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 (