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Calculate the work done against surface tension in blowing a soap bubble from a radius $10 c m$ to $20 c m$ if the surface tension of soap solution is $25 x 10^{- 3} N / m$ .

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$5 π x 10^{- 3} J$

$π x 10^{- 3} J$

$4 π x 10^{- 3} J$

$6 π x 10^{- 3} J$

answer is A.

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Detailed Solution

$Δ A = 2 x (4 {πr}_{2}^{2} - 4 {πr}_{1}^{2}) = 8 π [{(0.2)}^{2} - (0 . 1^{2})] W = TΔA = 25 x 10^{- 3} x 0.24 π = 6 π x 10^{- 3} J$

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