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.

Comments

Post a Comment

Popular posts from this blog

PHYSICS-15-10- 11th (PQRS) SOLUTION

XI (PQRS) PHYSICS REVIEW TEST-5 Select the correct alternative. (Only one is correct) [15 × 3 = 45] There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. Q.31 Two trains, which are moving along different tracks in opposite directions towards each other, are put on the same track due to a mistake. Their drivers, on noticing the mistake, start slowing down the trains when the trains are 300 m apart. Graphs given below show their velocities as function of time as the trains slow down. The separation between the trains when both have stopped, is: (A) 120 m (B) 280 m (C) 60 m (D*) 20 m 1 [Sol: x1 = 2 1 × 10 × 40 = 200 m; x2 = – 2 × 8 × 20 = –80 m  xreletive = x1 + x2 = 280  separation = 300 – 280 = 20 m (D) ] Q.32 One end of a thin uniform rod of length L and mass M is riveted to the centre of a uniform circular disc of radius r and mass 2 M so that the rod is normal to the disc. The centre of mass of the combination from the centre of the disc is at dist
Read more

8-Circular Motion

8.1 (D) The maximum angular speed of the hoop corresponds to the situation when the bead is just about to slide upwards. The free body diagram of the bead is For the bead not to slide upwards. m2 (r sin 45°) cos 45° – mg sin 45° < N (1) where N = mg cos 45° + m2 (r sin 45º) sin 45° (2) From 1 and 2 we get.  = rad / s. 8.2 (C) Let v be the speed of particle at B, just when it is about to loose contact. From application of Newton's second law to the particle normal to the spherical surface. mv2 = mg sin  (1) r Applying conservation of energy as the block moves from A to B.. 1 mv2 = mg (r cos  – r sin ) (2) 2 Solving 1 and 2 we get 3 sin  = 2 cos  8.3 (A) As the mass is at the verge of slipping ∴ mg sin37 –  mg cos37 = m2r 6 – 8 = 4.5   = 3 16 8.4 (B) As when they collide vt  1  72v 2  2 2  25R   R = vt  t =   5R 6v Now angle covered by A =   vt R 11 Put t  angle covered by A = 6 8.5 (C) The acceleration vector s
Read more

4. THEORY-Current Electrictricity

CURRENT ELECTRICITY  1 . ELECTRIC CURRENT (a) Time rate of flow of charge through a cross sectional area is called Current. if q charge flows in time interval t then average current is given by q Iav = t and Instantaneous current i =. lim q  dq t0 t dt (b) Direction of current is along the direction of flow of positive charge or opposite to the direction of flow of negative charge. But the current is a scalar quantity. i i q+ velocity q– velocity SI unit of current is ampere and 1 Ampere = 1 coloumb/sec 1 coloumb/sec = 1A 2 . CONDUCTOR In some materials, the outer electrons of each atom or molecule are only weakly bound to it. These electrons are almost free to move throughout the body of the material and are called free electrons. They are also known as conduction electrons. When such a material is placed in an electric field, the free electrons drift in a direction opposite to the field. Such materials are called conductors. 3 . INSULATOR Another class o
Read more