Question

Problem 4. (a) Surface tension of soap solution is 0.072
N/m. How much work is done in blowing a soap bubble of
radius1/3√π ?
(b) How much work is done to form an air bubble of
radius 1/3√π inside the solution?
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Answer 1

Question 1

Given

surface tension of given soap = 0.072

radius of bubbles= 1/3sqrtπ

increases in area of bubbles =4 πr^2 =4 π(1/3sqrtπ)^2

Now formula

work = Surface tension * total increases in area

On putting above value in formula

work = 0.072*π/9*4π= 0.02joule