VJC 2022 Prelim P1

1 2022 VJC Prelim H2 P1 Suggested solution Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans 1 D 6 D 11 A 16 A 21 C 26 D 2 B 7 B 12 A 17 C 22 A 27 D 3 B 8 B 13 D 18 A 23 C 28 A 4 A 9 B 14 D 19 B 24 B 29 C 5 C 10 B 15 B 20 D 25 D 30 D 1. Ans: D Base units of self-inductance = 22 2 2 2 1 1 1 11V W kgm s kgm s AAs A As As As − −− − − − = = =   2. Ans: B For the rod to be in equilibrium, the 3 forces acting on the rod should extrapolate and intercept at a common point. The 3 forces should form a vector triangle with the forces pointing in a single direction. 3. Ans: B Total momentum = 2Mv Mv Mv−= Option A is a correct statement (so don’t choose it). The spheres cannot come to rest at the same time as that will give a total momentum of 0 and not Mv. Option C is a correct statement. For total momentum to be conserved, the change in momentum of A and B must be equal and opposite. Option D is a correct statement. Newton’s 3rd law. T R F R T

2 4. Ans: A Work done by F on object = area under the graph = K + L + M Work done against frictional force = L + M (friction is constant, = 5 N) Net work done by F = work done by F – work done against frictional force = K + L + M – (L + M) = K = gain in kinetic energy of the object 5. Ans: C Output power, 0.25 (0.25)(10) 2.5oiPP= = = W (20)(0.50) 4.0 s2.5 o oo EP t E mght PP = = = = = 6. Ans: D l r   T mg For circular motion, Horizontally, 2sinT mr= where T = tension in string  = angle between string and vertical. m = mass of bob r = radius of circular motion. ( )( ) 2 22 2 sin sin 2 4 1 T m l f Tl mf l f     = =  Note: T is constant as it is equal to the weight of the brass weights.

3 7. Ans: B Magnitude of gravitational acceleration = potential gradient At the height of 370 km, 3 -1 3 [ 617.0 ( 649.6)] 10 1.63ms[380 360] 10g − − − == − 8. Ans: B At a higher orbit, r is increased, Gravitational force decreases 2G GMmF r= Gravitational potential energy increases GMmU r=− Linear speed decreases 2 2 GMm mv GM vr r r=  = Kinetic energy decreases 2 2 2 1 22 GMm mv GMm mvr r r=  = 9. Ans: B Heat was removed from it at a constant rate when it was totally liquid, and when totally solid. 400 300 300 200 100 50 2 liquid solid liquid solid liquid solid liquid solid liquid solid TTmc mc tt cc cc     =       −−    =       = 10. Ans: B ( ) ( ) 2 22 1 1 1 3 since 22 211 rms rms rms RTC T PV PV nRTM C PV C PV =   = = = = 11. Ans: A At U, velocity is negative (negative gradient of x -t graph), acceleration is positive (since displacement is negative). At Y, velocity is positive, acceleration is negative.

4 12. Ans: A Max force = m amax = m 2 x0 = 20 x 10-3 (3)2 (6 x 10-3) = 0.011 N 13. Ans: D The peak of B occurs at t = 0, while the peak of A occurs one quarter of a cycle later. So A lags B by 90o. 14. Ans: D vmax = x0 = 2f x0 = 2 v  x0 = 2 x 4.0 0.15 x 0.030 = 5.0 m s-1 15. Ans: B High resolving power means objects close together will still be well resolved. To improve the resolving power, the Raleigh’s criterion angle  should be made smaller. Since sin = a  ,  can be reduced by making a, the slit width, bigger. 16. Ans: A For each order, red (longest wavelength) will diffract more than violet (shortest wavelength). 2nd and 3rd orders overlap means the 2nd order red diffracts more than 3rd order violet 2nd order red  White light 3rd order violet 2r 3v

5 Slit separation d = 31 10 300 − = 3.333 x 10-6 m Using sin = n d  , For red: sin2r = 9 6 2 700 10 3.333 10 − −   2r = 24.8o For violet: sin3v = 9 6 3 400 10 3.333 10 − −   3v = 21.1o  = 2r - 3v = 24.8 – 21.1 = 3.7o 17. Ans: C The electric force on the positive charge is pointing in the direction of the E -field. So the external force exerted to move the charge is pointing in the opposite direction to the E - field. Work done by external force, W = F s cos  = qEs cos 0 = (2.6  10−8)(3.0  105)(4.0  10−3) = + 3.1  10−5 J 18. Ans: A When the electron moves downward, its gravitational PE decreases. (The Earth’s gravitational field is pointing vertically downward.) The electron has negative charge, so it wi ll accelerate towards the positive plate due to the downward electric force on it. Its gain in KE comes mostly from its loss of electric PE. So its electric PE decreases. (H2 students can also reason this way: when a negative charge moves from a region of lower electric potential to another region of higher electric potential, its electric PE decreases.) 19. Ans B With another identical lamp added parallel to X, the effective resistance across X is halved of its initial val ue. Total resistance in the circuit drops. By potential divider principle, potential difference across X is smaller and across Y is larger. Brightness of the lamp depends on the power (P=V2/R) delivered to lamp, hence X is less bright and Y is brighter.

6 20. Ans D When switch is open, VPQ = (0.70/1.0) x 2.0 V = 1.4 V E = 1. 4 V When switch is closed, VPQ = (0.50/1.0) x 2.0 V = 1.0 V By potential divider principle, p.d. across 2.0 Ω , V2.0 = (2.0/(2.0+r)) x 1. 4 V 1.0 = (2.0/(2.0+r)) x 1. 4 r = 0.80 Ω 21. Ans: C By right hand grip rule, magnetic field produced by wire at compass, Bwire, points to east. Bwire = 0 2 I r   = 7 3 4 10 3.0 2 9 10   − −   = 6.667 x 10-5 T tan = wire Earth B B  = tan-1 6.667 3.5 = 62o X BEarth = 3.5 x 10-5 T Bwire = 6.667 x 10-5 T Bresultant Current I  North East

7 22. Ans: A Going through crossed fields undeviated: FE = FB qE = B1qv Speed of particle, v = 1 E B Going through 2nd field, undergo circular motion of diameter 60 cm: 2 2 mv B qvr = m = 2B qr v = 12B B qr E = 5 19 2 2.0 10 1.2 2 1.60 10 0.30 6.0 10 −−       = 3.8 x 10-27 kg 23. Ans: C Magnitude of emf induced = rate of change of magnetic flux linkage ( ) ( ) 2 2 2 0.30(120) (0.080 0.020)2 4.0 0.127 V   =  =   =  − = = E t NA B t dNB t  130 mV (Even when the ends of the coil are not connected to form a closed circuit, there is still an emf induced in it due to the decreasing magnetic flux linkage.) 24. Ans: B Using the right -hand grip rule, the current in the vertical wire produces a magnetic field which is directed into the plane of the paper at the location of the loop. Hence, this produces a magnetic flux linkage  with the loop. As the loop moves to the right,  decreases. By Faraday’s law, an emf is induced in the loop and an induced current flows

8 in the loop because it is a closed circuit. By Lenz’s law, the current flows in such a direction so as to oppose the d ecrease in . Using the right -hand grip rule, the current flows clockwise around the loop, so as to produce a magnetic field which is directed into the plane of the paper (within the loop). Also, by Lenz’s law, the induced current will produce a magnetic force which opposes the motion of the loop. So the net magnetic force on the loop will be to the left. 25. Ans: D Mean power Pave = ½ Pmax 21 2 V R= 26. Ans: D 34 8 21 7 2 18 34 8 32 7 1 18 18 31 18 6.63 10 3.00 10 1.78 10 1.12 10 J 6.63 10 3.00 10 6.22 10 0.32 10 J Hence E (1.12 0.32) 10 1.44 10 J = 9.00 eV hcEE hcEE E   − − − − − − − −   