2023 RI H2 Physics Prelims P3 Answers

1 © Raffles Institution [Turn over 2023 Preliminary Examinations H2 Physics Paper 3 Solutions 1 (a) (i) Gravitational force provides the centripetal force 2 2 2 3 3 2 GMm mr r GM r GM r    − = = = 3 2k =− = – 1.5 (ii) 3 22 2 11 3 2 2 () 2 6700( )( )24 3600 42000 r r    − − = =  2 = 1.14  10−3 rad s−1 (b) (i) 2 2 2 2 22 k GMm mv GMm mvrrr mv GMmE r =  = == 2 2 T K PE E E GMm GMm rr GMm r =+ = + −  =− (ii) 21 12 11 24 7 5 7 7 ()22 11()2 (6.67 10 )(6.0 10 )(700) 1 1 ()2 4.2 10 (3.0 10 4.2 10 ) 2.37 10 J T GMm GMmE rr GMm rr −  =− − − =− =−   +  = (c) ET = EP + EK (ET and EP are negative values, EK is positive.) Moon’s orbit Earth x

2 © Raffles Institution [Turn over ET is constant, when Moon is nearest to the Earth, its EP is smallest (more negative), hence EK and speed is the largest. OR By conservation of energy, when Moon is nearest to the Earth, its EP is smallest (more negative), hence EK and speed is the largest. 2 (a) (i) By constantpV T = , CC BB CB pV pV TT = ( ) 5 5 7.0 10 0.0015 960 1260 0.0040 2.00 10 Pa CBB C BC TpVp TV=  = = (ii) ( ) ( ) 53 Area under the graph 1 7.0 2.0 10 4.0 1.5 102 1130 J W − = = +   −  = (b) 1. The process is carried out slowly. 2. The walls of the engine are good conductors of heat. 3. The engine is surrounded by a heat reservoir at constant temperature. Maximum 2 marks (c) change work done on gas / J heat supplied to gas / J increase in internal energy / J A → B −1000 2625 1625 B → C −1125 500 −625 C → D 0 −1000 −1000 D → A 555 −555 0 (d) During an explosion, the pressure at different points of the gas is not uniform. Hence, pressure is undefined and cannot be represented as a point or a curve on the p−V diagram. 3 (a) For a to be minimum, the minimum intensity detected directly below it must be the 1st minimum from the central bright fringe. i.e., 0.5n = Path difference = 0.5  222.00 2.00 0.5a + − = ( ) ( ) 222 22 2.00 2.00 0.5 4 4 2 0.25 a a   + = + + = + + ( ) ( )( ) 2 9 9 6 33 4 0.25 750 10 2 0.25 750 10 1.50 10 1.22474 10 1.22 10 a  − − − −− =+ =  +  =  =  = 

3 © Raffles Institution [Turn over  has to be substituted in at the last step and not earlier as calculator has insufficient number of digits. OR Since the interference pattern is symmetrical about perpendicular bisector between S1 and S2, 0.5 0.5a x x x=  +  = OR Distance (on the screen) between the central maximum and the 1 st minimum is 22 xa= ( )( ) 2 9 6 33 750 10 2.00 1.50 10 1.22474 10 1.22 10 m DDa xa aD a   −− −− == = =  =  =  =  (b) (i) ( ) ( )( ) 9 3 31 3 1.0 3 3 750 10 2.00 1.2247 10 3.67 10 m s xv D a  − − −− = =  =  = (ii) The frequency will decrease. Dx a = . D increases as the detector moves. Hence, x increases and the detector has to move a corresponding longer distance to detect consecutive maxima. (c) For images to be just resolved, 9 4 min 3 750 10 1.875 10 rad 4.0 10b  − − − = = =   min 3 4 min 1.22 10 6.507 m 1.875 10 sR aR   − − = = = =  OR ( ) ( ) 3 min 4 11 1.22 101 22tan 6.507 m 12 1.875 102 a RR − −  =  = =  4 (a) The electric potential at a point in an electric field is defined as the work done per unit positive charge by an external force in bringing a small test charge from infinity to that point. min for this aperture S1 S2 4 mm aperture a R 0.5

4 © Raffles Institution [Turn over (b) (i) 0 6 0 3 1 4 1 0.060 10 4 0.27 2.00 10 V R QV r  − = = = (ii) At x = 27 cm, V is 40 small squares. Using 1V r , When x is 2  27 = 54 cm, V is 20 small squares. When x is 3  27 = 81 cm, V is 13.3 small squares. When x is 4  27 = 108 cm, V is 10 small squares. (iii) The directions of EA and EB are both towards the right, in the same direction. 6 31 22 00 1 1 0.060 10 1.101 10 N C44 0.70 A A QE r  − −= = =  6 21 22 00 1 1 0.035 10 6.423 10 N C44 0.70 B B QE r  − −= = =  E = EA + EB = 1.101  103 + 6.423  102 = 1.74  103 N C−1 (iv) The charges on sphere A will redistribute such that there are more positive charges on the right side of sphere A. Hence, the electric field strength due to sphere A will be greater in magnitude. 0 x / cm V / V VR    

5 © Raffles Institution [Turn over 5 (a) 0.150 20 60 180 CQt= =   =I (b) V = IR = 0.150  12 = 1.8 V (c) Using the potential divider principle, 12 6.0 1.812 12 40 28 eff eff eff R R R =+ += = Since 33  resistor and resistor R are arranged in parallel, 1 1 1 33 28 185 R R += = (d) As the temperature of the thermistor increases, the resistance of the thermistor decreases. The equivalent resistance across the parallel combination of thermistor and 33  resistor will decrease and the effective resistance of the circuit will decrease. The ammeter reading will increase. 6 (a) ( )( ) 328 0 10 0 17 30 0 0218 Wb BAN .. .  − = =    = (b) Since induced e.m.f. =− dE dt , the max. e.m.f. is induced when the flux linkage  is changing at the highest rate, and this corresponds to when time t = 1.0 s. = − = −          max max Max d dBE AN dt dt 0 2 4 6 8 10 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 B / 10−3 T t / s

6 © Raffles Institution [Turn over where   max dB dt is the gradient of the B vs t graph at t = 1.0 s (As shown above) draw a tangent to the curve at t = 1.0 s. Taking the points (1.35, 6.0) and (0.65, 2.0) on the tangent, gradient = 336 0 2 0 10 5 714 10 1 35 0 65 .. . .. −−−  =  −   2 30 17 30 5 714 10 0 0156 V maxE . . .  −= −     =− (c) From t = 0 to 2.0 s, the magnetic flux density is increasing. Thus, by Lenz’s Law, the induced current produces an induced magnetic field upwards to oppose the increase in the magnetic flux in the coil. By right hand grip rule, the direction of the induced current is anti-clockwise. (d) 7 (a) ( )( ) max 19 19 KE 1.60 10 1.1 1.76 10 J seV − − = = = (b) ( ) ( )( ) ( )( ) 8 9 34 8 19 9 19 0 19 0 34 14 0 3.00 10 450 10 6.63 10 3.00 10 1.6 10 1.1 450 10 2.66 10 J 2.66 10 6.63 10 4.01 10 Hz s cf hf eV hf f f      − − − − − − −  ==  =+  =  +  = = =  = induced e.m.f. t / s 0 1 2 3

7 © Raffles Institution [Turn over (c) The energy of the photons remains the same because as E = hf and f is the same. Hence, by the photoelectric equation, the maximum kinetic energy of photoelectrons emitted will not be affected. (d) Fig. 7.3 Additional information: Photoelectrons will be emitted from both plates. Since plate A has a lower threshold frequency than plate B, photoelectrons emitted from plate A have higher maximum kinetic energy than plate B. Stopping potential of plate A, VSA will be larger than that of plate B, VSB. 8 (a) (i) ( )F kx=− (ii) By Newton’s second law of motion, netF ma kx ma kax m = −= =− Since both k and m are constants ( k m is a constant), a and x satisfy the defining equation of simple ha