2021 RI Prelims H2 Phy Paper 1 Soln

© Raffles Institution [Turn over 2021 Raffles Institution Preliminary Examinations H2 Physics Paper 1 Solutions 1 B  3 33 80 2.3873 g cm44 2.033 mm V r      3 o r 3 20 . 1380 4.0 md mr md mr             0.24 0.2 (1 s.f.)   32.4 0.2 g cm  2 C Between t1 and t2, displacement increases at increasing rate. Between t2 and t3, displacement increases at decreasing rate. After t3, the displacement remains constant as velocity is zero. 3 B Gradient of the first graph gives resultant force in the y-direction, which is a constant. Hence, acceleration in the y-direction is constant. Gradient of the second graph gives velocity in the x-direction, which is a constant. This is similar to a scenario of an object moving in a projectile (parabolic path) under constant acceleration of free fall. 4 C Gradient = 14 4  1 81 4 4t  t1 = 2.29 s change in momentum = area under graph = ½ (2.29)(8)  ½ (4  2.29)(6) = 4.03 N s 5 B For equilibrium, the lines of action of three non-parallel forces must intersect at a single point. The three forces should form a closed triangle. 6 C      0.015 1000 0.005 1000 5000 60 0.0272 m balloon blockUU k x W gg x x      7 D 0.80 9.81 0.80 1.0 8.648 N 8.648 2.0 17.296 J 10Energy Input 17.296 28.86 Fm gm a F F WF d J         8 D 2 2 1 2 2 2 0.22 9.81 0.40 7.02 10 m350 kx mgh mghx k     

2 © Raffles Institution 9 A  2 24 2 275 9.14 10 m s30 60ar      10 D 2 2 mvFm r m v r   '2 2 vFm F   11 B 22Gain in KE Loss in PE GM GMmhmgh m h RR     12 A 3 22 44 33 GM G GgR RRR     Hence, 0.50 0.70 0.35MM M EE E gR gR     Therefore, 0.35MM E Wm g Wm g 13 A  2 2 22 2 00 0 When the platform is above the equilibrium level, and moving down When the coin loses contact with the platform, 0 9.81 0.0243 m 23 . 2 mg N m x N mg m x x g gx             14 A This is derived from 21 3pV N m c  and is not an assumption. 15 B Work done on gas = 25  103  (80  40)  103 = 1000 J U = Q + W = 2500 + 1000 = 1500 J Avoid using 3 2UP V as the gas may not be monatomic. 16 A 2 0 cos I I  200.25 cos 60  2 0 new cos 90 0I I 17 B 2 2 2 2 , 4 1 A= 44 P kA r PPkA krr      II Amplitude is inversely proportional to distance (first graph). Intensity is proportional to amplitude2 (second graph). mg N equilibrium level

3 © Raffles Institution [Turn over 18 C 3 37 37.5 10 9.5 7.5 10 0.45 10 5.08 10 m 0.70 9.5 x xx d dD D              19 D 3 max 7 sin , where sin 1 at max. order 10 500 3.16 3 633 10 number of maximas 7 (3 maximas on both sides of the central maxima ) dn nn d              20 C Electric field lines point from higher potent ial to lower potential and must be perpendicular to equipotential lines. 21 B When negative charge is moved to Y, distance to Z is shorter. In addition, the vector sum of the two perpendicular electric field strengths will result in a resultant field strength of higher ma gnitude as shown above. 22 D total energy delivered by the battery  2500  500  3000 J total energy 3000e.m.f. 2.00 Vtotal charge 1500  23 C Increasing R4 increases the effective resistance of the combination of R2 and R4 in parallel. By potential divider rule, the p.d. across the parallel combination increases. Since R2 remains constant, the ammeter reading will increase. 24 D In the driver circuit, the p.d. across XY s hould be minimized, so the resistance of the NTC thermistor needs to be maximised, according to potential divider rule. Hence, the temperature should be low. The terminal p.d. of the test circuit should be maximised, so the resistance of the LDR needs to be maximised. Hence, the environment should be dark. W A X Y Z when negative charge is at Y W A X Y Z when negative charge is at X

4 © Raffles Institution 25 C For coil to remain horizontal,    3 6.8 2.5torque due to magnetic force 100 15 6.8 2.5 100 100 3.0 15 5.0 4.360 10 9.81 100 100 1000 100 7.81 A mg BL m g                           I I I 26 D The coil is connected to an ammeter (a closed circuit), so there is induced current in the circuit. By Lenz’s law, the magnetic force acting on the induced current should oppose the rotation of the coil. Hence, the rotation will slow down, the period of rotation will increase. So, the answer is either B or D. According to Faraday’s law, the emf induced is proportional to the rate of change of the flux linkage. If the rotation slows down, the rate of change of the flux linkage decreases, hence the amplitude of the graph will decrease. The answer is D. Mathematically, emf cos sindd NBA t NBA tdt dt     As the rotation of the coil slows down, the period T increases and 22 f T  (hence the amplitude NBA) decreases. Note: The magnetic force damps the rotation of the coil. But take note of the difference between this damping and the other exam ples of damping of SHM, such as a spring-mass system in water. For a spring- mass system in water, its am plitude and period of oscillation are unrelated. When damped, its amplitude will decrease exponentially in time, but its period (or frequency) can be taken to be unchanged. 27 A In the initial orientation, the component of B perpendicular to the area is B  0.15  103  sin10  2.60  105 T initial flux linkage, ini  BnA  2.60  105  5  0.12  1.56  105 Wb. The final flux linkage is 0 when the angle becomes 0. Hence, the change in flux linkage is 1.56  105 Wb. 28 D mean power 2 0 2 RP  I new mean power  2 2 0 02 4'4 22 R RPP I I 29 D 2 11 (0.0010) 0.0005022 0.00050 k k k pm E Ep pE pp      20000.00050 hh hx p pp   30 B If a red filter is used, only red light inci dents on the metal. Red light photons have lower energy than yellow. Hence, no photoelectric effect will be observed.

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