9. PRACTICE TEST-2 (Solution-Paper-1)

PAPER - 1 PART-I (PHYSICS) 1. Let a be the retardation produced by resistive force, ta and td be the time of ascent and descent respectively. If the particle rises upto a height h then h = 1 (g + a) t 2 and h = 1 (g – a) t 2 2 a ta  td = = 2 d = Ans. 2. Method - I As cylinder will remain in contact with wedge A Vx = 2u As it also remain in contact with wedge B u sin 30° = Vy cos30° – Vx sin30° Vy = Vx sin 30 cos 30 + u sin 30 cos 30 Vy = Vx tan30° + u tan 30° Vy = 3u tan30° = u V = = Method - II u Ans. In the frame of A 3u sin 30º = Vycos30º  V = 3u tan 30º = u and Vx = 2u  V = = 7 u Ans. 3. We draw axes for each block along the incline and normal to incline. The component of acceleration for each block are as shown, where a is acceleration of wedge in figure2. fig.-1 fig.-2 It is obvious that vertical component of acceleration is larger for block in figure 2.  T1 > T2 4. x = x1 and x = x3 dU are not equilibrium positions because dx  0 at these points. x = x2 is unstable, as U is maximum at this point. 5. As forces = rate of change in linear momentum. Torque = rate of change in angular momentum.  F t = mv (Linear) (1) F  𝑙  m𝑙2 and    t =    12 (Angular) (2) Dividing : (1) and (2) 12v 6v 2 = 𝑙   = 𝑙 Using : S = ut : Displacement of COM is   6v  = t =    t and x = vt 𝑙  2x 𝑙 Dividing :  = 6  x = 𝑙 12  Coordinate of A will be  𝑙  𝑙 , 0 Hence (D). 12 2  6. P = 1 2A2V using V = 2 P = 1 2 A2 2  =  f = 2 1 = 2 using the given data, we get f = 30 Hz. 7. Let h be the height of water surface, finally a2h = a . a . 2 a ; h = 2 a 4  a  a   3a 5a  C.M. gets lowered by a –    Work done by gravity = mg 5a 8 8  = a – 8 = 8 8. On the curve y = x2 at x = 1/2 1 y = 4 Hence the coordinate  1 , 1     2 4  Differentiating : y = x2  vy = 2xvx  1  vy = 2 2  (4) = 4 m/s   Which satisfies the line 4x – 4y – 1 = 0 (tangent to the curve) & magnitude of velocity : → | v | = 4 m/s As the line 4x – 4y – 1 does not pass through the origin, therefore (D) is not correct. 9. Equivalent circuit : Induced emf e = Br 2  2  =   ( Radius = a)  X  e       X  0  4   +     r  = 0  5X = 4e 4e 2Ba2 X 2Ba2  X = 5 = 5 and  = r = 5r also direction of current in ‘r’ will be towards negative terminal i.e. from rim to origin. Alternatively; by equivalent of cells (figure (ii)) : I = 10. At t = 0 e r  r 4 4 e = 5 r  Displacement x = x1 + x2 = 4 sin 3 Resulting Amplitude A = = 2 m.   = 2 m Maximum speed = A = 20 m/s Maximum acceleration = A2 = 200 m/s2 1 Energy of the motion = 2 m2 A2 = 28 J Ans. 11. V0 = 0 R = 10 × 10 = 100 volts (since,  = 10amp from figure) Also :  =  e – t/RC  0  t t Taking log ; log   = RC  C = Rlog( / )   0 At ; t = 2 sec,  = 2/5 A 2 C =  10  10 log 2.5    2 2 1 C = 10 log4 = 10  2log2 = 10𝑙n2 1 1 2 1  1  2 500 C = Heat produced = CV =   (100 ) = joules. 10𝑙n2 Hence (C) is correct 2 2  10𝑙n2  𝑙n2 12. If the particle is released at the origin, it will try to go in the direction of force. Here du is positive and dx hence force is negative, as a result it will move towards – ve x-axis. 13. When the particle is released at x = 10 –  it will reach the point of least possible potential energy (–15 J) where it will have maximum kinetic energy. 1 2 2 max = 35  v = m/s 14. Stable and unstable equilibrium for stable equilibrim U = Umin and for unstable equilibrium U = Umax 15. In transition  E2 = – 3.4 eV, E4 = – 0.85 eV E = 2.55 eV E = hc   = hc  E  = 487 nm. 16. Wavelength of radiation = 103 nm = 1030 Å  E = 12400 1030 Å ~– 12.0 eV So difference of energy should be 12.0 eV (approx) Hence n1 = 1 and n2 = 3 (–13.6)eV (–1.51)eV  Transition is V. 17. In an adiabatic expansion, TV–1 = constant  V 1 2 T V–1 = T   = 1 + 0  5  5  T = (273). (5)2/5 < (KE)rotational > = kT = 1.38 × 10–23 × 273 × (5)2/5 = 7 × 10–21 J (approx). 18. After switch is closed 9C passes through 'B'. 19. Excess pressure 4T 2 = – r 20 = 0.2 N/m2 = 2 × 10–1 N/m2  P = 2 20. From energy conservation,  3 GM  1   2 R  m0 =  3GM m V2 2 V = R = V = 5 × 104 m/sec. 21. Using newton’s formula xy = f2  20 y = (10)2  y = 5 cm. 22. Velocity before collision = velocity after collision = e = = 1 2 23. Initial seperation is large so 1 1  1  q2 mu2 + 0 = m.(2u)2 –  4   r 2 2 2q2  0  r = 4 0 .3mu2 9 109  2 (2 106 )2 = 3  1 (200)2 6 r = 107 .