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## Homework Statement

(Q)Air is pushed into a soap bubble of radius r to double its radius.If the surface tension of the soap solution is S, the work done in the process is

<a>8(pie)(r)

^{2}(S) <b>12(pie)(r)

^{2}(S) <c>16(pie)(r)

^{2}(S)

<d>24(pie)(r)

^{2}(S)

## Homework Equations

Surface energy of a liquid = (S)(A)

where S is surface tension and A is the area of the surface of liquid.

## The Attempt at a Solution

Taking the bubble as system.

Isn't it like this:-

>Work done in process i.e. work done by external agent = change in Potential energy = change in surface energy(U

_{f}- U

_{i})

> U

_{i}= 4(pie)(r)

^{2}(S)

> U

_{f}= 4(pie)(2r)

^{2}(S) = 16(pie)(r)

^{2}(S)]

> U

_{f}- U

_{i}= 12(pie)(r)

^{2}(S) = Work done we require

But the answer is <d>

Can any1 help me where i am getting wrong :)

Thanks for reading ^.^