Paper-3-PART- II (PHYSICS)

1. (C) PART- II (PHYSICS) Applying momentum conservation, 0 = mv1 – 2mv2  v = v1 (i) 2 From energy conservation, ki + Ui = kf + Uf 0   G(2m)   = 1 mv 2 + 1 (2m) v 2 +  3 G(2m)  (m) ..........(ii)  2R  2 1 2 2  2 R  Solving eqn.(i) & (ii) get, 8Gm v1 = 3R (A) COM will be fixed so, m1s1  m2s2 Scm = m1  m2 (m)(x)  (2m)((2R  x)) 0 = m  2m  x = 4R 3 (B) F = 0  a = 0  G(2m)   3 G(2m)  (D) Wgr = U   Wgr =   m – – 2R 2 R  m .     2. for t = 0 x = A sin  (1) v = A cos  (2) a = – A2 sin (3) By (1) and (3)  =  a = x  32m / s2  0.08m = 20 rad/s By (1) and (2) x tan   v =    tan  = 1 5 or  = 4 or 4 5 At t = 0 , x is negative hence  = 4 x from (1) , A = sin  = 0.08 × m = 11.3 cm 3. Consider a ring of radius x and thickness dx.   Q Equivalent current in this ring = 2 × charge on ring = 2 × (2  x dx) R2    2xQ  R   Q dB (due to this ring) = 0 2x   2 R2 dx   B =  2 0 dx. R2 4. ' = V  Vs f 332  32 = 1000 = 0.3 m 0 = 2R2 . R = 0 . 2R f (V  V0 ) V  Vs = 1000 × 332  64 332  32 = 1320 Hz '' = V  V0 f' = 0.2 m. 5. d sin =  For first bright fringe. (There will be one bright fringe on both sides of central bright fringe and both are called first bright fringe.) sin =   1 d  d  . If d = 2, two maxima will be formed.  6. EC   OC  30 OB  BC 30 Kq  q AC AC3 EC  30  AC3 AC Which is dependent on C hence field is non uniform inside cavity.  EA  0  OA 30   R / 23 30 R2 .BAˆ  EA  OA 30  R 240 R .BA   30 R OAˆ  2  240 R.OAˆ   R R  ˆ  ˆ 2OA OA EA   6  OA  24 R 8 R.OA  8  4  0 0  0 0 0 similarly for EB Hence C and D 7. A B Molecular wt. = 16 M0 mass = 2 m0 Molecular wt. = M0 mass = m0 nA = n0 n 8 B = n0 (A) K.E./atom = f k.T. = 2 f k.T. for both the gases. 2 (B) (B) CrmsA  , CrmsB  , (Crms)B = 4 (Crms)A (C) (P)A = (n0 / 8)RT , (P V B) = n0 RT V , (P B) = 8 (PA) (D) nB = 8 nA 8. The H-type atom is in the third excited state i.e. n = 4. 6200 Energy corresponding to wave length 51 This is the E2 – E1 for H and E4 – E2 for He+. we get z = 2 for 4  2 radiation Hence the atom is Helium ion. Ans. He+ , 12400  51 nm = 62000 = 10.2 eV (b) Let u be the speed of neutron before collision At end of the deformation phase (when the kinetic energy of (neutron + He+) system is least) Where ucm is velocity of centre of mass. From conservation of momentum ucm = mu u m  4m = 5 1 1  u 2 1  u 2 4  1 2  The loss of kinetic energy = mu2 – m  5  – 4m  5  = 5  2 mu  2 2   2     It K is the kinetic energy of electron then the maximum loss in K.E. of system is 4 51 5 5 K = 51 eV or K = 4 = 63.75 eV Kmin = 255 = 63.75 eV 4 9. Let ‘u’ be the required minimum velocity. By momentum conservation : mu = (m + m)v  v = u/2. Energy equation : 1 mu2 = 2 1 (2m)v2 + mgH. 2 Substituting v = u/2 : u = 2 10. By work-energy theorem on the system : – (mg) (BE) – mgh = – 1 mu2 4 1 H  h = – 4 m (4 gH)  BE =  11.12. When connected with the DC source R = 12 = 3  4 When connected to ac source  = V Z 12  2.4 =  L = 0.08 H V 2 V 2 R Using P =  V cos  = rms cos  = rms = 24 W rms rms Z R2  (L – 1 )2 C 13. B = 0 2iR2 2 2 3 / 2 = 0i = B(say) 1 4 (R  3R ) 16R  2i.4R2 B B2 1 B2 = 0 4 (4R2  12R 2 )3 / 2 = 2  B1 = 2 . 14. B net = = 2 = 32R . 15. Total energy released from Au198  Hg198 in ground state = (m loss ) c2 = (197.9682 – 197.9662) (930) = 1.86 MeV Energy released from 198Hg in first excited state  Hg in ground state = (–1.6) – (–2) MeV = 0.4 MeV  Energy released from Ag198  Hg198 second excited state = 1.86 – 0.4 = 1.46 MeV = max. K.E. of 2 particle 16. Similarly maximum kinetic energy of  particle = 1.86 – 1 = 0.86 MeV 17. By snell law n = sin i sin r Since for 1st case angular incedence is same for all ray. So r will be less for red. 19. in (A), V is on vertical axis. As V is icreasing, W is positive. V is decreasing, W is negative. As negative work in part-II is greater than positive work in part-I, net work during the process is negative. Using PV = nRT and as Vremains same for initial and final points of the process, it is obvious that final temp. is greater than initial temperature as pressure has increased. Therefore dU is positive. Similar arguments can be applied to other graphs.