Title 21.Magnetism and Matter Hard Ans.pdf ✅

1. (b) 3 7 3 0 1 2 2 2 1 10 2 r 2M . 4 B        =          = − ; 3 7 3 0 2 2 2 1 10 2 r M 4 B       =          = − ; B (2 10 ) (10 ) 5 10 T 7 2 7 2 7 net − − − =  + =  2. (d) Suppose distances of points X and Y from magnet are x and y respectively then According to question Baxial = Bequatorial  1 2 . 4 2 . 4 1 / 3 3 0 3 0 =  = y x y M x M     3. (d) By using tan  4 0.38 10 tan 60 −  =  = o V H V B B B 0.38 10 3 0.62 10 . −4 −4  BV =   =  4. (c) By using    cos tan tan  = ; where o o  = 40 ,  = 30 As cos 30o < 1 1 cos 30 1   o Hence           =    1 tan tan tan tan or o   40 5. (b) When a magnet is freely suspended in earth’s magnetic field, it's north pole points north, so the magnetic field of the earth may be suppose to be due to a magnetic dipole with it's south pole towards north and as equatorial point is on the broad side on position of the dipole. 6 3 4 7 3 0 e (6.4 10 ) M 0.3 10 10 r M. 4 B    =    = − − 22  M = 7.810 A-m 2 . 6. (c) At neutral point magnetic field due to magnet = Horizontal component of earth's magnetic field  BH r M = 3 0 2 . 4   4 3 7 0.3 15 (0.2) 10 2 1 =     − M  1.2 1200 . 2 2 M = amp m = ab − amp cm 7. (b) For equilibrium of the system torques on M1 and M 2 due to BH must counter balance each other i.e. M 1  B H = M 2  B H . If  is the angle between M1 and BH then the angle between M2 and BH will be (90 – ) so sin sin(90 ) M1BH  = M2BH −        = = =  = − 3 1 tan 3 1 3 tan 1 1 2   M M M M 8. (d) By using ; 2 2 2 2 2 1 d s d s T T T T M M − + = where T sec s 5 12 60 = = and T sec d 15 4 60 = = 4 5 (15) (5) (15) (5) 2 2 2 2 2 1 = − +  = M M 9. (b) Time period decreases i.e. field due to magnet (F) assist the horizontal component of earth's magnetic field (see theory) Hence by using 1 '  −      = T T B B H  3 1 1 3 1 2 2  − =  =      = F H H F . 10. (b) By using 2 o 1 1 2 1 2 1 tan tan 45 i / 3 i tan tan i i i tan   =      = o 2 2 2 30 3 1  3 tan  =1 tan  =   = So deflection will decrease by o o o 45 − 30 = 15 11. (d) When needle oscillates in horizontal plane Then it's time period is MBH I T = 2 ......(i) When needle oscillates in vertical plane i.e. It oscillates in total earth's total magnetic field (B) Hence M I T' = 2 .....(ii) Dividing equation (ii) by (i) 2 1 cos 60 ' cos = = = = B B B B T T H   2 ' T T = 12. (c) In vertical plane perpendicular to magnetic meridian. MBV I T = 2 ......(i)
In horizontal plane MBH I T = 2 .....(ii) Equation (i) and (ii) gives BV = BH Hence by using o H V B B tan  =  tan  = 1   = 45 13. (c) By using heat loss = VAnt ; whre V = volume = 10–3 m 3 ; A = Area = 0.1m 2 , n = frequency = 50 Hz and t = time = 1sec Heat loss = 10–3  0.1  50  1 = 5  10–3 J = 1.19  10– 3 cal 14. (b) By using B = H =  0  r H and (1 m )  r = +   H HA B r 0  0    = = (4 10 ) 1600 (0.2 10 ) 2.4 10 7 4 5 − − −      =   r = 596.8. Hence  m = 595 .8  596 15. (c) 0.08Amp / m 0.075 8 10 7500 m Md V M I 7 =   = = = − 16. (a) 3 10 Amp / m 1 1.5 10 2 10 V N V M I 3 23 26 =     =  = = − 17. (c) H = ni  A n H i 8 500 4 10 3 =  = = 18. (c)  E.d = – r 2 dt dB d E r E × 2r = + r 2 × dt dB as dt dB = –ve E = +ve  assume direction is right so q charge move along 3. 19. (c) dt d =  E.d d R E d R 2 dt dB = E × 2d E = 2d R 2 dt dB = 2d R 2 t B  Change in angular momentum of whole system =  × t × 2 = qE × 2 d × t × 2 = 2 qBR 2 20. (b) i1 = dt d1 R 1 = Rdt d[B Area] 1  = dt d           4 a 5t 2 × R 1 = 4R 5 a 2  i2 = dt d2 × R 1 = dt d[5a t] 2 × R 1 = R 5a 2 i3 = dt d3 × R 1 = dt d[0.5a ] 2 × R 1 = 2R a 2  i2 > i1 > i3 21. (b) M M Meq = 2 M 22. (d) V = 2 r KM cos , Vaxis = 2 r KM   = 0 Vequator = 0   = 900  equator axis V V =  23. (b)  = 8 × 10–3 , H = 160, B = H = 1.28 wb/m2 24. (c) Some domains increased which are in the direction of applied field, other decrease 25. (a) In tan A position 3 r 2kM = BH tan
3 2 1 r r         = 1 2 tan tan     = 300 26. (b) T = 2 MB I on breaking n equal parts perpendicular to length I' = 12 1       n m 2 n        of each part I' = 3 n I , M' = n M T T' = I M M' I'  = n n 1 3  = n 1  T' = n T 27. (d) In series current is same. In tangent galvanometer B = BH tan  2R ni 0 = BH tan  i = Same  n  tan  2 1 n n =   tan 45 tan 60 = 1 3 28. (d) d S N S N p d At P Bnet = 2 2 2 B1 + B = 2 3 2 3 d kM d 2kM        +      = 3 d kM ( 5 ) 29. (c) H = 2 × 103 , for solenoid H = n i n = –2 15 10 150  = 1000 i = n H = 3 3 10 210 = 2A 30. (c) O x dx An element is assumed at distance x from center whose width is dx. No. of turns in width b–a = N  No. of turns is width dx = n = dx b a N       − dB = 2x ni 0 = x dx 2(b a) 0 iN −   B =  −  b a 0 x dx 2(b a) iN =       −  a b log 2(b a) Ni e 0 31. (a) tan  =   cos tan  tan 450 = cos30o tan   = tan–1         2 3 32. (b) Magnetic field = BH T = 12 60 = 5 sec. Magnetic field = BH + BT = 15 60 = 4 sec. Let after reversing magnetic needle makes x oscillation/sec.  Magnetic field = BH – B, T = x 60 sec. T  Magnetic field 1  4 5 = H H B B + B  16 9 B B H =  (60 / x) 5 = H H B B − B  x = 63 33. (b) By using B = 0rH and A B  = , 4 5 7 4 0 r 1.25 10 2 10 4 10 2000 6.28 10 A H =        =     = − − − 34. (c) In C.G.S. 3 2 9 x M Baxial = = .....(i) equaterial 3 3 8 2 x M x M B =       = .....(ii) From equation (i) and (ii) Bequaterial = 36 Gauss. 35. (b) mass/densi ty M V M I = = , given mass = 1gm = kg 3 10 − , and density = 3 3 2 3 3 3 3 5 10 / (10 ) 5 10 5 / kg m m kg gm cm =   = − − Hence 3 10 6 10 5 10 3 7 3 =    = − − I 36. (b) Net magnetic field at mid point P, B = BN + BS
where BN = magnetic field due to N- pole BS = magnetic field due to S- pole T r m BN BS 7 2 7 2 0 4 10 2 0.1 0.01 10 4 − − =        = = =    8 10 . 7 Bnet T −  =  37. (b) Relation for dipole moment is, M = I  V , Volume of the cylinder , 2 V = r l Where r is the radius and l is the length of the cylinder, then dipole moment, (0.5 10 ) (5 10 ) 7 22 (5.30 10 ) 2 3 −2 2 −2 M = I r l =      2.08 10 J / T −2 =  38. (b) Work done, W (cos cos ) = −MB  2 − 1 (cos180 cos0 ) o o = −MB − = −MB(−1− 1) = 2MB = 2  2.5  0.2 = 1J 39. (d) In equilibrium Magnetic torque = Deflecting torque  MB sin  = F.d or d mlB F sin  = N o 6.25 0.12 24 0.25 0.25 sin 30 =   40. (a) The weight of upper magnet should be balanced by the repulsion between the two magnet gm wt r m  . = 50 − 4 2 2    50 10 9.8 (9 10 ) 10 3 6 2 7 =     − − − m  m = 6.64amp m 41. (a) By using 2 3 cos 45 cos 30 (cos ) (cos ) ( ) ( ) cos 2 1 2 1 =  = = =    H H H B B B B 42. (b) B B B T o H 4 5 tan 0.34 10 tan 30 1.96 10 − − =   =  =  43. (c) By using H BH T MB I T 1 = 2   6 H B H A H B B A 36 10 (B ) 60 / 20 60 /10 (B ) (B ) T T −   =  = (B ) 144 10 T. 6 H B −  =  44. (b) 1 2 2 1 1 2 M M T T M T MB I T H =     = If 100 M1 = then (100 19) 81 M2 = − = . So, 2 1 1 2 1 11% 9 10 10 9 100 81 T T T T T = =  = = 45. (b) By using MB H I T = 2  1 2 2 1 ( ) ( ) H H B B T T =  5 2 0.1 10 ( ) 2,5 60 / 40 −  = BH  BH T 6 ( )2 0.36 10 − =  . 46. (d) By using 6 3 1 tan 60 2 tan 30 tan tan tan 2 2 2 1 2 1   =  = =  i = i i i i o o    amp. 47. (c) As we move towards equator BH increases and it becomes maximum at equator. Hence 2 , MBH I T =  we 48. (d) 2 ; MBH I T =  When a bar magnet is broken in n equal parts so magnetic moment of each part become n 1 times and moment of inertia becomes of each part becomes 3 1 n times. Hence time period becomes n 1 times i..e. 4 ' T T = In this question n = 2 so, sec T T 1 2 2 2 ' = = = 49. (b) By using    cos 2 2 MB I MB I T H = = 0.1 m S N P BN Bs