Title 01. Units and Dimensions Hard Ans.pdf ✅

1. (c) [X] = [Force] × [Density] = [ ] [ ] −2 −3 MLT  ML = 2. (d) (n) = Number of particle passing from unit area in unit time = A  t No. of particle [ ] [ ] [ ] 2 0 0 0 L T M L T = = [ ] −2 −1 L T [n1 ] = [n2 ] = No. of particle in unit volume = [ ] −3 L Now from the given formula [ ] [ ][ ] [ ] 2 1 2 1 n n n x x D − − = [ ] [ ][ ] 3 2 1 − − − = L L T L [ ] 2 −1 = L T . 3. (a) [E] = energy = [ ] 2 −2 ML T , [m] = mass = [M], [l] = Angular momentum = [ ] 2 −1 ML T [G] = Gravitational constant = [ ] −1 3 −2 M L T Now substituting dimensions of above quantities in 5 2 2 m G El = 5 1 3 2 2 2 2 2 1 2 [ ] [ ] [ ] [ ] − − − −   M M L T ML T ML T = [ ] 0 0 0 M L T i.e., the quantity should be angle. 4. (b) According to principle of dimensional homogeneity       = v x [k] = [ ] 1 T LT L =       − . 5. (b) From the dimensional homogeneity [ ] [ ] 2 x = B  [B] = [L 2 ] As well as   [ ] [ ] [ ][ ] 2 1 / 2 x B A x U + =  [ ] [ ][ ] [ ] 2 1 / 2 2 2 L A L ML T = − [ ] [ ] 7 / 2 −2  A = ML T Now [ ] [ ] [ ] 7 / 2 2 2 AB = ML T  L − [ ] 11 / 2 −2 = ML T 6. (c) Energy density = Volume Energy 2 1 2  0 E = [ ] 1 2 3 2 2 − − − =         = ML T L ML T 7. (c) Let x = length  [X]=[L] and [dx]=[L] By principle of dimensional homogeneity =      a x dimensionless  [a] = [x]=[L] By substituting dimension of each quantity in both sides: [ ] [ ] [ ] 2 2 1 / 2 n L L L L = −  n =0 8. (b) F = BIL [ ][ ] [ ] [ ][ ] [ ] Dimension of[ ] 2 I L MLT I L F B −  = = = [ ] −2 −1 MT I Now dimension of [ ] [ ] [ ] [ ] 2 2 2 1 2 2 M MT I L m B l P  = = − − [ ] 2 −4 −2 = ML T I 9. (b) Unit of velocity = m/sec ; in new system = 100 sec sec 100 m m = (same) Unit of force 2 sec kg  m = ; in new system 100 sec 100 sec 100 10 1  =  m kg 2 1000 sec 1 kg  m = Unit of energy 2 2 sec kg m = ; in new system 100 sec 100 sec 100 100 10 1   =  m m kg 2 2 10 sec  kg  m = Unit of pressure 2  sec = m kg ; in new system sec sec kg m 100 100 1 100 1 10 1  =   2 7 10 m sec kg  = − 10. (a) [ ] [ ] 2 −2 E = ML T 1 eluoj 6 2 4 2 100 10 10 sec − − = kg  m  4 2 2 10 sec − = kgm  Joule 4 = 10 11. (d) 1 - − gm cm s 3 2 1 10 10 − − − = kg  m  s 5 1 10 − − = kg m  s = 10–5 Ns 12. (a) GM R T 3 = 2 2 3 2 gR R =  g R = 2 Now by substituting the dimension of each quantity in both sides. 1 / 2 2 [ ]       = − LT L T = [T] L.H.S. = R.H.S. i.e., the above formula is Correct. [ ] 2 −2 −2 M L T 2 2 [100 ] [1 ] [100 sec]− = kg  km 
13. (d) Given m = mass = [M],  = coefficient of rigidity = [ ] −1 −2 ML T , L = length = [L] By substituting the dimension of these quantity we can check the accuracy of the given formulae 1 / 2 [ ][ ] [ ] [ ] 2         = L M T   = 1 / 2 1 2       − − ML T L M = [T]. L.H.S. = R.H.S. i.e., the above formula is Correct. 14. (a) Given T v = terminal velocity = [ ] −1 LT , m = Mass = [M], g = Acceleration due to gravity = [ ] −2 LT r = Radius = [L],  = Coefficient of viscosity = [] By substituting the dimension of each quantity we can check the accuracy of given formula r mg vT    [ ][ ] [ ][ ] [ ] 1 1 2 1 ML T L M LT LT − − − − = = [ ] −1 LT L.H.S. = R.H.S. i.e., the above formula is Correct. 15. (a) Given V = Rate of flow = [ ] sec Volume 3 −1 = L T , P = Pressure = [ ] −1 −2 ML T , r = Radius = [L]  = Coefficient of viscosity = [ ] −1 −1 ML T , l = Length = [L] By substituting the dimension of each quantity we can check the accuracy of the formula l P r V   8 4 =  [ ][ ] [ ][ ] [ ] 1 1 1 2 4 3 1 ML T L ML T L L T − − − − − = = [ ] 3 −1 L T L.H.S. = R.H.S. i.e., the above formula is Correct. 16. (b) Let r KA V A x 0  = By substituting the dimension of each quantity in both sides [ ] [ ].[ ][ ] [ ] 3 L L L L L x  = [ ] [ ] 3 x L L +  = ;  3 + x = 1 or x = −2 −2  A   17. (a) Relative error in measurement of length is minimum, so this measurement is most accurate. 18. (c) Total mass = 2.3 + 0.00215 + 0.01239 = 2.31 kg Total mass in appropriate significant figures be 2.3 kg. 19. (d) Area 2 = 1.5 1.203 = 1.8045 cm 2 = 1.8 cm (Upto correct number of significant figure). 20. (b) Total surface area = 2 2 6 (5.402) = 175 .09 cm 2 = 175 .1cm (Upto correct number of significant figure) Total volume 3 3 = (5.402) = 175 .64 cm 3 = 175.6cm (Upto correct number of significant figure). 21. (a) 9.99m + 0.0099m = 9.999m = 10.00 m (In proper significant figures). 22. (b) 3.124  4.576 = 14.295 =14.3 (Correct to three significant figures). 23. (c) The number of significant figure is 5 as 6 10 − does not affect this number. 24. (b) Value of current (3.23 A) has minimum significant figure (3) so the value of potential difference V(= IR) have only 3 significant figure. Hence its value be 35.0 V. 25. (c) Volume 3 3 = a = (7.023) 3 = 373.715m In significant figures volume of cube will be 3 373 .7 m because its side has four significant figures. 26. (d)     a = b c /d e So maximum error in a is given by 100 e e 100 . d d 100 . c c 100 . b b 100 . a a max    +    +    +    =         = (b1 + c1 + d1 + e1 )% 27. (a) Electric flux = Electric field × Area, dipole moment = charge × length. 28. (b)
hf = I  or T 1 T T I f h 2 = =  = 29. (c) MOI = Mr2 = [ML2 ], moment of force = torque = I = r × F = ML2T -2 30. (b) astronomical unit of distance 31. (d) a = 4 2 3 MLT t F MLT ,b t F − − = = = 32. (b) g = GM/R2 or g/G = M/R2 33. (b) [M0LT-1 ] = [MT-2 ] a [ML-1T -1 ] b [ML-3 ] c = [Ma + b + c L -b-3c T - 2a – b – 3c] on solving a = 1, b = - 1, c = 0, v = s/ 34. (a) V = r 2 l l dl r 2dr V dV = + V = 0.5 2 8.5 2 8.5 7 22    dV = 2.83        = +      +  10 1 21 1 2.83 .050 .005 8.5 2 .02 = 0.296 cm3 Thus V = 2.83  0.3 cm3 35. (c) 1450 × 1.61 = 2334 kmh-1 1 mile = 1.61 km 2334 × ms 1 18 5 − = 647.5 ms-1 36. (a) Least count = numberof divisiononcircularscale pitch 0.5 10 cm 2 0.1 −3 =  37. (a) Exponential must be dimensionless. Therefore b shall have dimension L-1 and K = Fr2 = (MLT-2 ) (L2 ) = ML3T - 2 38. (b) 100 p dp r b db m a da 100 n X dX          = + + = 2 × 1 + 2(0.5) + 4(0.75) = 6% 39. (b) Energy = ML2T -2 = [MLT-2 ] a [LT-2 ] b [T]c = MaL a + b T 2a – 2b + c a = 1, a + b = 2  b = 1, -2a – 2b + c = - 2 and c = 2 40. (a) (b) and (c) are dimensionally correct. 41. (c) There are 1440 traditional minutes in a day. The watch is showing 650 decimal minutes. Equivalent traditional minutes = 1440 1000 650  Equivalent traditional hours = 15.6 60 1440 1000 650  = hours 0.6 hours = 0.6 × 60 = 36 minutes Therefore the time is 1536 hrs. or 3 : 36 PM. 42. (b) A measurement is reported in significant figures which contain all certain and one, that is last, uncertain digit. In 0.3655 the last digit is uncertain, it comes from approximation or using mental division. The numbers before it are multiple of least count i.e. 0.365 is multiple of 0.001 m. 43. (a) Thickness of 1 sheet of paper = 1.16/75 cm = 0.0154666 mm When a measured quantity is multiplied or divided by a exact number, the result should not contain significant digits more than the measured quantity, i.e. it should contain significant figures equal to the significant figures of the measured quantity. Here 75 is a exact number and measured quantity 1.16 cm contains three significant figures. Therefore after rounding off properly answer should be expressed in three significant figures. 44. (c) Any mathematical operation like squaring; square root etc. cannot affect the number of significant digits in the result. Or in other words mathematical operations can neither increase or decrease accuracy of any measurement. Therefore answer in this case should be
expressed in two significant digits which are equal to the minimum significant digits contained by any participating quantity, (in this case by number 5.1). 45. (c) In addition or subtraction, the last significant digit of the sum or difference occupies the same relative position as the last significant digit of the quantities being added or subtracted. In the result there should only be one uncertain digit. Also sum of a certain and a uncertain digit becomes uncertain. As there can only be one uncertain digit in the result, therefore 3 which comes by addition of certain digit 2 and uncertain digit 1, is rounded off to 4. It is kept as the uncertain digit. 46. (b) An accurate measurement is very close to actual value. An instrument with lower least count is more precise and a precise measurement uses more number of significant figures. Clock B is not precise as it is not showing seconds measurement but it is accurate. A high quality measurement is both precise and accurate. 47. (c) Systematic error is due to faculty instrument, parallax etc. and is always unidirectional. It can not be reduced by increasing number of readings. Random error or probable error can be positive or negative in each reading and large number of reading reduced its effect. 48. (d) In land IV, at the starting and at the end of the motion, acceleration dv/dt is infinite. It is not practically possible to produce infinitely large accelerations. 49. (a) For horizontal motion, as component of velocity in this direction is not changed with time t = 2.5 s = 25m/ux  ux =10 m/s Also at the moment of landing, velocity is V, then V cos 60o = ux  V = (1/2) 10 = 20 m/s and Vy = V sin 600 = 2 3 20 = 10 3 m/s For the vertical motion, using v = u + at and taking upward direction as positive - 10 3 = uy + (-10)(2.5)  uy = 25 - 10 3 = 8 m/s, upward. Angle of projection tan  = 5 4 10 8 u u x y = = 50. (d) Speed of projection u = u u 8 10 12.8m/s 2 2 2 y 2 x + = + = 51. (c) N m2 kg–2 52. (a) Energy 53. (d) [M L T ] 1 2 -2 54. (a) Light year [M L T ] 0 1 0 = 55. (b) Work 56. (d) [M L T ] 1 -1 -2 57. (d) f = cmx k y [M L T ]= 0 0 -1 [M ]x [M L T ]y 1 0 0 -2 0 0 -1 x y 0 - 2y M L T M L T + = x + y = 0 ...(1) - 2y = -1 ...(2) 58. (a) 0 1 2 0 1 0 M L T M L T g 2h − =         = 0 0 2 M L T [T1 ] = Time 59. (b)