7.PRACTICE TEST-1 (Solution-Paper-1)

PAPER - 1 PART-I (PHYSICS) 1. Let x be the distance of the top of window from the top of building and t be the time taken by the ball from the top of building to the top of window. (i) Since, acceleration is constant = g u  v So, S = 2 t (across the window) 3 = vT  vB 2 t  3 = vT  vB 2 0.5 Aliter : So, vT + vB = 12 m/sec. For motion from O to A V 2 = u2 + 2gx = (0)2 + 2gx V 2 = 2gx (i) VT = 2 + gx = 0 + gt VT = gt (ii) For motion from O to B V 2 = u2 + 2g (x + 3) V 2 = (0)2 + 2g (x + 3) V 2 = 2g (x + 3) (iii) VB = u + g (t + 0.5) VB = 0 + g (t + 0.5) VB = g (t + 0.5) (iv) From equations (ii) and (iv) VB – VT = g (0.5) (v) From equations (i) and (iii) V 2 – V 2 = 2 g (3) (vi) B T From equations (v) and (vi) 2 2 B T VB – VT 2g (3) = g(0.5) (VB – VT )(VB  VT ) (VB – VT ) = 12  V + VB = 12 ms–1 (A) Ans 2. From wedge constraint aAY (aA ) = (aB ) aAX cos 53° – aAY cos 37° = aB cos 53º aB = – 5 m/s → ˆ aB  5i 3. 1.8 t – µk 15 = 1.5 a for t = 2.85 sec. a = 0 so  = 0.24 k(1.5g) 1.8 × 2.85 4. Initial extension will be equal to 6 m. 1  Initial energy = 2 (200) (6)2 = 3600 J. 1 Reaching at A : 2 mv2 = 3600 J  mv2 = 7200 J From F.B.D. at A : mv2  7200 N = R 5 = 1440 N 5. Velocity of sound is inversely proportional to the square root of density of the medium. V1 i.e., V = constant  V2 = = = Ans. 6. Since tension is the two rods will be same, hence 1 1 1  = A Y    A Y  = A Y  . F N 7. Acceleration of two mass system is a = FBD of block A 2m leftward F 60° N cos 60° – F = ma = mF solving, 2m 30° 8. Applying NLM on 40 kg block 400 – 4T = 40 a For 10 kg block T = 10.4 a Solving a = 2m/s2 T = 80 N  9. 4 = 0.1   = 0.4 m a from graph  T = 0.2 sec. and amplitude of standing wave is 2A = 4 cm. Equation of the standing wave  2   2  y(x, t) = – 2A cos  0.4 x . sin  0.2 t  cm     y (x = 0.05, t = 0.05) = – 2 cm y(x = 0.04, t = 0.025) = – 2 cos 36°  speed = T = 2 m/sec. dy 2 cos  2 x    2t  Vy = dt = – 2A × 0.2  . cos   0.4 0.2     1 Vy = (x = 15 m, t = 0.1) = 20  cm/sec. P 2 10.  P2RT = k  = k PM  kM   PT =  R  ..........(i)   P 2 P´2  =  / 2 P’ = and P  K´/ T Hence from (i) T’ = T 2 , PT = constant hence P–T curve is a parabola. 11. For cylinder ; 0 ir B = 2 R2 0 i = 2 r ; r < a ; r > a We can consider the given cylinder as a combination of two cylinders. One of radius 'R' carrying current  in one direction and other of radius R carrying current 2 I 3 in both directions. At point A : B = 0 (I / 3) + 0 = 2(R / 2) 0  4 I / 3   R   0 I 3R 0 I At point B : B =  2   + 0 = 12. F F max min. = kx +  mg = kx –  mg 2  R   2  3R max – Fmin. = 2  mg or 2 = 2  10  m = 0.1 13. Fmax + Fmin. = 2 kx (1) from graph Fmax + Fmin. = 5 and x = 0.1 Putting in equation (1) 5 = 2 k(0.1) k = 25 N/m. 14. When x = 0.03 kx = 25 × 0.03 = 0.75 N, which is less than  mg = 0.1 × 10 = 1 N  The block will be at rest, without applying force F. 15. The equation of wave moving in negative x-direction, assuming origin of position at x = 2 and origin of time (i.e. initial time) at t = 1 sec. y = 0.1 sin (4t + 8x) Shifting the origin of position to left by 2m, that is, to x = 0. Also shifting the origin of time backwards by 1 sec, that is to t = 0 sec. y = 0.1 sin [(4t –1) + 8(x – 2)] 16. As given the particle at x = 2 is at mean position at t = 1 sec.  its velocity v = A = 4 × 0.1 = 0.4  m/s. 17. Let V´ be volume of mercury and V the volume of vessel. V × 27 × 10–6 × T = V´ × 1.8 × 10–4 × T V´  3 V 20  x = 3 18. Consider a ring at angular position  Charge on ring dq = .2R2.sind  1  dq R  dV =  4   R = 2  sin d  0  0  Net potential at centre O R V = 2 0  / 3 sin d = 0  R 2 0 cos  / 3 R = 4 0 2 0 2 = 4 0 V = 1 V. 1 1  3    1 1  19.  =  2 1    x  x   2  1 .    1  300   x = 1200 cm = 12m 20. x so 3m 2 300 to reach  = 270º, it has to cross the potential energy barrier at  = 180º and to cross  = 180º angular velocity at  = 180º should be 0+ ki + Ui = kf + Uy 1  3 MR 2  2  2  2  =  + (–Mi AB cos 0º) = 0 + (–NiAB cos180º)   9 rad/sec. 21. Total heat produced= KE of the rod = 1 x 2x 32 = 9 J 2   13.6  1  1  22. Energy equivalent to line of lyman series  12 42  13.6  1  1      10.2 So according to the question  12 42  –     2.55 or max = 2.55. So the next integer is 3. (2,1,9) (2,1,9) 23. V  V =  y2dx  2xydy = d(xy2 ) = xy2 (2,1,9) = 2. (0,0,0) (2,1,9) ( 0,0,0 ) ( 0,0,0 ) ( 0,0,0 )