Title 01. Units and Dimensions Easy Ans.pdf ✅

1. (c) Light year is a distance which light travels in one year. 2. (b) Because magnitude is absolute. 3. (c) Watt=Joule/second = Ampere×volt = Ampere2×Ohm 4. (c) Impulse = change in momentum = F × t So the unit of momentum will be equal to Newton-sec. 5. (c) Unit of energy will be 2 2 kg - m /sec 6. (d) It is by standard definition. 7. (c) nm m cm 9 7 1 10 10 − − = = 8. (d) micron m cm 6 4 1 10 10 − − = = 9. (c) Watt = Joule/sec. 10. (c) ; 2 1 2 d Gm m F =  2 2 1 2 2 Nm / k g m m Fd G = = 11. (a) 12. (c) Angular acceleration 2 Time sec Angular velocity rad = = 13. (c) Stefan's law is ( ) 4 E =  T  4 T E  = where, E = Area Time Energy  2 Watt m = 2 4 4 2 Watt- − − − = = Watt − m K K m  14. (b) Kg-m/sec is the unit of linear momentum 15. (d) 2 ct must have dimensions of L  c must have dimensions of 2 L /T i.e. −2 LT . 16. (d) dt dL  =  dL =   dt = rFdt i.e. the unit of angular momentum is joule-second. 17. (c) 18. (a) Volume of cube = 3 a Surface area of cube = 2 6a according to problem a 3 = 6a 2  a = 6  V a 216 units 3 = = . 19. (b) 6 10 60 10 60 microns 5 6  =  = − − 20. (d) 21. (d) Because temperature is a fundamental quantity. 22. (a) 23. (a) 1 C.G.S unit of density = 1000 M.K.S. unit of density  3 0.5 gm/cc = 500 kg/m 24. (b) 25. (d) 26. (b) Mach number Velocity of sound Velocity of object = . 27. (d) 28. (d) dx dV E = − 29. (d) 30. (b) Surface tension = Length Force = Newtons / metre 31. (a) 32. (b) Henry A Wb I L = = =  . 33. (a) R L is a time constant of L-R circuit so Henry/ohm can be expressed as second. 34. (b)       = sec m mv kg 35. (a) Quantities of similar dimensions can be added or subtracted so unit of a will be same as that of velocity. 36. (b) MeV eV 6 1 = 10 37. (a) Energy (E) = F × d  d E F = so Erg/metre can be the unit of force. 38. (b) Potential energy 2 2 sec sec        =      = = cm cm g cm mgh g
39. (b) volt ampere watt = 40. (b) 41. (d) 42. (c) 43. (c) Energy = force  distance, so if both are increased by 4 times then energy will increase by 16 times. 44. (b) 1 Oerstead = 1 Gauss = Tesla 4 10 − 45. (a) Charge = current  time 46. (c) ohm cm L RA A L R =    = =  47. (c) 48. (a) Astronomical unit of distance. 49. (a) Physical quantity (p) = Numerical value (n)  Unit (u) If physical quantity remains constant then n  1/u  n1u1 = n2u2 . 50. (b) eV coulomb volt J 19 19 1 1.6 10 1 1.6 10 − − =   =  . 51. (b) kWh W J 3 5 1 = 110 3600 sec = 36 10 52. (c) According to the definition. 53. (c) 54. (c) As I 2 2 = MR = kg − m 55. (c) Area Force Stress = = 2 m N 56. (b) 4 −2 −1 −4 = AT  = Jm s K t Q   57. (a) M = Pole strength  length = amp −metre metre 2 = amp − metre 58. (c) Curie = disintegration/second 59. (a) 60. (c) Pico prefix used for 12 10 − 61. (c) 62. (d) Unit of e.m.f. = volt = joule/coulomb 63. (d) 64. (b) 65. (c) L L A F Y  = . = 2 4 2 5 2 0.1 / 10 10 N m m N cm dyne = = − − 66. (a) Pressure Dimensionless Force/Area Strain Stress Y = =  Y  . 67. (b) 1 yard = 36 inches = 36  2.54 cm = 0.9144 m . 68. (c) metre -15 1 fermi = 10 69. (b) 70. (d) 71. (b) 72. (b) 73. (d) 1 Newton = 105 Dyne 74. (c) 2 2 2 [x] = [bt ]  [b] = [x / t ] = k m / s 75. (b) Units of a and PV2 are same and equal to dyne × cm4 . 76. (b) 77. (b) 78. (c) Impulse = Force  time (kg-m/s ) s kg-m/s 2 =  = 79. (c) 80. (a) K = C + 273 .15 81. (a) 82. (d) 83. (c) 84. (b) 85. (d) Watt is a unit of power
86. (d) meter 15 1light year = 9.46 10 87. (b) m W V = so, SI unit = kg Joule 88. (a) 89. (c) 2 2 1 2 1 1 2 1 2 1 −                         = T T L L M M n n = 1 1 2 sec 100 −                     m min cm k g gm = 1 2 2 1 3 60 sec sec 10 10 100 −                     cm cm gm gm n2 = 3.6 10 3600 3 = 90. (a) [L/R] is a time constant so its unit is Second. 91. (d) Poission ratio is a unitless quantity. 92. (b) 93. (a) 94. (d) u P nu n 1 =   95. (a) 1 Faraday = 96500 coulomb. 96. (b) 97. (a) 98. (d) 99. (b) 100.(d) 2 2 1 2 1 2 2 1 2 4 1 4 1 − − = =  = C m N Fr q q r q q F   101.(d) Joule-sec is the unit of angular momentum where as other units are of energy. 102.(a) −1 = = Nm l F T 103.(a) Because in S.I. system there are seven fundamental quantities. 104.(d) 1 1 [ ] − −  = ML T so its unit will be kg/m-sec. 105. (b) 106. (b) 107.(b) According to the definition. 108. (b) Pyrometer is used for measurement of temperature. 109.(a) Pressure 1 2 Area Force − − = = ML T Stress 1 2 Area Restoring force − − = = ML T 110.(c) Strain   = L L dimensionless quantity 111.(b) Power 2 3 2 2 Time Work − − = = = ML T T ML T 112.(a) Calorie is the unit of heat i.e., energy. So dimensions of energy 2 −2 = ML T 113.(b) Angular momentum = −1 2 −1 mvr = MLT  L = ML T 114.(c) R L = Time constant 115.(c) Impulse = change in momentum so dimensions of both quantities will be same and equal to MLT–1 116.(b) RC = T ∵ [ ] [ ] 2 −3 −2 R = ML T I and [ ] [ ] 1 2 4 2 C M L T I − − = 117.(a) m Q Q = mL  L = (Heat is a form of energy) = [ ] 0 2 2 2 2 − − = M L T M ML T 118.(d) Volume elasticity = Volume strain Force/Area Strain is dimensionless, so = [ ] Area Force 1 2 2 2 − − − = = ML T L MLT 119.(b) 1 2 2 2 1 2 m m Fd G d Gm m F =  =  [ ] [ ] [ ][ ] [ ] 1 3 2 2 2 2 − − − = = M L T M MLT L G
120.(a) Angular velocity = , t  [] [ ] [ ] [ ] 1 0 0 0 − = = T T M L T 121.(a) Power = [ ] Time Work done 2 3 2 2 − − =         = ML T T ML T 122.(a) Couple = Force  Arm length = [ ][ ] [ ] −2 2 −2 MLT L = ML T 123.(b) Angular momentum = mvr [ ][ ] [ ] −1 2 −1 = MLT L = ML T 124.(b) Impulse = Force  Time = [ ][ ] [ ] −2 −1 MLT T = MLT 125.(d) Modulus of rigidity = = Shear strain Shear stress [ ] −1 −2 ML T 126. (a) 127.(c) [ ] [ ][ ] [ ] [ ] 2 −2 −1 2 −1 E = hv  ML T = h T  h = ML T 128.(b) Moment of inertia [ ][ ] 2 2 = mr = M L Moment of Force = Force  Perpendicular distance = [ ][ ] [ ] −2 2 −2 MLT L = ML T 129.(a) Momentum = [ ] −1 mv = MLT Impulse = Force  Time = [ ] [ ] [ ] −2 −1 MLT  T = MLT 130.(b) Pressure = Volume Energy Area Force = = −1 −2 ML T 131.(d) [ ] [Angular momentum] [ ] 2 −1 h = = ML T 132.(a) By principle of dimensional homogenity P V a =      2  [ ] [ ][ ] [ ] [ ] 2 1 2 6 a = P V = ML T  L − − = [ ] 5 −2 ML T 133.(d) = 2 2 1 CV Stored energy in a capacitor = [ ] 2 −2 ML T 134. (a) 2 2 1 Li = Stored energy in an inductor = [ ] 2 −2 ML T 135.(d) Energy per unit volume = [ ] [ ] [ ] 1 2 3 2 2 − − − = ML T L ML T Force per unit area = [ ] [ ] [ ] 1 2 2 2 − − − = ML T L MLT Product of voltage and charge per unit volume Volume Power Time Volume Volume  = =  = V Q VIt  [ ] [ ] [ ][ ] 1 2 3 2 3 − − − = ML T L ML T T Angular momentum per unit mass = [ ] [ ] [ ] 2 1 2 1 − − = L T M ML T So angular momentum per unit mass has different dimension. 136.(d) Time constant  = [T] and Viscosity [ ] −1 −1  = ML T For options (a), (b) and (c) dimensions are not matching with time constant. 137.(d) By putting the dimensions of each quantity both the sides we get x y [T ] [M] [MT ] −1 −2 = Now comparing the dimensions of quantities in both sides we get x + y = 0 and 2y = 1  2 1 , 2 1 x = − y = 138.(c) m = linear density = mass per unit length =       L M A= force = [ ] −2 MLT  [B]= [ ] [ ] [ ] [ ] 1 2 − − = ML MLT m A [ ] 2 −2 = L T This is same dimension as that of latent heat. 139.(c) Let .    x y z v = kg Now by substituting the dimensions of each quantities and equating the powers of M, L and T we get  = 0 and x = 2,y = 1,z = 1 . 140.(a) Farad is the unit of capacitance and V Q C = = [ ] [ ] 2 −2 −1 ML T Q Q = 1 2 2 2 M L T Q − − 141.(a) l RA  = i.e. dimension of resistivity is [ ] 3 −1 −2 ML T Q 142.(b) From the principle of homogenity       v x has dimensions of T. 143.(b)       = − dx d KA dt dQ   [K] = [ ][ ] [ ] [ ] [ ] 2 2 2 L K L T ML T  − = −3 −1 MLT K 144. (c) Stress = [ ] [ ] [ ] Area Force 1 2 2 2 − − − = = ML T L MLT 145. (c) 146.(a) [C] = [ ] 1 2 4 2 2 2 2 2 2 M L T A ML T A T W Q V Q − − − =         =          =     