Tiêu đề 09. Elasticity Easy Ans.pdf ✅

1. (c) FL 1 l l YA A =   2. (b) Stress  Strain  Stress l L  3. (d) 2 F L L L Y l A A A d  =     2 L l d  [As F and Y are constant] The ratio of 2 L d is maximum for case (d) 4. (c) 2 FL L l l AY d =    2 2 1 1 2 2 2 1 1 1 1 2 2 8 l L d l L d     =  =  =         5. (b) Young's modulus of wire does not varies with dimension of wire. It is the property of given material. 6. (c) Depression in beam 3 3 4 WL Ybd  =  1 Y   7. (c) 2 FL 1 l l AY r =   (F, L and Y are constant) 2 2 1 2 2 1 1 2 (2) 4 4 4 l r l l cm l r   = = =  = =     8. (c) 9. (b) 2 1 l r  . If radius of the wire is doubled then increment in length will become 1 4 times i.e. 12 3 4 = mm 10. (c) 11 6 1 10 1.1 0.1 1.1 10 10 mgL l m mm AY −   = = =   11. (d) 11 6 4 2.2 10 2 10 5 10 2 1.1 10 2 YAl F N L − −      = = =  12. (a) Interatomic force constant K Y r =  0 11 10 2 10 3 10 60 / N m − =    = 13. (b) To double the length of wire, Stress = Young's modulus  12 2 2 10 . F dyne A cm =  If A = 1 then F = 2 × 1012 dyne 14. (d) 15. (a) Because due to increase in temperature intermolecular forces decreases. 16. (c) Breaking Force  Area of cross section of wire (r 2 ) If radius of wire is double then breaking force will become four times. 17. (a) Y K Y = − = + 3 (1 2 ) and 2 (1 )    Eliminating  we get 9 3 K Y K   = + 18. (a) 10 6 9 10 4 10 0.1 360 100 YAl F N L   −      = = = 19. (d) Energy stored per unit volume 1 Stress Strain 2 =   1 1 2 2 Young's modulus (Strain) 2 2 =   =   Y x 20. (c) l L  i.e. if length is reduced to half then increase in length will be . 2 l 21. (b) 2 2 2 11 8 (8 10 ) 1.5 9.8 9.6 10 2 2 5 10 L dg l m Y −    − = = =    22. (b) force Stress Area =  2 1 Stress r  2 2 (2) 4 B A B A A B S r S S S r   = =  =     l W b  d
23. (b) Breaking force  Area of cross section of wire i.e. load hold by the wire does not depend upon the length of the wire. 24. (d) If length of wire doubled then strain = 1 Y = stress  12 12 F Y A dyne =  =  =  10 0.5 0.5 10 25. (b) Due to elastic fatigue its elastic property decreases. 26. (d) FL l AY =  2 1 l r  (F,L and Y are same) 2 2 1 2 4 A B B B A B l r r l r r     = = =          4 A B l l = or 4 A B l l = 27. (c) 28. (b) S cu cu S FL l Y l AY l Y =  = (F,L and Y are constant)  11 11 1.2 10 3 2 10 5 s cu l l  = =  29. (b) If length of the wire is doubled then strain = 1  Y = Force Stress Area = = 5 5 2 2 10 10 2 dyne cm  = 30. (b) 2 FL L l l AY r =   (F and Y are same)  2 2 2 2 1 1 1 2 1 1 2 2 2 l L r l L r     = =  =          1 2 0.5 . 2 2 l l l mm = = = 31. (b) F = force developed =  YA  ( ) 11 4 5 4 10 10 10 100 10 N − − =    = 32. (c) F YA =   αΔθ F A 33. (d) / strain strain F A F Y A Y =  =  = 4 9 10 7 10 0.002   = 1 2 4 2 10 7.1 10 14 m − −  =  34. (b) strain stress F A   Ratio of strain = 2 2 2 2 1 1 4 16 1 1 A r A r     = = =         35. (a) 6 2 F N L m l cm A m 2000 , 6 , 0.5 , 10− = = = = 12 2 6 2 2000 6 2.35 10 / 10 0.5 10 FL Y N m Al − −  = = =    36. (b) 4 3 9 9.8 1.96 10 / 4.5 10 F F Kx K N m x −  =  = = =   37. (a) 2 FL l r Y   2 1 l r  ( , and are constant F L Y ) 2 2 1 2 2 2 1 1 2 ( ) l r n l n l l r   = =  =     38. (b) Longitudinal strain 6 5 11 stress 10 10 10 l L Y − = = = Percentage increase in length 5 10 100 0.001% − =  = 39. (d) It is the specific property of a particular metal at a given temperature which can be changed only by temperature variations. 40. (a) 9 11 2 10 3.6 10 / 1.2 10 / 3 10 N Å Y N m m − −  = =   41. (c) 42. (d) 2 2 YA Y r Yr K K L L L  = =   i.e. force constant of a wire depends on young's modules (nature of the material), radius of the wire and length of the wire. 43. (c)
44. (a) 45. (b) 46. (d) Increase in tension of wire =  YA  6 11 2 4 8 10 2.2 10 10 10 5 8.8 N − − − =       = 47. (a) A small change in pressure produces a large change in volume. 48. (c) 2 1 2 YAl W L =  6 2 2 1 1 (0.2 10 ) 0.4 2 1 Y − −    =   Y 11 2 = 2 10 / N m 49. (d) 50. (c) Y K = − 3 1 2 (  ) 3 6 K Y K  − = = 10 10 10 3 11 10 7.25 10 0.39 6 11 10    −   =   51. (c) 52. (a) If density of the material increases then more force (stress) is required for same deformation i.e. the value of young's modulus increases. 53. (c) 4 2 4 2 5 Y N m A m F dyne N 10 / , 2 10 , 2 10 2 − = =  =  = 4 4 2 2 10 10 FL L l L AY −  = = =    Final length = initial length + increment = 2L 54. (a) 55. (b) Y is defined for solid only and for powders, Y = 0 56. (a) 57. (a) 2 2 FL FL FL l l AY r Y r  = =   (Y = constant)  2 2 2 2 2 1 1 1 1 2 1 2 2 1 2 l F L r l F L r     =  =   =          2 1 l l = i.e. increment in its length will be l. 58. (d) Breaking stress = strain × Young's modulus 11 10 2 0.15 2 3 10 Nm− =   =  59. (a) In accordance with Hooke’s law. 60. (a) F A Y =  strain = 4 11 6 1 10 2 10 0.1 2 10 N −     =  61. (c) 62. (b) Because strain is a dimensionless and unitless quantity. 63. (d) Force Stress area = . In the present case, force applied and area of cross-section of wires are same, therefore stress has to be the same. Stress Strain Y = Since the Young’s modulus of steel wire is greater than the copper wire, therefore, strain in case of steel wire is less than that in case of copper wire. 64. (b) Initial length (circumference) of the ring = 2r Final length (circumference) of the ring = 2R Change in length = 2R – 2r. change in length strain original length = 2 ( ) 2 R r r   − = R r r − = Now Young's modulus / / / ( )/ F A F A E l L R r r = = −  R r F AE r   − =     65. (a) 2 FL l  r r =  2 F l r  (Y and L are constant) 2 2 2 1 2 1 1 2 2 (2) 8 l F r l F r   =  =  =      2 1 l l mm = =  = 8 8 1 8 66. (c) 2 2 FL L l l r Y r =   (F and Y are constant) ( ) 2 2 1 1 2 2 2 1 1 2 2 l L r l L r   = =      1 2 1:1 l l = 67. (d) 2 1 l r  (F,L and Y are constant)
2 2 2 1 1 2 2 1 2 1 2.4 0.6 2 4 4 l r l l l cm l r     = =  = =  =         68. (b) l F Y A L =    2 F r  (Y,l and L are constant) If diameter is made four times then force required will be 16 times. i.e. 16 × 103 N 69. (a) l F Y A L =    2 r F L  (Y and l are constant)  2 2 1 1 2 2 2 1 2 1 1 1 4 F r L F r L         = = =                  1 2 1:1 F F = 70. (d) Increment in length 2 2 L dg l Y =  2 l L d  71. (b) Adiabatic elasticity E P =  For argon 1.6 E P Ar = ....(i) For hydrogen 2 1.4 ' E P H = ....(ii) As elasticity of hydrogen and argon are equal  1.6 1.4 ' P P =  8 ' 7 P P = 72. (d) 73. (c) 2 2 ( ) FL FL FL l AY AL Y VY = = = . If volume is fixed then 2 l L  74. (c) ( ) 11 6 5 F YA t N  2 10 3 10 10 20 10 60 − − =  =      − = 75. (a) Because dimension of invar does not varies with temperature. 76. (a) 2 FL l r Y =  2 L l r  (Y and F are constant) ( ) 2 2 2 2 1 1 1 2 1 1 2 2 2 l L r l L r     =  =  =          1 2 0.01 0.005 2 2 l m l m = = = 77. (d) Poisson’s ratio varies between – 1 and 0.5 78. (d) L l 2 2 2 = +  (1   ) and L l 1 1 1 = +  (1   )  − = − +  − (L L l l l l 2 1 2 1 2 2 1 1 ) ( )    ( ) Now (L L l l 2 1 2 1 − = − ) ( ) so, 2 2 1 1 l l   − = 0 79. (a) Thermal stress = Y  ( ) 11 5 7 2 1.2 10 1.1 10 20 10 1.32 10 / N m − =     − =  80. (a) 2 FL 1 l l AY r =   (F,L and Y are constant) 2 2 1 2 2 1 1 2 (2) 4 4 3 12 l r l l mm l r   = =  = =  =     81. (d) 2 2 3 2 6 (8) 1.5 10 10 9.6 10 2 2 5 10 L dg l m Y    − = = =    82. (d) 2 L l r  (Y and F are constant) Maximum extension takes place in that wire for which the ratio of 2 L r will be maximum. 83. (a) 84. (d) 85. (a) 6 3 250 9.8 2 50 10 0.5 10 MgL Y Al − −   = =    10 2 =  19.6 10 / N m 86. (c) 87. (c) 2 FL L l l AY r =   (F and Y are constant) 2 2 2 2 1 1 2 1 1 2 1 1 2 2 2 2 l L r l l l L r     =  =  =  =         i.e. the change in the length of other wire is 2 l 88. (b) 89. (c) 11 6 1 10 1 0.05 2 10 10 MgL l mm YA −   = = =  