Title 10. Surface Tension Easy Ans.pdf ✅

1. (a) 2. (b) 3. (b) 4. (a) 5. (d) Soap helps to lower the surface tension of solution, thus soap get stick to the dust particles and grease and these are removed by action of water. 6. (a) 7. (b) 8. (b) 9. (b) 10. (c,d) At critical temperature ( T C K o c = 370 = 643 ), the surface tension of water is zero. 11. (d) 12. (a) Weight of spiders or insects can be balanced by vertical component of force due to surface tension. 13. (b) 14. (b) 15. (c) Force on each side = 2TL (due to two surfaces)  Force on the frame = 4(2TL) = 8TL 16. (d) 17. (a) 18. (c) This happens due to viscosity. 19. (c) 20. (d) The total length of the circular plate on which the force will act = 2R Force to pull = 2RT = 2  5 75 = 750 dynes 21. (b) 22. (d) (1 ) 0 T = T − t 23. (a) Due to force of attraction it is not easier to separate the two glass plates. 24. (a) Soluble impurities increases the surface tension. 25. (c) 2 5 728 2  = = l F T  T = 72.8 dyne/cm 26. (d) Cohesive force > Adhesive force, so shape of liquid surface near the solid would be convex. For example mercury surface in glass capillary is convex. 27. (d) Surface tension decreases with increase in temperature. 28. (b) 29. (b) 30. (d) 31. (d) Because surface tension of water > surface tension of oil 32. (c) Surface tension pulls the plates towards each other. 33. (d) Sphere has the minimum surface area for the given volume of the liquid. 34. (c) Weight of metal disc = total upward force = upthrust force + force due to surface tension = weight of displaced water + T cos  (2 r) = W + 2 rT cos  35. (a) N m l F T 0.1 / 2 10 10 2 10 2 2 2 =    = = − − F T F = T × 2l T  T  T
36. (b) Surface tension of water decrease with rise in temperature. 37. (b) 38. (a) Force required to separate the plates F = 2TA t = 2×70×10 −3×10 −2 0.05×10−3 = 28N 39. (a) 40. (c) The cohesive force is the force of attraction between the molecules of same substance. 41. (d) 42. (c) T = F l = [MLT −2] [L] = [ML 0T −2 ] 43. (d) Net force on stick F F (T T )l = 1 − 2 = 1 − 2 = (0.07 − 0.06)l = 0.01  2 = 0.02 N 44. (a) Because film tries to cover minimum surface area. 45. (a) Force required, F = 2rT = 2  2 70 = 280 .Dyne 46. (a) 47. (a) Energy needed = Increment in surface energy = (surface energy of n small drops) – (surface energy of one big drop) 4 4 4 ( ) 2 2 2 2 = n r T − R T = T nr − R 48. (d) 49. (a) When two droplets merge with each other, their surface energy decreases. W = T(A) = (negative ) i.e. energy is released. 50. (d) E = 4πR 2T(n 1⥂/⥂3 − 1) = 4 × 3.14 × (1.4 × 10 −1 ) 2 × 75(125 1/3 − 1) = 74 erg 51. (d) W = 8πT(R2 2 − R1 2 ) = 8πT[(2r) 2 − (r) 2 ] = 24πr 2T 52. (b) Work done in splitting a water drop of radius R into n drops of equal size= 4πR 2T(n 1/3 − 1) = 4π × (10 −3 ) 2 × 72 × 10 −3 × (10 6/3 − 1) = 4π × 10 −6 × 72 × 10 −3 × 99 = 8.95 × 10 −5 J 53. (c) W = 4πR 2T(r 1/3 − 1) = 4πR 2T(8 1/3 − 1) = 4πR 2T 54. (d) W = T × 8π(r2 2 − r1 2 ) = T × 8π ( D 2 4 − d 2 4 ) = 2π(D 2 − d 2 )T 55. (c) Work done to increase the diameter of bubble from d to D W = 2π(D 2 − d 2 )T = 2π[(2D) 2 − (D) 2 ] T = 6πD 2T 56. (c) 8 ( ) 2 1 2 2 W = T r − r                 −         = 2 2 2 1 8   T  W =   3 8   30  = 720 erg 57. (c) W = T × ΔA = 5 × 2 × (0.02) (Film has two free surfaces) = 2 × 10 −1 J 58. (c) W R T 2 = 8  2 W  R (T is constant) If radius becomes double then work done will become four times. 59. (c) W = 4πR 2T(n 1/3 − 1) = 4π × 1 × 50(10 3/3 − 1) = 1800π erg 60. (a) 61. (b) Surface energy of combined drop will be lowered, so excess surface energy will raise the temperature of the drop. 62. (b) Surface energy = surface tension × increment in area = T  A 63. (d) W = 8πR 2T = 8 × π × (10 −2 ) 2 × 2 × 10 −2 = 16π × 10 −6 J 64. (a) E = 4πR 2T(n 1/3 − 1) = 4 × 3.14 × 10 −4 × 35 × 10 −1 (10 6/3 − 1) = 4.4 × 10 −3 J 65. (a)
66. (b) = 67. (b) Surface energy = T  A = 0.5  2 (0.02) = J 2 2 10 −  68. (d) Volume of liquid remain same i.e. volume of 1000 small drops will be equal to volume of one big drop n 4 3 πr 3 = 4 3 πR 31000r 3 = R 3R = 10r ∴ r R = 1 10 surface energy of one small drop surface energy of one big drop = 4πr 2T 4πR2T = 1 100 69. (a) E = T × ΔA = 3 × 10 −2 × 2(100 × 10 −4 ) = 6 × 10 −4 J 70. (a) W = 8πR 2T = 8 × 3.14 × (10 × 10 −2 ) × 3 100 = 7.536 × 10 −3 J 71. (b) Work done = 4πR 2T(n 1/3 − 1) = 4π ( D 2 ) 2 σ(n 1/3 − 1) = πD 2σ(27 1/3 − 1) = 2πD 2σ 72. (d) As volume remain constant therefore R = n 1/3 r surface energy of one big drop surface energy of n drop = 4πR 2T n × 4πr 2T R 2 nr 2 = n 2/3r 2 nr 2 = 1 n1/3 = 1 (1000) 1/3 = 1 10 73. (b) W = T × ΔA  T = W ΔA T = 3×10 −4 2×(110−60)×10−4 (Soap film has two surfaces) =3 × 10 −2N/m 74. (d) 4 3 πR 3 = 1000 × 4 3 πr 3 (As volume remains constant) R 3 = 1000r 3 R = 10r ⇒ r = R 10 75. (c) Because energy is liberated 76. (a,d) 77. (c) As volume remains constant R 3 = 8000r 3 ∴ R = 20r Surface energy of one big drop Surface energy of 8000 small drop = 4πR 2T 8000 4πr 2T = R 2 8000r 2 = (20r) 2 8000r 2 = 1 20 78. (b) Surface energy = T  A = 5  2 (0.15) = 1.5 J 79. (c) As volume remains constant therefore R = n 1/3 r Energy of big drop Energy of small drop = 4πR 2T 4πr 2T = R 2 r 2 = (8) 2/3 = 4 80. (a) N m A W T 2 10 / 2 (50 10 ) 2 10 2 4 4 − − − =     =  = 81. (a) W = TΔA = 4πR 2T(n 1/3 − 1) = 4 × 3.14 × (10 −2 ) 2 × 460 × 10 −3 × [(10 6 ) 1/3 ⥂ −1] = 0.057 82. (a) 83. (b) Increment in area of soap film = A2 − A1 = 2 × [(10 × 0.6) − (10 × 0.5)] × 10 −4 = 2 × 10 −4m2 Work done = T × ΔA = 7.2 × 10 −2 × 2 × 10 −4 = 1.44 × 10 −5 J 84. (a) Increase in surface energy or work done in splitting a big drop = 4πR 2T(n 1/3 − 1) ⇒ W = 4π × (2 × 10 −3 ) 2 × 0.465(8 1/3 − 1) = 23.4 μJ 85. (b) The ratio of the total surface energies before and after the change = n 1/3 : 1 = 2 1/3 : 1 86. (a) W = 8πS(R2 2 − R1 2 ) = 8πS[(2R) 2 − R 2 ] = 24πR 2S 87. (a) W = 8πr 2 × T = 8π × (0.2) 2 × 0.06 = 192π × 10 −4 J 88. (b) Increment in Potential energy = T × ΔA = 0.02 × 2 × 0.05 = 2 × 10 −2 J 89. (a) E = T × ΔA = 75 × 0.04 = 3J 90. (c) r = r1r2 r2−r1 = ∞ sin ⥂ ce r1 = r2 2 2 2 2 8 8 (1 10 ) 1.9 10 − − W= R T =      J 6 15.2 10 − 
91. (b) 92. (a) 93. (b) Cohesive force decreases so angle of contact decreases. 94. (d) 95. (b) 96. (b) 97. (d) 98. (b) 99. (a) 100.(c) Angle of contact is acute. 101. (a) 102. (c) 103. (b) 104.(b) Since for such liquid (Non-wetting) angle of contact is obtuse. 105. (b) Both liquids water and alcohol have same nature (i.e. wet the solid). Hence angle of contact for both is acute. 106.(d) Tangent drawn at point of contact makes 0° with wall of container. 107. (c) 108.(c) Since R P 1   109.(b) Excess pressure r T P 4  = 2 3 1 10 4 2 25 10 − −     = 20 N/m 20 Pa 2 = = (as r = d/2) 110. (c) 111. (c) 112. (c) rdg T h r T hdg 2 2 =  = 113. (b) 2 40 / 4 N m r T P = = 114. (b) 115.(b) 4 4 rhdg hdg T r T P = =  = 4 0.35 0.8 1 10 3    = = 70 dyne/cm  68.66 dyne/cm 116.(c) Outside pressure = 1 atm Pressure inside first bubble = 1.01 atm Pressure inside second bubble = 1.02 atm Excess pressure P1 = 1.01 −1 = 0.01 atm Excess pressure P2 = 1.02 −1 = 0.02 atm 1 2 0.01 1 1 0.02 1 2 2 1 = =    =      P P r r P r r P Since 1 8 1 2 3 4 3 3 2 1 2 3 1  =      =         =  = r r V V V r 117.(b) =  2cos rhdg S Pressure difference cos 2 r S = hdg = 118. (c) 119.(c) Excess pressure inside soap bubble is inversely proportional to the radius of bubble i.e. r P 1   This means that bubbles A and C posses greater pressure inside it than B. So the air will move from A and C towards B. 120.(c) P1V1 = P2V2 (H + h)ρg × 4 3 πr 3 = H × 4 3 π(2r) 3 ⇒ H + h = 8H h = 7H