Paper-1-PART-I (PHYSICS)

PART-I (PHYSICS) 1. From wedge constraint aAY (aA ) = (aB ) aAX cos 53° – aAY cos 37° = aB cos 53º aB = – 5 m/s → ˆ aB  5i 2. For spherical surface using n2 – v n1 = u n2  n1 R  n – 2R 1 n  1  = R  n = 2n – 2  n = 2. 3. Zero error (excess reading) = 0.3 mm. observed thickness of block = 13.8 mm. Actual thickness = 13.8 – 0.3 = 13.5 mm 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. Since tension is the two rods will be same, hence 1 1 1  = A Y    A Y  = A Y  . 6. Figure shows the slope of the final reflected ray is tan 45º = 1 Hence the most appropriate option is (D) 7. Rate of heat transfer is same through all walls K1.A.(10) 40 cm K 2A(20) = 20 cm = K 2 A (40) 10 cm  K1 = K 4 2 = 4K3  K1 = 4K2 = 16 K3 . hc 8.  hc = 5 eV0 +  2 hc hc 3 = eV0 +   3 = 4eV0   = 6 Ans. 9. Here path difference will be : x = ( –  ) t   = 2  (2 –  ) t Hence (C) 10. 1 mV 2  3 kT 2 2 1 2  2 kT 2 2 V   11. (A) If both are positive, then null point lies in between them where direction of the electric fields due to both are opposite in direction. (C) If qA is positive and qB is negative then in both regions x < xA and x > xB, field directions are opposite. But null point would lie nearer to the charge of the lesser magnitude.  12. 4 = 0.1   = 0.4 m 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 y(x = 0.04, t = 0.025) = – 2  cm 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. 13. (A) When dielectric is removed then the capacitance  but 'Q' does not change momentarily  V  Now by KVL you can see that potential of point 'a' will be higher them that of 'b'  Current will be from 'a' to 'b'. (B) Finally the charge on both A and B will be same (C) Wext + Wbattery = Heat + Ucap. Ucap  Wbattery Wext = Heat + ––––  0 (We can check that | Wb | > | Ucap | by taking some numerical values for fast calculation). (D) Positive charge flows into positive terminal of the cell.  Work is done on the cell which  its internal energy. → → 14. Using – e (V B) for the region outside the plates, direction of magnetic field can be found. Inside the plates, net force on the electron is zero hence electric force is opposite to that of magnetic force. Direction of electric field between the plates is opposite to that of direction of force on the negative (electron) charge. 15. The free body diagram of block + tube inclusive of water is as shown below Since the block + tube system shown in figure is in equilibrium  Net weight of system = buoyant force 2𝑙 2d Vg + r2 3 dg = (V + r2𝑙) dg r 2𝑙d r2𝑙 where V is the volume of block B  (2d – d) V = 3 or V = 3 The pressure of the air trapped inside the tube is same as pressure at point A in the water as shown in figure. h  𝑙  h  𝑙   P = P +    dg Ans. P  a = P0 + dg    3  16. When the structure becomes inverted, there is no decrease in the potential energy of the ring. therefore, Decrease in PE of the rod = Gain in rotational kinetic energy of the structure  M.g. 4R = 1 2 2 sy (As COM of the rod comes down by distance 4R) MR2  M.4R2  2  Now sy = 2  12 M.4R   MR2  13MR2 29MR2 = 2 3  = = = 6 = 8 rad/s   13.6  1  1  17. Energy equivalent to line of lyman series   12 42   13.6  1  1  So according to the question  12 42  –   10.2     2.55 or max = 2.55. So the next integer is 3. 18.  6 × 2.2 – 3 × 1.8 = (6 + 3) v momentum conservation v = 7.8  13 9 15 1 1 1  13 2 1  6  2.22   3 1.82 = 9     512x2 2 2 2  15  2  x = 0.25   = 5 19. From given graphs : ax = 3 t and a =  3 4 y  4  1  v =  x 3 8 t2 + C At t = 0 : v = – 3  C = – 3 3  v = 8 t2 – 3  3 t2    3 dt  .... (1)  3 t2  t   Similarly ; dy =   8 4 dt  .... (2) As dw = F. ds = F.(dx ˆi  dy ˆj) W 4  3 ˆ  3  ˆ  3 2  ˆ  3 2  ˆ  dw    4 t i   4 t  1 j .  8 t  3  i   t 8  t  4  j dt 0 0           W = 10 J Alternate Solution : Area of the graph ; ax dt = 6 = V(x) f  (3)  V = 3. and ay dt = –10 = V(y) f  (4)  V = – 6. Now work done =  KE = 10 J 20. 1  2 = 1N  2N decay rate of  decay = decayrate of decay probability of  decay = probability of  decay 75 100  3 = 25 100