RVHS H2 Physics P1 Soln

River Valley High School Pg 1 of 7 JC 2 H2 Physics 9749 Preliminary Examinations 2024 2024 J2 Prelim P1 Q1 D Q6 C Q11 C Q16 B Q21 A Q26 D Q2 B Q7 B Q12 D Q17 A Q22 A Q27 C Q3 D Q8 B Q13 C Q18 A Q23 A Q28 D Q4 B Q9 D Q14 D Q19 B Q24 C Q29 B Q5 C Q10 C Q15 C Q20 A Q25 A Q30 D 1. Ans: D Option A is wrong. 2 2 − −=   = kgF ma N kg m s ms Option B is wrong. 11 −=  f Hz sT Option C is wrong. =  W Fd J N m Option D is correct. 2 12 sin sin   − −−=  =   = F kg m sF BIL B T kg A s IL A m 2. Ans: B distance travelled between 0 to 20 s distance travelled between 5 to 15 s distance travelled between 0 to 10 s distance travelled between 5 to 10 s= area under v-t graph from 0 to 10 s area under v-t graph from 5 to 10 s 1 0.75 13. = = = 3. Ans: D Since the acceleration on the way up will be larger (air resistance and weight acting in the same direction) than the acceleration of the way down (air resistance and weight acting in opposite directions), the time taken on way up will be less. 4. Ans: B 1 1 1 (62)(0.21) 6.512 6.51 43.40.150 impulse kg m s pv m sm − − == = = = 5. Ans: C 2(10)(9.81 ) 4.45910 12acceleration m s −== +

River Valley High School Pg 2 of 7 JC 2 H2 Physics 9749 Preliminary Examinations 2024 By considering the 12 kg block, tension = resultant force Hence, (12)(4.459) 53.5tension N== 6. Ans: C 2 2 2 2 tan(3.8 ) 9.81 0.65158 0.65158 57.8 3.6 3960.65158 centripetal acceleration centripetal acceleration m s v msr rm − − = = =  == 7. Ans: B For satellite in geostationary orbit, period = 24 hours = 86400 s 2 86400  = gravitational force provides for centripetal force 2 2 1 3 1 11 243 7 22 71 (6.67 10 )(5.97 10 ) 4.2227 10 2 86400 2 (4.2227 10 ) 307186400 GMm mrr GMrm v r v r m s      − − =    = = =      = = =  = Alternatively, a shorter method would skip calculating the radius r. 2 2 32 3 3 3 2 3 3 2 3 11 24 13 3 ( )( ) 2(6.67 10 )(5.97 10 ) 3071 24 3600 GMmmr r r GM to both sides r GM v GM v GM m s         −− = =  = = = =   =  

River Valley High School Pg 3 of 7 JC 2 H2 Physics 9749 Preliminary Examinations 2024 8. Ans: B increase in GPE of space probe = GPEfinal − GPEinitial 11 22 77 8 11(6.67 10 )(1.30 10 )(5500) 1.19 10 (1.19 0.24) 10 6.73 10 final initial GMm GMm rr J − =− − − =   −   +  = 9. Ans: D Statement D is not an essential assumption of kinetic theory of gas and can be derived from pressure of a gas 10. Ans: C heat gained by ice in melting + heat gained by 20 g of water (total mass of water) = heat lost by 130 g of water 0.020 × 3.3 × 105 + [0.020 × 4.2 × 103 × (T – 0)] = 0.130 × 4.2 × 103 × (50 – T) 6.6 × 103 + 84 T = 27.3 × 103 – 546 T 630 T = 20700 T = 33 C distractor: T = 38 C (heat gained by melted ice water omitted) 11. Ans: C Resonance occurs when the natural frequency of water molecules matches the frequency of the incident microwave. By changing its frequency, resonance will not occur, resulting in smaller amplitude of vibration of water molecules and therefore, lesser amount of energy absorbed by the food for it to be cooked. To warm up the food faster, the water molecules need to vibrate more to transfer more energy to the food. This can be achieved by increasing the amplitude of the incident microwave which will result in more energy being transferred from the microwave to the water molecules. 12. Ans: D I 1/x2 IQ/IP = (1/2)2 = ¼ IQ/IP = (AQ/ AP )2 = (6 / AP2 ) = 1/4 AP = 12.0 μm 13. Ans: C At P, acceleration is maximum, OS represents two periods, Kinetic energies at Q and R are the same 14. Ans: D To obtain a steady interference with minimum intensity at point P, both the sources must be coherent and be out of phase by π radians. Both sources can be of the same amplitude. The next minimum produced at Q corresponds to a path difference of 1 m and this should be equal to one wavelength. Thus, the wavelength of the waves must be 1 m and not 2 m.

River Valley High School Pg 4 of 7 JC 2 H2 Physics 9749 Preliminary Examinations 2024 15. Ans: C Separation between the horizontal threads is determined by the vertical bright fringes. 0.0548 / 2tan 0.01372.00 == O0.7849 = sindm = O9sin 0.7849 2(685 10 )d −= d = 1.00 x 10-4 m = 0.100 mm. For comparison, the spacing of the vertical threads is determined by the horizontal bright fringes: O9sin 0.7849 1(685 10 )d −= d = 5.00 x 10-5 m = 0.0500 mm. 16. Ans: B Initially, since the oil droplet drops with a reduced acceleration, it means the electric force due to the two metal plates is acting in the opposite direction. 𝑊 − 𝐹𝐸 = 𝑚𝑎 Subsequently, when the polarity is reversed, we have 𝑊 + 𝐹𝐸 = 7𝑚𝑎 Thus by subtracting the first expression from the second 2𝐹𝐸 = 6𝑚𝑎 2 𝑞𝑉 𝑥 = 6𝑚𝑎 𝑞 = 3𝑚𝑎𝑥 𝑉 17. Ans: A The electric field is directed upwards in order to produce an upward electric force to balance the downwards weight. Therefore, the lower plate will be at higher potential and upper plate at lower potential. Since the particle is positively charged, when it moves towards the upper plate (lower potential), its electric potential energy will decrease. 18. Ans: A For NTC thermistor, as current increases, the amount of heat generated increases and the equilibrium temperature T increases. As T increases, bonded electrons break free from bonds, increasing the number of ‘mobile charge carriers’, hence decreasing resistance. Since resistance is the ratio of V to I, the graph should show an increasing ratio of I to V.

River Valley High School Pg 5 of 7 JC 2 H2 Physics 9749 Preliminary Examinations 2024 19. Ans: B 31effective resistance of the outer most loop 0 75 31 effective resistance across XY 0 75 1 1 2 75 R . R R( . ) . R ==  + = + + = 20. Ans: A magnetic flux density at centre of coil Q, 00 22 QQ Q NB rr == II magnetic flux density at a distance 2r from wire P at centre of coil Q, ( ) 00 2 2 2 PP PB dr  == II For the resultant magnetic field to be zero at the centre of coil Q, BQ and BP a the centre must be equal in magnitude and opposite in direction. ( ) 0 0 2 2 2 2 2 QP Q P P Q PQ BB rr      = = = = I I II II Since BQ is out of the page at the centre of coil Q, BP must be into the page at the centre. Hence direction of current in wire P flows to the right. 21. Ans: A Since the reading on the balance increased, there must be a downward force acting on the horseshoe magnet and an upward force on the wire according to Newton’s 3rd law. Using Fleming’s left hand rule, since magnetic force on wire is upwards and current is from Y to X, the magnetic field must be from A (North) to the opposite face (South) 22. Ans: A As solenoid X moves away from Y, Y will experience a decreasing magnetic flux through it, therefore, the direction of the induced e.m.f. will be so as to create a magnetic field in the same direction to oppose the decrease. By right hand grip rule, the cur rent will flow from N to M. Since the magnetic North pole for Y is facing the magnetic South pole for X, the force will be attractive in nature. This can also be explained by Lenz’s law as the solenoid Y will need to be attracted so that the motion of solenoid X away from it can be opposed. Lastly, the rate of change of magnetic flux linkage will be decreasing as the magnetic flux density will not be dropping at a linear rate but follows an inverse cube r elationship. Therefore current is diminishing.

River Valley High School Pg 6 of 7 JC 2 H2 Physics 9749 Preliminary Examinations 2024 23. Ans: A 24. Ans: C 2 120000 206000 (20) (3.0) 1200 Pcurrent A V power dissipated in wire W = = = == 25. Ans: A 00 22sin( ), 1.5 12.6x x t where and x T 