2024 JPJC H2 Prelim P1 Solutions - Printed

[Turn over Answers to 2024 JC2 Prelim Exam Paper 1 (H2 Physics) 1 C 6 B 11 B 16 C 21 D 26 A 2 A 7 C 12 C 17 D 22 B 27 A 3 C 8 A 13 A 18 B 23 A 28 B 4 B 9 D 14 D 19 C 24 D 29 C 5 A 10 A 15 B 20 D 25 B 30 A Suggested Solutions: 1 Assume that the diameter of the cross-section of the wire in a paper clip is 1 mm. Cross-sectional area of the wire in a paper clip  −−= =    2 7 2 7 20.001 7.85 10 m 8 10 m2 Answer: C 2 22 1 Using 2 0 2(9.81)(2.5) 7.0 m s v u as v − = + = + →= =+ → = + →= 21 2 21 2 Using 0.12 7.0 (9.81) 0.017 s s ut at tt t Answer: A 3 Applying conservation of momentum, ( )1 1 2 2 1 1 2 2 12 21 0 m v m v m v m v vm vm = + − = = Answer: C 4 By resolving each forces into its vertical and horizontal components, only option B is most likely to be in equilibrium. Answer: B

2 2024/JPJC/PHYSICS/9749 5 Using conservation of energy, loss in G.P.E. = gain in K.E. + work done against frictional force ( ) ( ) ( ) ( )( ) ( ) =+ =+ = 2 2 1 2002 160 50 60 20 2002 87 N mgh mv f gf f Answer: A 6 For circular motion at the highest point, 2 2 2 2 3 3 2 3 2 3 mvmg N r W mvW r mg mvmg r mg mv r grv −= −= −= = = Answer: B 7 −=gcF N F −=  2 2 2GMm N mR TR Since N = 0, =  2 2 2GMm mR TR = 3 2 RT GM Answer: C

3 2024/JPJC/PHYSICS/9749 [Turn over 8 acceleration, = 0 sina a t 0 0 0 0 displacement, sin 2 sin 3 2 sin 3 0.87 x x t Tx T x x    =−   =−      =−  =− Answer: A 9 A and B are correct as they are valid assumptions in the simple kinetic theory of gases, i.e. how ideal gases particles should behave. C is correct as the total energy of the system must remain constant due to the system being isolated. D is not correct – given that the initial momentum of a molecule before collision is p, there will be a change in momentum of 2p because the final momentum of the molecule will be –p (same magnitude, opposite in direction). Answer: D 10 The pressure of a gas is derived from the forces of collision between the molecules and the container walls, and not intermolecular collisions nor the energy transfer to the walls. C will lead to a decrease in pressure instead, as the time interval between collisions will increase. Answer: A 11 Using the equation Q Tmc=  2 2 1 2 3 3 8 4 mcmv T vT c  = = Answer: B 12 Phase difference   =  = 5.0 sin30 210 1.6 rad Answer: C

4 2024/JPJC/PHYSICS/9749 13 By Malu’s law, cosfAA = Since 2AI , 2 2 2)( cos cosf k A kA ==I Since power is proportional to intensity, power is proportional to 22cosA  . Answer: A 14 In order to get destructive interference at Q, the path difference of waves moving between P and Q should be (m – 0.5)  . Hence, path difference 2 311 24 21Q PS Q ) (S 2P( ) m −− = + − + = =  l l l l Answer: D 15 Using Rayleigh Criterion, min angle of resolution between the two pixels,  − − −  = =   9 4 3 550 10 2.75 102.0 10 radb   − −  =  ==    = = 4 4 2 2 2.75 102 2(0.76)tan22 2.1 1 tan tan m 0.2 m 0 1m x D xD Answer: B 16 At equilibrium, mgd qVF ==E When d is twice, FE is reduced by half, mgF 2 1 E = ga mamgmgmgF 5.0 5.02 1 resultant = ==−= Oil drop accelerates downwards as weight is now greater than upward electric force. Answer: C

5 2024/JPJC/PHYSICS/9749 [Turn over 17 The forces due to the charges at different positions are shown. 12 2 qqF r At A, the force due to the negative charge −2Q is much larger than the vector sum of forces due to the 2 positive charges +Q. Hence, there is a net downward force. At B and C, the direction of resultant force on test charge is upwards. Answer: D 18 ( ) ( )I = = = =  b nq nqVAnvq ac n q abct t t Answer: B 19 From Fig. 19.1, when the current is 5 mA, the p.d. across the diode is 0.8 V The p.d. across the 50  resistor is given by V = (5 mA)(50 ) = 0.25 V So the p.d. across the supply = 0.8 + 0.25 = 1.05 V Answer: C 20 Potential at X is 12 V The p.d. between X and Y = p.d. across the 3  resistor = (1)(3) = 3 V So the potential at Y is 12 − 3 = 9 V Answer: D 21 By Fleming’s left-hand rule, F is perpendicular to B. Answer: D −2Q +Q +Q A B C D

6 2024/JPJC/PHYSICS/9749 22 Work done is required on the loop to maintain uniform speed, when the right edge of the square loop is out of the magnetic field and only the left edge is in the field. This is because that there will be a resistive magnetic force acting on the left edge o f the loop, and this occurs over a distance of L, the length of each side of the square loop. Hence, total work done against resistive force, I = = =  =  = induced 2induced 2 2 3 ( )( ) BW F L B L L EBL R BLvBL R BL vR where B is the flux density of the field, R is the resistance of each side of the wire. Hence W is directly proportional to v. Answer: B 23 Average e.m.f. induced, 2 cos 45 300 cos0 1.0 (1 co 8 (0.02 s 45 )0.060 8.3 V 0)4 fi t t NBA E t NBA  =− −= −   = = −  =     Answer: A 24 When the current is flowing from X to Y, the effective resistance of the circuit is 2.0 Ω, and when the current is flowing from Y to X, the effective resistance of the circuit is 3.0 Ω. Hence, the peak currents will be different when the current is flowing from X to Y and from Y to X. Answer: D

7 2024/JPJC/PHYSICS/9749 [Turn over 25 Peak voltage in secondary coil is 90 V. Resistor R2 and R3 are in parallel connection,  total resistance in circuit connected to the secondary coil is 120 Ω. R.m.s. voltage in secondary coil, r.m.s 90 V 2 V = R.m.s. current in secondary coil, r.m.s 90 2 A120=I Average power dissipated in R1 2 r.m.s ) 80 22.5 23 W (  =  = I Answer: B 26 A and B seems to be the closest options. hfo should be a form of energy rather than the threshold frequency. Answer: A 27 Velocity decreases → de Broglie’s wavelength increases → more diffracted → larger diameter of circles Answer: A

8 2024/JPJC/PHYSICS/9749 28 − = −−−−−31 3 (1)hcEE − = −−−−−21 2 (2)hcEE − = −−−−−32 1 (3)hcEE (1) – (2) – (3) = 0   − − = 3 2 1 0hc hc hc   =+ 3 2 1 1 1 1 Answer: B 29 A: Not true, mass-energy is conserved. B: Not true, the larger nucleus has lower binding energy per nucleon, hence has lower binding energy. C: True, from conservation of momentum, particles Y and Z has the same magnitude of momentum p. From p2 = 2mE, KE is inversely proportional to mass. D: Not true. Answer: C 30 At time t, there are equal number of nuclides No of X and Y. After one half-life, number of X nuclide NX = No/2 number of Y nuclide NY = No/2 + No = 3No/2 i.e. the ratio of NX : NY = 1 : 3 Answer: A E2 E3 E1