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 ch