2025 RVHS H2 Physics P2 Soln Prelim

River Valley High School Page 1 of 10 J2 H2 Physics 9749 2025 JC2 Preliminary Examinations 2025 RVHS JC2 H2 Physics Preliminary Examinations Paper 2 Mark Scheme Questions Answers Marks 1 (a) (i) The gas molecules move randomly (and rapidly) B1 (ii) The volume of their molecules is negligible compared to the volume of the containing vessel. B1 (b) (i) The gas molecules move in three dimensions (or x- , y- and z- direction.) B1 (ii) Using 𝑝 = 1 3 𝜌 < 𝑐2 >, we get 101 000 = 1 3 × 1.25 × √𝑣 M1 𝑣 = 492 m s−1 A1 (iii) Kinetic energy of ideal gas ∝ 𝑇, and Kinetic energy of ideal gas ∝ 𝑣𝑟𝑚𝑠2 so 𝑣𝑟𝑚𝑠2 ∝ 𝑇 M1 𝑣 492.34 = √273.15 + 127 273.15 𝑣 = 596 m s−1 award M1 mark if temperature not given in thermodynamic temperature, but A0 A1 Questions Answers Marks 2 (a) Its acceleration is directly proportional to its displacement from a fixed point (equilibrium position) and is always directed towards that fixed point. B1 (b) gain in gPE = mgh C1 (0.150)(9.81)(1.0 × 10−3) = 1.4715 × 10−3 ~ 1.5 × 10−3 J A1

River Valley High School Page 2 of 10 J2 H2 Physics 9749 2025 JC2 Preliminary Examinations (c) (i) (ii) B1 B2 (d) From calculation in (b), total energy is 1.5 × 10−3 J So the horizontal displacement should be 40 mm B1 (e) Direction of displacement must be opposite to that of velocity, i.e. if line is in positive displacement, velocity must be in negative region Subsequent quarter of oscillation must decrease in amplitude and magnitude of velocity. B2 TE KE

River Valley High School Page 3 of 10 J2 H2 Physics 9749 2025 JC2 Preliminary Examinations Questions Answers Marks 3 (a) f v= m185.0 1780 330 = = A1 (b) 22 1 412 +=DS = 12.649 m Path difference = 12.649 – 12 18539.0 649.0= 5.3= M1 A1 (c) Since the two sources are in antiphase, the waves meet at D exactly in phase (Phase difference is 0 rad) and there is constructive interference. Hence a maximum intensity is detected. B1 A1 (d) Since path difference remains unchanged and there are two more changes in sound intensity, Path difference = (3.5 + 2) new = 5.5 new (new is the new wavelength) 5.5 new = 3.5  )18539.0(5.3)(5.5 = newf v fnew = 2797 Hz = 2800 Hz (3s.f.) B1 C1 A1 Questions Answers Marks 4 (a) Resistivity () is the proportionality constant between the dimensions of a specimen of a material and its resistance (that is constant at constant temperature) such that A LR = B1 (b) (i) N.B. If the soil is acidic, it reacts with copper and this produces an e.m.f. This occurs due to a galvanic reaction, where two dissimilar metals (in this case, copper plates) in an electrolyte (the acidic soil) create a voltage. V = 1.398 – 0.281 = 1.117 V I = 0.31x10-3 A R = V/I = 3603 ~ 3600 Ω M1 (ii) R = ρl/A, thus ρ = RA/l = 3600 x 0.800 x 0.210 / 0.900 = 670 Ω m (2 s.f.) M1 A1 (iii) Any one of these:

River Valley High School Page 4 of 10 J2 H2 Physics 9749 2025 JC2 Preliminary Examinations • Among all the readings given, the least significant or most imprecise is the current reading (only 2 s.f.). The current reading will be subject to significant random errors. EITHER: Increase the area of the copper plates in the soil OR descrease the distance between copper plates. This will decrease the resistance of the sample of soil to be measured and increase the current readings for the same voltage applied. Main point is to increase measured current, so that it will be more than 2 s.f., as given in the question. M1 A1 (c) (i) Power = 246.0 1222 ==R V W A1 (ii) 1 mark for correct graph (at least two periods) for ac supply 1 mark for labelling of values on axes B2 (iii) Power input = Power output 0.7V1 I1 = V2 I2 0.7(230) I1 = 24 I1 = 0.15 A M1 A1 (iv) Voltage can be easily stepped up or down using a transformer in order to reduce power loss in transmission wires. B1 24 48 P / W t / s a.c. 0.02

River Valley High School Page 5 of 10 J2 H2 Physics 9749 2025 JC2 Preliminary Examinations Qn Answer Mark 5 (a) • electric field due to point charges are radial. B1 • So vectors can only cancel away completely if they are collinear / parallel. Hence ... Alternative • electric field due to point charges are radial. • So vertical components of vectors can only cancels away completely along the line joining the two charges. B1 (b) Show understanding that electric field for point charges ∝ 1 𝑟2 B1 Deduce that P is at left of A B1 1 4𝜋𝜀0 2.4 𝑥2 = 1 4𝜋𝜀0 2.9 (0.15 + 𝑥)2 M1 • 𝑥 = 1.51 m A1 Example of alternatives define P distance x to be right of B 1 4𝜋𝜀0 2.4 (0.15 + 𝑥)2 = 1 4𝜋𝜀0 2.9 𝑥2 𝑥 = −1.66 m and left of B C1 B1,M1 A1 (c) B1: line cuts AB nearer to lesser charge B1: line curves towards lesser charge. B1 ; B1 Questions Answers Marks 6 (a) (i) (ii) B1 B1 B F

River Valley High School Page 6 of 10 J2 H2 Physics 9749 2025 JC2 Preliminary Examinations (iii) F = BIL = 𝜇0𝐼1𝐼2𝐿 2𝜋𝑟 M1 = 𝜇0(5.0)(7.0) 2𝜋(3.0) =2.33310−6 ~ 2.310−6 N m−1 A1 (b) Faraday's law of electromagnetic induction states that the e.m.f. induced in a conductor is directly proportional to the rate of change of magnetic flux linkage (or the rate of cutting of magnetic flux) B1 (i) For solenoid → 𝐵 ∝ 𝐼 B1 (ii) From Faraday’s Law ( ) ( )cosd N d NBAE dt dt = − = − → 𝐸 ∝ − 𝑑𝐵 𝑑𝑡 B1 (iii) =(4  10–7)(10100)(I) = 1.0  10–3 M1 → I = 0.7958 A ~ 796 mA A1 InB 0= InB 0=  B

River Valley High School Page 7 of 10 J2 H2 Physics 9749 2025 JC2 Preliminary Examinations 7 (a) A photon is defined as a quantum of electromagnetic energy. (b) (i) 1.0 V By C.O.E, EPE = KE_max EPE = (1.0)( 1.6  10–19) = 1.6  10–19 J (ii) From principle of conservation of energy, hf = Φ + KEmax KEmax = eVs = ½ mv2 [M1] v =√ (2(1.6  10−19)(1)) (9.11  10−31 ) = 592673 ~ 590 km s-1 [A1] (iii) For Φ, E = hc/λ = (6.63  10−34 )(3.00  108 ) 365  10−9 =5.4493 10 − 19 [M1] = 3.4058 ~ 3.4 eV Φ = 3.4058 – 1.0 eV ~ 2.4 eV [A1]

River Valley High School Page 8 of 10 J2 H2 Physics 9749 2025 JC2 Preliminary Examinations Questions Answers Marks 8 (a) After fission, products have smaller nucleons with higher binding energy per nucleon ( more stable) , so energy will be released. Accept other equivalent statement. B1 (b) • Δ𝑚𝑎𝑠𝑠 = 235.123 + 1.009 − (94.945 + 138.955 + 2 × 1.009) = 0.214𝑢 B1 • 𝑬 = 𝒎𝒄𝟐 = 0.214 × 1.66 × 10−27 × (3.0 × 108)2 = 3.197 × 10−11J M1 • 𝐸 = 3.197×10−11 106×1.6×10−19 = 199.8 MeV = 200 MeV A1 (c) • number of reactions = 6.02×1023 mol-1 235 g mol-1 ×1.0 g = 2.56 × 1021 reactions or • number of reactions = 1×10−3 kg 235.123×1.66×10−27 kg = 2.56 × 1021 reactions M1 • 𝐸 = 2.56 × 1021 × 3.197 × 10−11 = 8.19 × 1010 J A1 (d) • energy required from nuclear reactions 500×106×60×60 0.3 = 6.0 × 1012 J M1 • 6.0×1012 J 8.19×1010 J g-1 = 73.3 g A1

River Valley High School Page 9 of 10 J2 H2 Physics 9749 2025 JC2 Preliminary Examinations Questions Answers Marks 9 (a) (18 000 − 3 000) kg 5 kg km−1 = 3 000 km B1 (b) 3 000 kg 5 kg km−1 = 600 km B1 (c) 42 000 + 18 000 + 15 000 = 75 000 kg total mass = mass of plane + mass of fuel (at full capacity) + mass of 150 passengers B1 (d) Fig. 9.2 vertical forces are balanced lift force + normal contact force + weight = 0 B1; B1 (e) Using 𝑣2 = 𝑢2 + 2𝑎𝑠, 𝑢 = 0, gives 752 = 0 + 2(𝑎)(1500) M1 𝑎 = 1.88 m s-2 A1 (f) Using 𝐹 = 𝑚𝑎 gives