit, gravitational force provides centripetal force, and since centripetal force decreases, so is the centripetal acceleration (Thus D is wrong). From Kepler’s Third Law, T2 r3 T increases as r increases.

3 Hence angular velocity decreases with increasing distance (Thus C is wrong). B is correct, as gravitational potential energy –GMm/r is a scalar and it becomes less negative (hence increases) with increasing distance. 12 D Let the distance from M1 be r from where it experiences no resultant gravitational force. At r, 13 A A is incorrect as the amount of energy absorbs is equal to the latent heat rather than the specific latent heat. 14 A × M1 d r M2

4 15 C As temperature is constant, the internal energy of the gas is unchanged. Hence A is wrong. Work is done by the gas during expansion, hence B is wrong. C is correct as D is wrong as temperature is unchanged thus root-mean-square speed is unchanged. 16 C Positions of the KE = 0 ( at t = 0, T/2 and T ) corresponds to x = amplitude positions. Positions of max KE corresponds to x = 0 positions. Hence C is correct. 17 B Let the height of the cylinder be h. h – 0.835 = / 4 ------(1) h – 0.171 = 3/ 4 ------(2) (2) – (1): 0.664 = 0.5  = 1.328 m f = v/ = (340)/(1.328) f = 256 Hz 18 A 19 B 20 A Weight of oil drop = Electric force (vertically upwards) + Force due to air resistance

5 Mg = QE + kv, Rearranging, v = Mg k – QE k A graph of v against E will give a straight line of gradient, −Q k and y- intercept, Mg k 21 A From the definition of current, 22 C 23 B Due to symmetry, A and C will have the same effective resistance, so they cannot be correct answers. D has one identical resistor, say R, in parallel with the rest, so the effective resistance will be <R. 24 C Effective resistance of circuit = Current drawn from cell = Voltmeter reading = 25 C By conservation of moments, magnitude of moment due to force acting on current carrying conductor =

6 magnitude of moment due to paper rider     26 D Component of weight parallel to slope = Component of magnetic force acting of rod in the opposite direction Thus,  27 D The equation is I = I0 sin(2 πt T ) = I0 sin( 2 πt 0.0025 ) = I0 sin(800 πt) 28 B <P> = Irms 2R = (I0/√ 2)2R = (2.0/√ 2)2(200) = 400 W 29 A Since the target is unchanged, the wavelengths of the peaks will remain the same. However, with an increase in the p.d. between the target and cathode, the target will be bombarded with electrons of higher energy. These electrons will have a higher chance to remove an inner shell electron from the

7 target, resulting in more de-excitations between energy levels and hence higher intensity of the peaks. 30 A ~ THE END ~