RVHS H2 Physics P1 Soln

River Valley High School Pg 1 of 6 H2 Physics 9749 JC2 Preliminary Examinations 2022 Solutions to J2 Preliminary Examination Paper 1 1 A 11 C 21 D 2 D 12 A 22 C 3 B 13 B 23 B 4 D 14 D 24 A 5 A 15 C 25 A 6 C 16 C 26 A 7 B 17 D 27 C 8 C 18 B 28 B 9 B 19 A 29 C 10 D 20 B 30 D

River Valley High School Pg 2 of 6 H2 Physics 9749 JC2 Preliminary Examinations 2022 Qn Key guide 1. A Considering units of the equation, we have s2 = kg m2 units of K , so units of K = kg m2 s-2 = (kg m s-2) m = N m 2. D Since air resistance is negligible, the gradient of both graphs will be the same. From the v-t graph, it can be seen that at any time, the difference in the area under graph for X and Y will be increasing hence the separation will be increasing. 3. B Note that the carriages and the engine have a common acceleration, say a. Let the mass of each carriage be m. Consider both carriages, Newton’s 2nd law gives 𝑇 = (2𝑚)𝑎, so 𝑎 = 𝑇 2𝑚 Consider the back carriage. Tension pulling the back carriage is then 𝑚𝑎 = 𝑚 ( 𝑇 2𝑚) = 𝑇 2 4. D During acceleration, 𝑊 − 𝑚𝑔 = 𝑚𝑎, so 𝑊 = 𝑚(9.81 + 2.0) During deceleration, 𝑚𝑔 − 𝑊′ = 𝑚𝑎, so 𝑊′ = 𝑚(9.81 − 2.0) So 𝑊′ 𝑊 = 9.81−2.0 9.81+2.0 = 0.661 5. A Since the same load is applied, for X, if extension is L, then extension for Y will be 2L For Z, the extension will be the greatest as the top two springs will stretch by the same amount of L while the lower spring will have twice the extension due to the weight being supported by only one spring 6. C Work done = 120 cos 37o x 5.0 = 479 = 480 W 7 B Initially, when no force is applied to the piston, force on piston by air in container = force on piston by air in atmosphere = pressure x area = ( )( ) 33100 10 3.5 10 − = 350 N ( )( )( ) 3 3 3100 10 3.5 10 80 10 T −−   = ( )( )( ) 33new pressure 3.5 10 160 10 0.5T −− new pressure = 25 kPa new force on piston by air in container = pressure x area = ( )( ) 3325 10 3.5 10 − = 87.5 N v t x y

River Valley High School Pg 3 of 6 H2 Physics 9749 JC2 Preliminary Examinations 2022 For piston to remain stationary, Force on piston by air in container + F = Force on piston by air in atmosphere 87.5 + F = 350 F = 260 N (262.5) 8 C Let X be the gas compressed isothermally (no change in temp). There is heat exchange between the gas and the surroundings. Let Y be the gas compressed and is isothermally isolated from surroundings. There is no heat exchange between gas and the surroundings. For X: Since temperature does not change due to heat exchange between gas and the surroundings, p1V = p2(0.5V), p2 = 2 p1 For Y: Using pV = nRT, when V decreases, pressure increases and leads to larger speed of the molecules. This leads to higher temperature of gas. A: No heat is given to both gases during compression. B: Internal energy of Y is higher due to higher temperature of gas that leads to higher average KE. C: Since mass and volumes of gases are the same, there is no change in density of gases. D: Since work done on gases depends on the pressure and the change in volume and there is no information on how pre ssure varies for both gases during compression, the work done on gases may not be the same. 9 B The following options are wrong. A: There is no thermal energy supplied to the system. C: If the air becomes hot, there should be increase in internal energy. D: The statement does not lead to the effect of higher temperature of gas. 10 D Using vr = , dv dr dt dt = Since dr dt and  are constant, it should be a straight line graph for a graph of v against t. 11 C Use v = u + at 0 = 45 – gx(5.2) gx= 8.65 m s-2 gy = ΔΦ/Δx = 6.0(4.0) = 1.5 m s-2 gy : gx = 1.5/8.65 = 0.17 12 A Using 2 E k Gm mE r= , is larger than ABmm and since both satellites have same kinetic energy,

River Valley High School Pg 4 of 6 H2 Physics 9749 JC2 Preliminary Examinations 2022 Using total energy 2 EGm m r=− , satellites A and B have the same total energy, (Option A is false) Orbital radius for satellite A is larger than that of satellite B. (Option B is true) Using T2 is proportional to r3, satellite A has a larger period (Option C is true) Since angular velocity is inversely proportional to T, satellite A as smaller angular velocity (Option D is true) 13. B Total energy of mass in S.H.M: ( ) 22 2 2 0 3 1 1 8 5 2 (40)2 2 1000 1000 6.32 10 J mx −    =        = 14. D With increased damping, amplitude will generally be lower throughout and peak will shift left. 15. C 16 C  = L Points between two adjacent nodes are in phase. Points in adjacent segments are anti-phase. 17 D Using Dx a = , ( )( ) 9600 10 1 x a − = ( ) 9400 10 2 D x a − = D = 3.00 m 18 B Current in circuit, 10 2.5 A4.0 VI R= = = , Potential difference across connecting wires = 12 – 10.0 – (2.5)(0.20) = 1.5 V Power loss in connecting wires = (2.5)(1.5) = 3.75 W

River Valley High School Pg 5 of 6 H2 Physics 9749 JC2 Preliminary Examinations 2022 19 A Using Fa m= and F Eq= , Eqa m= proton: proton Eqa m= alpha particle: (2 ) 42 proton alpha aEqa m== Using v = u + at, 0protonv at =+ 0 2 alpha avt =+ Ratio = 0.5 20 B Electric potential decreases along the direction of electric field strength. 21. D By conservation of charges, 𝐼1 = 𝐼2 + 𝐼3 ---- (1) Also for the parallel circuit, we note that 𝐼2 𝐼3 = 𝑅3 𝑅2 ----- (2) Using (1) and (2) to eliminate 𝐼2, (1) becomes 𝐼1 = 𝐼3 ( 𝑅3 𝑅2 ) + 𝐼3, or 𝐼1 = 𝐼3 ( 𝑅3 𝑅2 + 1), giving 𝐼3 𝐼1 = 𝑅2 𝑅2+𝑅3 22. C 0 2B d  = I Since we want to find the distance at which the PEAK flux density would be, that would occur when the PEAK current is flowing through the circuit. ( )06 2000 2 100 10 2 d   −  = 5.66d = m 23. B Current-carrying conductors will attract when currents are in same direction, but repel when in opposite directions. Hence, PS will attract each other while PQ and PR will repel each other, since PR are further apart, the repelling force will be weaker than the force between PS and PQ, 24. A There is maximum cutting of the magnetic field when the coil is horizontal therefore the induced e.m.f. will be the greatest. OR Since dNE dt =− , the graph of magnetic flux linkage and that of induced e.m.f. will be out of phase by π/2, thus when magnetic flux linkage is zero, induced e.m.f. will be maximum.

River Valley High School Pg 6 of 6 H2 Physics 9749 JC2 Preliminary Examinations 2022 25. A According to Faraday’s law, dNE dt =− . Magnitude of induced e.m.f. in coil = 22 3 5.00( (8.0 10 ) ) 10.1 V10.0 10 BA t  − − = = = Since resistance of coil is 4.00 Ω, current through coil = 10.1 / 4.00 = 2.51 A According to Lenz’s law, direction of induced e.m.f. is to oppose the change causing it. Thus the direction of induced e.m.f. will produce a current to oppose the decreasing flux into the coil. Hence by right hand grip rule, the current will be flowing clockwise in order to produce flux going into the page. 26. A Using turn ratio, voltage across the secondary coil is 120 × 10 500 = 2.4 V So current in secondary coil is 2.4 15 = 0.16 A. For ideal transformer, 120 × 𝐼𝑝 = 0.16 × 2.4 so 𝐼𝑝 = 0.0032 A 27. C Diffraction is a phenomenon exhibited by waves. Electron undergoing diffraction shows that it has wave property. 28. B Energy (= eV) transferred to the electron and proton is the same, since they have the same magnitude of charge, so since 𝐸 = 𝑝2 2𝑚, their momentum is given by 𝑝 = √2𝑚𝐸 , so their associate wavelength is 𝜆 = ℎ 𝑝 = ℎ √2𝑚𝐸. Hence 𝜆 ∝ 1 √𝑚 or 𝜆𝑝 𝜆𝑒 = √ 𝑚𝑒 𝑚𝑝 = √9.11×10−31 1.67×10−27 = 0.0234 29. C Original count rate per minute due to source = 532 – 24 = 508 per minute After two half-lives, the count rate would have dropped to 0.5*0.5*508 = 12