Tiêu đề 10. Surface Tension Med Ans.pdf ✅

1. (d) Force on one side of the stick = 0.07  2 = 0.14 N and force on other side of the stick = 0.06  2 = 0.12N So net force on the stick = F1 − F2 = 0.14 − 0.12 = 0.02N 2. (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  3. (a) Force on wire due to surface tension F = T  2l N m l F T 0.1 / 2 10 10 2 10 2 2 2 =     = = − − 4. (d) Suppose tension in thread is F, then for small part l of thread l = R and 2F sin / 2 = 2Tl = 2TR  F 2 (sin / 2 / 2) sin / 2 / 2       = = = TR  TR TR 5. (a) From the figure Curved part = semi perimeter 2 (a + b) =  and the plane part = minor axis = 2b  Force on curved part = 2 (a b) T +   and force on plane part = T  2b  Ratio b a b 4 ( + ) =  6. (b) Weight of the body hung from wire (mg) = upward force due to surface tension (2Tl )  g Tl m 2 =       = − d r R T v 6 1 1 7. (a) ( ) 2 1 2 2 W = 8T R − R 8 [(2 ) ( ) ] 2 2 = S R − R R S 2 = 24 8. (c) Excess pressure P = Pin − Pout = 0.01atm and similarly P2 = 0.02atm and volume of air bubble 3 3 4 V = r 3 3 ( ) 1 P V r     [as r P 1   or P r   1 ]  1 8 1 2 0.01 0.02 3 3 3 1 2 2 1  =       =      =           = P P V V 9. (b) Excess pressure inside a bubble just below the surface of water r T P 2 1 = and excess pressure inside a drop r T P 2 2 = P1 = P2 10. (b) rdg T h 2 cos = W Hg Hg W Hg W Hg W d d T T h h   cos cos  = [as r and g are constants] 1 13.6 cos135 cos 0 . 3.5 10 o Hg W T T  = 34 5 136 20 3.5 13.6 10 0.7 = =    = Hg W T T 11. (b) The height upto which water will rise cos h l = cos 60 2cm = = 4cm . [h = vertical height,  = angle with vertical] 12. (d) rdg T h 2 cos = [If diameter of capillaries are same and taking value of  same for both liquids]                  = 1 2 2 1 2 1 d d T T h h              = 0.8 0.6 50 60 10 9 40 36  =      = . 13. (b) F1 = T1  L F2 = T2  L = 1.01atm − 1atm a  (a+b) 2 2b  T  T
Mass of the liquid in capillary tube M = V = (r 2 h)  M  r h  r 2 [As r h 1  ] So if radius of the tube is doubled, mass of water will becomes 2M, which will rise in capillary tube. 14. (a) rdg T h 2 cos = g h 1   [If other quantities remains constant] moon earth earth moon h g h g = = 6  hmoon = 6h [As gearth= 6gmoon] 15. (d) g h 1  . In a lift going downward with acceleration (a), the effective acceleration decreases. So h increases. 16. (b) rdg T h 2 cos = , [ =0, r mm m 3 0.5 10 2 1 − = =  , T = 0.06 N / m , d = 3 3 10 kg / m , g = 9.8 m/s 2 ] 0.5 10 10 9.8 2 0.06 cos 3 3      = −  h = 0.0244 m = 2.44cm 17. (c) As r h 1  1 2 2 1 r r h h  = or 1 3 2.2 6.6 1 2 2 1 = = = h h r r SS 18. (d) The lower end of capillary tube is at a depth of 12 + 3 = 15 cm from the free surface of water in capillary tube. So, the pressure required = 15 cm of water column. 19. (b) When the tube is placed vertically in water, water rises through height h given by rdg T h 2 cos = Upward force = 2r  T cos Work done by this force in raising water column through height h is given by W = (2rT cos)h = (2rh cos)T r h dg rhdg rh 2 2 2 cos (2 cos )      =      = However, the increase in potential energy Ep of the raised water column 2 h = mg where m is the mass of the raised column of water m r hd 2  =  So, 2 2 ( ) 2 2 2 hg r h dg EP r hd    =       = Further, 2 2 2 r h dg W Ep   −  = The part ( ) W − EP is used in doing work against viscous forces and frictional forces between water and glass surface and appears as heat. So heat released = J r h dg J W Ep 2 2 2  =  −  20. (d) Weight of liquid = upward force due to surface tension 75 10 2rT 4  = − Circumference 0.125 6 10 75 10 75 10 2 2 4 4 =   =  = − − − T r = m 2 12.5 10 −  21. (c) Under isothermal condition surface energy remain constant  r T r T R T 2 2 2 2 1 8 + 8 = 8  2 2 2 1 2 R = r + r 22. (b) Radius of curvature of common surface of double bubble 2 1 2 1 r r r r r − = 20cm 5 4 5 4 = −  = 23. (a) 1 1 cos h l = and 3 / 2 1 / 2 cos 30 cos 60 cos cos 1 2 2 1  = = = o o l l   = 1 : 3 24. (b) Force required to separate the glass plates A A t AT F =  2 V TA A t TA 2 2 2 ( ) 2 =  = . 25. (c) If  < 900 , then liquid does not wet the solid surface 26. (b) Since the density of sea-water is more than that of river therefore lesser volume of sea water is required to be displaced to balance the weight of the boat. 27. (b) ( – ) 2 1 2 r r hg
v  r 2 ; 2 1 2 1 v v r r = 28. (c) P = 2T/R ; R is less for smaller bubble, P is more for larger bubble, R is more, P is less. Since air flows from higher pressure to lower pressure therefore the air shall flow from smaller bubble to larger bubble. 29. (c) Because film has two surface so U = T × (2A) = 2 × 10–1 J 30. (c) A become A/4 so radius become just half of its present value and h  r 1 so h become just double so h = 40 cm. 31. (b) From due to surface tension property. 32. (b) On mixing partially soluble impurities, surface tension decreases. 33. (d) From practical experience it is proved. 34. (b) detergent decreases the surface tension so level of water rise will be lesser. 35. (c) h = rdg 2Tcos  r1 h1 = r2 h2 and A = 2  r 2  r  A  A1 h1 = A2 h2  A × 4 = 4 A × h2  h2 = 8 cm 36. (b) Contact angle between mercury and glass plate is acute. 37. (d) Pin – Patm = excess pressure = R 4T  2 1 R R = 1 2 P P   = 2 3  2 1 V V = 3 2 3 1 R R = 8 27 38. (b) Meniscus will be concave from upside and in soap solution it should decrease. 39. (a) 2T cos  = mg = r 2 g  r = g 2T  40. (c) Because mercury meniscus is convex. The pressure just inside the hole will be less than the outside pressure by r 2T  hg = r 2T or h = r g 2T  41. (a) gh = 2T         1 2 r 1 r 1 – = 2T         1 2 2 l r r r – r  T = ( – ) 2 1 1 2 2 r r ghr r 42. (d) Net upward force = 2 R2S + 2R1 S contact angle = 00  Capillary rise is given by h = (R R ) g 2 S(R R ) 2 1 2 2 1 2  −   + = (R R ) g 2S 2 − 1  43. (a) Excess pressure is directed towards centre of curvature and inversely proportional to radius of curvature. 44. (c) Mass of liquid inside the capillary = r 2 h d = (r h d). r since, hr = constant  mass of liquid inside  r
45. (c) Surface tension does not depend upon area of surface. 46. (b) When break into eight drop radius of small drop is r = 1/3 n R  r = 1/3 8 R = 2 R For large drop P1 = R 2T for small drop P2 = r 2T = (R / 2) 2T = 2       R 2T 47. (c) 3 4 R 3 = n × 3 4 r 3 or r = 1/3 n R Now work done = increase in area × surface tension = (4r 2 n – 4R 2 ) × T = n R T n R 4 2 2/3 2           −  (n1/3 – 1) 48. (d) As temperature increases; surface tension decreases ; As h = R g 2Tcos    h also decreases. 49. (b) Figure shows one of the legs of the mosquito landing upon the water surface. Therefore, T.2a × 8 = W = weight of the mosquito. 50. (c) A  r 2  radius becomes half. As Hr = constt.  H becomes double. 51. (a) P3 – P2 = 1 r 2T , P2 – P1 = 2 r 2T Add both equations 52. (b) W = 2T (.x) = 2 × 40 × 10–3 × .2 × .1 = 1.6 mJ 53. (a)Molecule on the surface experiences a net molecular attraction force downwards. 54. (b)After the portion A is punctured’ the thread has 2 options as shown in the figures. or Clearly, due to surface tension, the soap film wants to minimize the surface area which is happening in option (ii). Hence the thread will become concave towards A. 55. (d)In the satellite, geff becomes zero but the surface tension still prevails. Hence capillary action will push if fully outward. 56. (c)We know that surface energy US = T × Area. Here there are two surface front and back US = T × 2 × (a) = 5 N m × 2 × 0.02 m2 = 0.2 J 57. (b)Water will rise to a height more than h when downward force (mgeff ) becomes lesser than mg. so in a lift accelerating downwards, geff is (g –a0 ). Hence capillary rise is more. On the poles geff is even more than g. Hence the capillary will even drop. 58. (d)Insects use the surface tension to keep floating. 59. (a)Surface tension is a property of surface there is no surface, at critical temperature, hence surface tension is zero. 60. (d) f = 2rT = 2 × 5 × 75 = 750 dyne 61. (a) f = 2 T 2 × 10–2 = 2 × T × 0.10 T = 0.1 N/m 62. (b)W = TA 2 × 10–4 = (60 – 30) × 10–4 T × 2 T = 2 30 2  = 1 30 = 3.3 × 10–2 N/m