(EJC) 2024 J2 H2 PRELIM Answers P1-3

©EJC 2024 9749/J2H2PRELIM/2024 EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2024 9749 PHYSICS MARK SCHEME Paper 1 – Multiple Choice Questions Question Key Question Key Question Key 1 A 11 A 21 C 2 D 12 D 22 A 3 C 13 C 23 A 4 D 14 D 24 C 5 C 15 C 25 B 6 A 16 D 26 B 7 A 17 B 27 C 8 C 18 D 28 A 9 B 19 A 29 C 10 C 20 D 30 C 1 Answer: A Notice that the options are all given in electrical units, hence we express power in terms of potential difference and current. 2 unit for ( ) unit for unit for unit for ( ) Vk = I I unit for L 2 unit for ( ) unit for ( )unit for ( ) V R    = = =  I 2 Answer: D Velocity is a vector, thus the direction sign of the velocity when the ball is going up is the negative of the direction sign of the velocity when the ball is coming back down. Thus, reject option A and option B. Given the asymmetry of the height-time graph where the time taken for upward journey of ball is shorter than downward journey, air resistance must be present. Since work is continually done against air resistance, the final kinetic energy of the ball (just before hitting ground) would be less than the initial kinetic energy of the ball (when it was first thrown up from the ground). Also, the area under graph for the first part of graph should be same as that in the second part since the distance travelled up and down are the same. So, reject option C.

2 ©EJC 2024 9749/J2H2PRELIM/2024 3 Answer: C : yys u t+  = 21 2 : xx gt s u t + + → = Combining the 2 equations gives 2 1 2 x y x ssg u =   or x x ys u s Thus, 21 2 1.4112 A A A B B B x u s x u s= = = = 4 Answer: D For Z, FYZ = 4ma ……(1) For Y, FXY – FZY = 3ma ……(2) From Newton’s 3rd Law: FYZ= FZY= 4ma ……(3) From (2) & (3), FXY – 4ma = 3ma FXY = 7ma = FYX Thus, 77 44 YX ZY F ma F ma== F Z Y X FYX FXY FZY FYZ a a a 4m 3m m

3 ©EJC 2024 9749/J2H2PRELIM/2024 5 Answer: C 1 Newton's 2nd Law gives When string breaks, tension 0 ( ) ( ) 0 1000(1.0)(9.81) (800)(1.0)9.81 (800)(1.0) 2.45 m s net wood wood wood water wood wood wood wood F m a U m g T m a T V g V g m a a a  − = − − = = − − = −= = 6 Answer: A ( ) 2 2 22 Apply Newton's 2nd Law: At bottom, .....(1) At top, assume points downwards .....(2) From bottom to top, conservation of ener gy gives 1(2 ) .....(3)2 B B T T T BT mvN mg r N mvN mg r mg r m v v −= += =− (1)-(2), then substituting (3) gives 6 5.5 6 0.5 Negative sign indicates should point up wards, opposite to assumption. BT T T T N N mg N mg N mg N −= −= =− mg T U a mg NT mg NB vB vT

4 ©EJC 2024 9749/J2H2PRELIM/2024 7 Answer: A The work done by the net force provides the change in kinetic energy, which is equal to the kinetic energy if the initial kinetic energy is 0. KE at displacement s = work done by the net force Fds== area under the F-s graph Hence, gradient of KE-s graph = F. (It’s easier to work with gradient.) From s1 to s2, F increases from 0, so the gradient of KE-s increases from 0. From s2 to s3, F decreases from a positive value to 0, so the gradient of KE-s decreases. Only A fits the above description. KE is always positive, so reject options C and D. Area under F-s graph = work done by F = Gain in KE. From s1 to s3, area always increasing, also F always positive, so KE always increasing, thus reject option B. 8 Answer: C 6 14 2 6 2 14 6 6 .......... (1)(6.371 10 ) 80000 9.81 3.98 10 .......... (2)(6.371 10 ) Sub (2) into (1): 3.98 10 62 10 62 MJ (6.371 10 ) 80000 GM GM r GM GMg GM r GM r   = − = − + = −  = −  =   = − = − =  = + 9 Answer: B Using 23Tr  (Can use in MCQ direct, but must prove starting from Fg provides Fc if in P2 or P3) Note to find “radius of orbit” given the satellite’s height above Earth. Height r  radius of orbit = 2r Height 3r  radius of orbit = 4r 23' '2 3 2 ' 4 2 2 22 Tr Tr TT TT    =       = =

5 ©EJC 2024 9749/J2H2PRELIM/2024 10 Answer: C ( ) ( ) ( ) 6 0 0 0 2 4 1 26 10 Hz 2 3222 10000 0 32 10 = =  −=  = f.a x x . 11 Answer: A From displacement-time graphs of P and Q in question, the amplitude of oscillation by Q decreases faster than that of P  Q undergoes more damping than P. Hence answer is A. For heavier damping, • Peak has lower amplitude. • Peak shifts to a lower frequency.

6 ©EJC 2024 9749/J2H2PRELIM/2024 12 Answer: D Question asked for “rate of heat loss”. [J s-1] • Eliminate A and C by unit analysis. Note that m is defined as rate of loss of mass. [kg s-1] -1 -1 J s J kgkg s EP t mm  = = =   • Option B: P1 – P2 removes the rate of heat loss term. Eliminate. • Mathematical proof for D: 21 1 2 2 1 1 vm m l Pm P m−= 2 1 2 vm h m m l+− ( )211 21 mmmh mm −− =− 21 h mm− 13 Answer: C For full cycle processes, the net change in internal energy is zero. 560 [( 4.2) (1.0 10 (20.0 5.0) 10 )] 2.7 J U Q W Q Q −  = + = + − +   −  = 14 Answer: D Note that  is measured with respect from the polarising axis of P1. Let the light emerging from P1, P2 and P3 have intensity I1, I2, I3 respectively. ( ) ( ) ( ) 1 22 3 22 2 2 cos 30 cos 30 14 30cos 30 cos 3 cos 30 0 68 II I   = −  = −  =  −  =  

7 ©EJC 2024 9749/J2H2PRELIM/2024 15 Answer: C For small angle, ( )( ) 9 2 4 100 550 10 10 1 m 0.55 10 mm 0 55 s rb rs b .   − − −    = =  = 16 Answer: D ( ) ( ) ( ) 99 2 1 1 2 2 1 2 1 12 11 9 480 3sin 480 10 640 10 640 4 min 4 and min 3 Using sin , 0.01 sin 4 480 105000 74       −− − = =   =   = = == = = = nd n n n n n nn dn 17 Answer: B Applying potential divider principle, the potential difference across the variable resistor, VR varies between: 100 30 5100 500 RV=  = + V to 500 30 15500 500 RV=  = + V camera lens θ θ b = 10−2 m sensor From circular measure s  rθ Rayleigh’s Criterion

8 ©EJC 2024 9749/J2H2PRELIM/2024 18 Answer: D A and B are wrong because the diode is connected in reverse-bias. When temperature of the thermistor is low, resistance is high and hence p.d. across the thermistor is high. The heater should be connected across the thermistor. 19 Answer: A change in potential energy = final potential energy − initial potential energy 1 2 1 2 0 1 2 0 3 0 1 0 2 3 12 0 1 2 1 0 2 3 3 ( ) ( )1 1 1 1 4 4 4 4 1 1 1 1 44 qQ q Q qQ q Q x x x x x x qQ qQ x x x x x x                  −−= + − +          ++              = − + −  ++  work done by electric field = −(change in potential energy) 12 0 1 1 2 0 3 2 3 1 1 1 1 44 qQ qQ x x x x x x  = − + −  ++  Note: The change in potential energy is the work done by the external force. The electric force is equal and opposite to the external force (in defining the electric potential), hence there is an overall negative sign between the work done by the two forces. 20 Answer: D The electric field is uniform between the plates, so FA = FB. Point A is closer to the negative plate, hence is at a lower potential, i.e. VA < VB. Since U = qV, and electron is negatively charged (q < 0), UA > UB.

9 ©EJC 2024 9749/J2H2PRELIM/2024 21 Answer: C Draw the magnetic forces acting on each wire due to the rest of the wires. Wire Y has a large magnetic force acting to the right due to wire X and its larger current. Wire Y is also attracted to the right by wire Z. OR Force on wire X due to wire Y = 2 2 2 where 2 XY XY Y X oY Y