TMJC 2025 H2 Physics P1 Soln

1 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics Tampines Meridian Junior College 2025 JC2 H2 Physics Preliminary Examination Paper 1 Suggested Solution 1 D 11 B 21 C 2 B 12 A 22 B 3 C 13 C 23 A 4 D 14 B 24 D 5 B 15 A 25 C 6 B 16 D 26 B 7 B 17 C 27 B 8 D 18 D 28 D 9 C 19 C 29 C 10 A 20 B 30 D Q1 Ans: D Q2 Ans: B Gradient of the graph started with maximum magnitude before slowing decreasing and becoming zero. Acceleration (given by gradient of V-t graph) of an object falling in the presence of air resistance will decreases until terminal velocity is reached and become zero. Thus option B is the answer. Q3 Ans: C To ensure rotational equilibrium, the line of action of the forces must pass through a single point. vbicycle relative to bus Vbicycle –vbus 3 N 4 N 5 N

Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics Q4 Ans: D ( ) ( ) Let be tension in string sin (1) (2) (2) (1) sin sinsin T T mg ma Mg T Ma Mg mg Ma ma MmMg mgag M m M m    −= −= + − = + −−== ++ Q5 Ans: B Torque by 30 N couple = 30 (2)(0.065) = 3.9 N m (anti-clockwise) Torque by 25 N couple = 25 (2)(0.12) = 6.0 N m (clockwise) Torque by 15 N couple = 15 (2)(0.10) = 3.0 N m (clockwise) Net torque = 3.0 + 6.0 – 3.9 = 5.1 N m (clockwise) Q6 Ans: B Area under the graph gives the elastic potential energy of the spring. As extension is doubled from x1 to x2, area changes from ½ F1x1 to ½ F2x2. Hence change in EPE= ½ F2x2 - ½ F1x1 Q7 Ans: B ( ) ( )( ) 1 8.4 16 0.525 rad s 0.525 8.0 4.2 rad vr t     − = = = = = = Q8 Ans: D Angular speed and the linear speed are different ways of describing the same motion; they do not cause each other. By Newton’s second law, net force causes acceleration and hence changes in both angular speed and the linear speed simultaneously. M θ a Mg T

3 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics Q9 Ans: C 2 2 2 2 For an object of mass orbiting around E arth, Gravitational force provides for centripetal force 1 Since they are both orbiting around same Earth, 1 Earth Earth ISS satellit m Mm mvG rr GM v r vr v v =  =   2 4 -1 3 4.2 103.1 7.7 km s6.8 10 satellite e ISS satellite ISS satellite ISS r r rvv r  =    = = =    Q10 Ans: A The potential energy – time graph should be a sinusoidal squared graph. As the displacement at t1, t2, t3 is zero, the speed is the maximum, kinetic energy is maximum, and hence potential energy is zero. Q11 Ans: B 2 kHz has a period of 0.5 ms. A typical CRO screen has 10 horizontal divisions of 1 cm each. The period will span over 5 divisions with a 0.1 ms cm−1 setting.

Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics Q12 Ans: A The two supports have to be node. Taking distance between 2 walls as L, the wavelength of the given wave is 2L, and the next possible stationary wave has wavelength L. = = ==' 2 2 vf vf L vff L Q13 Ans: C sin 35 5 3 xy x y dn     = = = Q14 Ans: B Using first law of thermodynamics U Q w = + , Q = 0 and W = +ve for adiabatic compression. Hence U and temperature increased. Q15 Ans: A Using =  =   =0.200 390 30 2300 JQ mc since final temperature at thermal equilibrium is 50 oC. Q16 Ans: D KE = 3/2 Nk T  = = = 2 22 2 2 1 2 ( ) 1.2 1.44 350 500 K rms rms rms cT cT T c T

5 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics Q17 Ans: C Force on a charge in the field is given by F= q E. Since it’s a uniform electric field, E is constant. Both proton and electron have the same magnitude of charge q. Therefore the proton should experience the same magnitude of force as the electron. The other options are correct statements because Option A: Since the work done by the electric force on both the electron and proton is positive, therefore change in electric potential energy is negative for both (Wby electric force = − ΔU) Hence they both lose electric potential energy. Option B: The proton experiences an upward electric force that is of same magnitude as that on the electron but has a larger mass. Therefore it will experience a smaller acceleration and undergo a smaller deflection. Hence the work done by electric force is smaller for proton, giving it a smaller U . Option D: The proton loses less electric potential energy than the electron. By conservation of energy, the gain in kinetic energy must be less than that of the electron. Q18 Ans: D The resistors are connected in parallel to the power supply. − = + =  == = + = 1 22 11 772 1300 1900 240 0.31 A772 240 240 75 W1300 1900 effective total R P I Q19 Ans: C At about 2.4 V, the graph of the diode and lamp intersects. Hence the V and I values are the same. Since R = V/I, the diode and lamp have the same resistance. Q20 Ans: B For the voltmeter reading to be highest, the combined resistance of the thermistor and LDR has to be lowest (by potential divider principle). High brightness (i.e. high intensity of light) would give LDR lowest resistance. High temperature would give thermistor lowest resistance.

Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics Q21 Ans: C As electron moves to the left, conventional current (middle finger) will be to the right. Using right hand grip rule for wire, magnetic field acting on electron will be out of paper. Applying Flemming’s left hand grip rule, force on electron Q22 Ans: B For wire to start to lift off, magnetic force= weight of wire. = = = sin 0.004(9.81) 0.040(0.30)(sin30 ) 6.54 o mg B L A I I I Q23 Ans: A To determine the induced e.m.f.: cosBLv= where v is the velocity of rod at the bottom of slope, and L is length of rod. To determine v: ( ) ( ) 22 2 2 0 2 sin 2 sin v u as v g y v y g   =+ =+ = Hence, ( ) ( )( )cos 2 sin cos 2 sin cosBLv BL y g BL yg      = = =    /  Induced emf / V A 35 0.620 2BL yg (largest) B 40 0.614 2BL yg C 45 0.595 2BL yg D 50 0.563 2BL yg

7 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics Q24 Ans: D Inserting an iron core concentrates the magnetic field lines and provides better linkage between coils P and Q. Hence the flux linkage through both coils will be larger and amplitude of the induced voltage in coil Q will increase. Option A is incorrect – although increasing the frequency will increase the amplitude in coil Q (due to a faster rate of change of flux linkage), the number of cycles per unit time (i.e. frequency) will also increase. Option B is incorrect – decreasing the cross-sectional area of coil P will have no effect on the flux density created by coil P. Hence no change to the flux linkage of both coils. Option C is incorrect – decreasing the number of turns in coil Q will result in a lower amplitude as QQ Pp VN VN= . Q25 Ans: C Option A: 22 0 60 22Mean power dissipated = 225 W8.0 8.0 V          == Option B: Mean power dissipated = ½ (225) = 112.5 W Option C: ( ) 22 0 60Mean power dissipated = 450 W8.0 8.0 V == Option D: Mean power dissipated = ½ (450) = 225 W Q26 Ans: B Current from the a.c. supply can flow in both directions around the circuit. But the setup of diodes result in current flowing through R to be only in the upward direction. Q27 Ans: B ( ) ( )( )  − −− − − − =  =  = = = =   2 16 31 2 16 71 11 31 7 1 9.3 102 1 9.11 10 9.3 102 4.52 10 m s 1.6 10 m 9.11 10 4.52 10 mv v v hh mv

Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Physics Q28 Ans: D ( )( ) ( ) −− −−           9 34 26 1 7.2 10 6.63 10 9.2 10 kg m s x p h p p Since the uncertainty in the position