2025 RVHS H2 Physics P1 Soln Prelim

River Valley High School Pg 1 of 4 H2 Physics 9749 JC2 Preliminary Examinations 2025 2025 H2 Physics Preliminary Examination Paper 1 Qn Ans Qn Ans Qn Ans 1 B 11 B 21 B 2 C 12 C 22 D 3 C 13 C 23 D 4 D 14 C 24 B 5 D 15 B 25 A 6 B 16 D 26 D 7 A 17 A 27 B 8 D 18 A 28 A 9 B 19 C 29 D 10 A 20 C 30 A Qn Key guide 1 B Area = [L][L] → 1 SI base unit Charge = [I] [T] → 2 SI base units Current = [I] → 1 SI base unit Force = [M] [L] [T]−2 → 3 SI base units 2 C An error is systematic if repeating the measurement under the same conditions yields readings with error of the same magnitude and sign. i.e. ALL measurements are either bigger or smaller than the true value consistently. Readings with systematic error change in a predictable manner depending on the conditions. So to reduce systematic error, the source of error has to be identified and removed. Zeroing an instrument is one of such examples. 3 C Jumper at rest when v = 0; jumper with zero acceleration when gradient of v – t graph is zero. 4 D Using conservation of momentum 0 = 𝑚𝛼𝑣𝛼 + 𝑚𝑋𝑣𝑋 giving 𝑣𝛼 𝑣𝑋 = 𝑚𝑋 𝑚𝛼 = 3.89 × 10−25 6.65 × 10−27 = 58 5 D acceleration of the system = 𝐹 𝑀1+𝑀2+𝑀3 Newton’s 2nd law for M3 ( 𝐹 𝑀1+𝑀2+𝑀3 ) 6 B Work done is the area under F-x graph. 7 A There are different modes of energy conversion in this scenario, e.g. change in GPE, change in KE as well as work done on the mass due to frictional force. Hence the heat generated is simply Fx, the distance travelled by the system, since the cord is inelastic. 8 D P = Fv Using v = u + at, v = at However, v is non uniform. Have to use average velocity. (v – u)/2 = v/2 Hence, P = ½ aFt Plotting P against t, gives gradient of ½ aF, through origin. 9 B Only actual forces should be presented on a FBD, so there should only be the weight of the aircraft and lift force on the aircraft, directed normal to the plane of the wings.

River Valley High School Pg 2 of 4 H2 Physics 9749 JC2 Preliminary Examinations 2025 10 A At steady state bulb X and Y have the same pressure, so 𝑉𝑋 𝑛𝑋𝑅𝑇𝑋 = 𝑉𝑌 𝑛𝑌𝑅𝑇𝑌 giving 𝑛𝑋 𝑛𝑌 = 𝑉𝑋 𝑉𝑌 𝑇𝑌 𝑇𝑋 , but for the same gas, 𝑛 ∝ 𝑚 ( since 𝑛 = 𝑚 𝑚𝑅 ) Hence 𝑚𝑋 𝑚𝑌 = 𝑉𝑋 𝑉𝑌 𝑇𝑌 𝑇𝑋 = 2 ( 400 200) = 4 so, 𝑚𝑌 = 𝑚 4 11 B work done = 𝑝Δ𝑉 = 𝑝𝑓𝑉𝑓 − 𝑝𝑖𝑉𝑖 = 𝑛𝑅𝑇𝑓 − 𝑛𝑅𝑇𝑖 Since the process happens at constant pressure when volume doubles, temperature doubles. so work done = 𝑝Δ𝑉 = 𝑛𝑅𝑇𝑓 − 𝑛𝑅𝑇𝑖 = 𝑛𝑅(2𝑇 − 𝑇) = 𝑛𝑅𝑇 work done = 𝑅𝑇 when n = 1 12 C Option A, B and D are not correct. Internal energy is also affected by the amount of substance in the system. 𝑈 = 3 2 𝑁𝑘𝑇 13 C Using g = GM/r2 and ρ = mass / volume = M / [(4/3)πr3] , g = G[ρ(4/3)πr3 ] / r2 = (4/3)Gρπr gE/gM = ρErE / ρMrM = ρErE / ρMrM 6 = (5/3) / (rE / rM) rE / rM = 3.6 14 C Change in GPE = mΔϕ = 50 (-60-(-20)) x 106 = - 2000 MJ Negative sign indicates a loss of GPE. The mass moves closer to the Earth 15 B In SHM, the restoring force is changing. → left with options B and C Since options B and C, the only difference is “total energy” and “angular frequency”, and that the total energy of any closed system should be conserved. 16 D Velocity variation with displacement is an ellipse about the origin. Since question is on speed, which is non-directional, the vertical axis should focus only on the positive region. 17 A Only transverse waves can be polarised. Intensity will change when view light through a polarised sheet. 18 A Since the wave is moving to the right, R will following the behaviour of the particles to its left, which currently have displacements to the right. B: The distance between P and T is two wavelengths. C: Q vibrates about its position. D: R is not a node. 19 C y1 = (600 x 10-9)(2) / (0.30 x 10-3) = 4.0 mm y3 = 12.0 mm y3 – y1 = 8.0 mm 20 C When R = r = 3 Ω, Commented [TYY(1]: Include potential energy

River Valley High School Pg 3 of 4 H2 Physics 9749 JC2 Preliminary Examinations 2025 P = I2R= ( 12 6 )2(3) = 12 W When R = 6 Ω, Pnew = ( 12 9 )2(6) = 10.67 W Pnew = 0.89P 21 B vd=neAI A2=2A1 vd2 = I/neA2 = I/ne(2A1) = vd1 /2 = 2.0 × 10−4 / 2 = 1.0 × 10−4 m s−1 22 D As oil drop rises at constant speed, no net force is acting on the oil drop and drag force acts downward, hence 𝑚𝑔 + 𝐷 = 𝑞 𝑉1 𝑑 . As oil drop falls at the same constant speed, no net force is acting on the oil drop, drag force acts upwards and has the same value as before, so 𝑚𝑔 = 𝑞 𝑉2 𝑑 + 𝐷 Adding the two equations, we can eliminate D on both sides, giving. 2𝑚𝑔 = 𝑞 𝑑 (𝑉1 + 𝑉2), so 𝑞 = 2𝑚𝑔𝑑 𝑉1+𝑉2 23 D By potential divider principle, voltmeter reading increase when effective resistance across thermistor is increased or resistance R3 is reduced. 24 B Current source at Q cannot produce the magnetic field at Q. Hence consider only P. Using right hand grip rule, the direction of current must be out of the plane at P. 25 A Due to the direction of current flow in X, the end of X facing Y will produce a magnetic south. As X moves away from Y, the solenoid in Y detects a reduction in magnetic south, and hence according to Lenz’s Law, will generate an induced e.m.f. that may produce an effect to oppose that reduction in magnetic south, i.e. generating more magnetic north at the end of Y facing X. The nature of force will be attractive, while the induced current in Y will be flowing from N to M. The induced current will be decreasing, since according to Faraday’s Law, ∆𝜑 ∆𝑡 is decreasing as a result of reducing flux linkage. 26 D B = μ0nI ∝ I E = -N Φ/t = -N BA/t = -NA B/t Identify the gradients of the I-t graph at various points and factoring in Lenz’s law 27 B Rms current =√(42×6)+ (62×2) 10 = 4.10 A

River Valley High School Pg 4 of 4 H2 Physics 9749 JC2 Preliminary Examinations 2025 28 A The average power of R without diode = Irms2R = 0.50 Io2R When the diode is connected in the circuit, the average power of R will be halved (0.25 Io2R) since it only received power for half the period. 29 D Derive using wave particle duality and KE 30 A Only change is nucleon number is only affected by 𝛼-particle. So there are 238−206 4 = 8. So conservation of proton number gives 92 = 82 + 8(2) + n(−1) where n is the number of 𝛽− particles. So n = 6