2023 JPJC Prelim H2 Phy P1 Solutions

1 2023/JPJC/Prelim/9749/01 Answers to 2023 JC2 Preliminary Examination Paper 1 (H2 Physics) Answer: 1 B 2 B 3 C 4 C 5 A 6 A 7 D 8 B 9 C 10 B 11 C 12 A 13 D 14 C 15 D 16 B 17 A 18 D 19 A 20 A 21 C 22 A 23 C 24 D 25 B 26 D 27 B 28 B 29 D 30 A Suggested Solutions: Qn Ans Solution 1 B R = σT4  4 R Tσ = Unit of 12 4 J s m Kσ −− = = (kg m2 s−2) s−1 m−2 K−4 = kg s−3 K−4 2 B Z = X − Y  X = Z + Y 3 C By conservation of energy, 22 fi 11 22mv mv mg h= +∆ Since, the magnitude of the loss in gravitational potential energy is the same for stones P and Q, with the same initial kinetic energy, the stones will end up with the same magnitude of kinetic energy, implying they will hit the ground with the same speed. 4 C Using 21 2s ut at= + , distance travelled in first 10 s is 24 m, ( ) 2 2 124 102 0.48 m s a a − = ∴= Distance travelled in the first 25 s, ( )( ) 21 0.48 252 150 m s = = Distance travelled from 10 st = to 25 st = 150 24 126 m = − = 5 A ( ) ( ) 2 For A: 2.0 ........... (1) For B: (5.0)(9.81) 5.0 ............ (2) Sub (1) into (2): 49.05 2 5 7.0 m s Ta Ta aa a − = −= −= =

2 2023/JPJC/Prelim/9749/01 Qn Ans Solution 6 A By principle of conservation of linear momentum, ( )2 3( ) ( 3)m v m v m mV+ −= + 1 4Vv=− The speed of mass m is 1 4 v after the collision. 7 D Weight of boat = upthrust = weight of fresh water displaced = 35.6 kN Weight of boat = weight of salt water displaced = mg = ρVg = 35.6 kN 33 3 35.6 10 35.6 10 (1024)(9.81) 3.54 m V gρ ××= = = 8 B Extension for spring X, x1 = 1 1.5 0.30 m5.0 F k = = Extension for spring Y, x2 = 2 1.5 0.15 m10.0 F k = = Elastic P.E. stored = 21 2 kx Total elastic P.E. stored = 1 2 (5)(0.3)2 + 1 2 (10)(0.15)2 J = 0.34 J 9 C Loss in G.P.E. by Y = gain in G.P.E. by X + gain in K.E. by X and Y ( ) 213.0 9.81 3.0sin40 2.0 9.81 3.0sin30 2.0 3.0 2 v×× ° = ×× ° + + 3.3v ≈ ms−1 10 B For mass m1: 2 1 2 1 () mv keLe ke L ev m =+ += For mass m2: 2 2 2 2 2 2 (2 2 )2( ) ( 2)2( ) 2 ( 2 )( ) mv kL eLLe mv kL eLe kL e L em v = +−+ = ++ ++= 1 1 2 ( 2 )( ) k( ) 2 ( 2) kL e L em eL e mL e e ++= + +=

3 2023/JPJC/Prelim/9749/01 Qn Ans Solution 11 C total energy at top of ramp = total energy at lowest point 2 2 1 mvmgh =  ghv 22 = By Newton’s second law of motion, at the lowest point, r mvmgR 2 = − ( ) mgr ghmR + =2 Since r = h, mgR 3= 12 A Gravitational potential at infinity is zero. Moving from −30 MJ kg−1 to −70 MJ kg-1 means moving closer to Earth. Potential energy change = 50 [−70 − (−30)] = −2000 MJ 13 D A is true: From pV = nRT, at constant pressure, when volume is doubled, temperature is doubled. B is true: work done W = p∆V = 4Po(2Vo − Vo) C is true: W = 0  ∆U = Q. From pV = nRT, since there is a decrease in pressure, there must be a decrease in temperature and hence a decrease in internal energy, i.e. ∆U is negative. This means Q is negative, i.e. heat is removed. 14 C From pV = nRT, when p and V are tripled, the temperature would be 9T New temperature = (200 + 273) × 9 = 4257 K = 4257 − 273 = 3984 = 4000 °C From 22 2 311 2 22 33kT kTm c mv kT v v mm= = → = →= , when the new temperature is 9T, the new r.m.s. speed = 3v 15 D Option A: The kinetic energy of the oscillator is proportional to square of the frequency of its motion. Incorrect. Option B: The potential energy of the oscillator is maximum when the oscillator is momentarily at rest. Incorrect. Option C: The kinetic energy of the oscillator is maximum when the oscillator is at the equilibrium position. Incorrect. Option D: When the kinetic energy of the oscillator is equal to its potential energy, the oscillator is neither at the rest position nor at the maximum displacement positions. Correct.

4 2023/JPJC/Prelim/9749/01 Qn Ans Solution 16 B Using 00vx ω= , 0 0 0 2 3.0 0.063 2 0.03 m vx v T ω π π = = ×= = 2 00 2 0 2 2 2 2 0.030.063 298 300 m s ax xT ω π π − = = = × = ≈ 17 A Let the speed of the wave be v. The wavelength 12.512.5 vv vfλλ== →= For the displacement to be zero at Q, the wave travels a distance 0.1258d λ λ= = . The time taken for this = 0.125 0.010 s12.5 d v λ λ= =

5 2023/JPJC/Prelim/9749/01 Qn Ans Solution 18 D At lowest frequency, 1 1 1 1 1 4 4 4 L L vvf L λ λ λ = = = = At next highest frequency, 2 2 21 2 3 4 4 3 3 3344 L L vv vff LL λ λ λ = = = = = = At third highest frequency, 3 3 31 3 5 4 4 5 5 5544 L L vv vff LL λ λ λ = = = = = = So the frequencies that can be produced by the instrument are 1111, 3 , 5 , 7 , ...ffff 19 A Option A: Correct 9 6 590 10 0.295 m 0.30 m 2.0 10 bb λλθ θ − − ×=⇒== = ≈ × Option B: Incorrect Red stars give out red light of wavelength of 700 nm. 9 6700 10 2.3 10 radians0.30b λθ − −×= = = × which is larger than the angular separation of 62.0 10 radians−× . Hence, the red stars cannot be resolved. Option C: Incorrect The resolving power of the telescope is dependent of the wavelength of light, according to the expression b λθ = . Option D: Incorrect The angular separation between the stars is 62.0 10 radians−× for the telescope to be able to resolve. Hence, the telescope cannot be used to distinguish the stars.

6 2023/JPJC/Prelim/9749/01 Qn Ans Solution 20 A For the first 2.0 m, the movement is along the direction of increasing potential which resulted in positive work done by the electric field. Positive work done by field = ( )( )( )5.0 4.0 2.0 40 = J Thus, work done against electric field = −40 J. For the next 3.5 m, movement is along the same potential, hence, there is zero work done. Therefore, total work done against electric field = −40 J. 21 C By conservation of energy, for the electron to just reach plate AB (that is, , work done against electric force = loss in kinetic energy of electron . 22 A Effective resistance is made up of copper and iron resistors in parallel. Per unit length, Copper resistance 8 6 3 1.7 10 8.5 10 2.0 10 − − − ×= = ×Ω × Iron resistance 7 4 3 1.0 10 1.0 10 1.0 10 − − − ×= = ×Ω × Effective resistance per unit length 1 6 64 11 7.8 10 8.5 10 1.0 10 − − −− = += × ×× Ω m−1 23 C The smallest reading on the ammeter means smallest current passing through it. Ammeter has a resistance of 3 Ω. Option A: Incorrect. Current passing through ammeter, 3 V=I , because ammeter is in parallel connection with the 1 Ω and 2 Ω resistors. Option B: Incorrect. Current passing through ammeter, 3.67 V=I . Option C: Correct. Current passing through ammeter, 6 V=I . Option D: Incorrect. Current passing through ammeter, 4 V=I . 24 D For zero deflection, magnetic force = electric force Bqv qE= Ev B= perpendicular to both B and E. ) vy 0= ( ) 2 2 1 θcos v meV e=

7 2023/JPJC/Prelim/9749/01 Qn Ans Solution 25 B The current in the wire will result in a magnetic field coming out of the coil according to the right hand grip rule. As the current in the wire increases, the B -field through coil increases. Applying Lenz Law and right hand grip rule to the coil, the induced B-field must be into the coil. Thus, induced current in the coil is clockwise. 26 D The a.c. is half-wave rectified with a single diode.. Voltmeter reads the r.m.s. voltage, V = 2.5 V So peak voltage across NB = 2(2.5) = 5.0 V By potential divider principle, 100 100 100 (5.0) 12.5 V40 40 40 AB AB AB o NB NB NB VL VV VVL  == →= = = =    Peak power = 2 212.5 15.6 16 W10 oV R = = = 27 B By de Broglie’s expression, h pλ = , Since 2 22 pE p mEm= ⇒= , 22 1 hh mE meV V λ λ = = ∴∝ 28 B The peaks will remain the same as these are characteristic of the