Tiêu đề 10. Surface Tension Hard Ans.pdf ✅

1. (a) W R T J 2 2 2 4 75.36 10 100 3 8 8 (10 10 ) − − =  =   =  2. (a) Increase in surface energy 4 (2 10 ) (0.465 )(8 1) 3 2 1 / 3 =  − −  = J 6 23.4 10 −  = 23.4J 3. (b) 2 1 A = 10  6 = 60cm 4 2 60 10 m − =  , 2 4 2 2 A 10 11 110 cm 110 10 m − =  = =  As the soap film has two free surfaces W = T  2 A 2 ( )  W = T   A2 − A1  2 4 4 4 3 10 2 50 10 3 10 2 50 10 − − − − =     =   = W T N/m 4. (b) As film have two free surfaces W = T  2A 2 3 7.2 10 2 0.1 1 10 − − =      5 1.44 10 − =  J 5. (d) As volume of the bubble 3 3 4 V = R  1 / 3 1 / 3 4 3 R  V      =   2 / 3 2 / 3 2 4 3 R  V      =   2 2 / 3 R  V Work done in blowing a soap bubble W R T 2 = 8  2 2 / 3 W  R  V  2 / 3 1 2 1 2         = V V W W 2 / 3 2       = V V W2 3 W 2 / 3 1 / 3 = (2) = (4)  = 4 6. (b) Energy released =       − r R TR 1 1 4 3         −      = r R R T 1 1 3 4 3 3        = − r R VT 1 1 3 7. (c) Pressure inside a bubble when it is in a liquid R T Po 2 = + 3 3 5 0.1 10 70 10 1.013 10 2 − −   =  +  5 = 1.027 10 Pa 8. (a) Excess pressure inside a soap bubble r T P 4  =  1 : 4 1 2 2 1 = =   r r P P 4 ( 1) 2 1/ 3 = R T n −