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Question

# 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 π r2 S

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B

12 π r2 S

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C

16 π r2 S

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D

24 π r2 S

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Solution

## The correct option is D 24 π r2 S Initial surface area of bubble =4πr2 Surface energy=2×S×4πr2=8πr2S {As there are two surfaces one inside and one outside and due to negligible thickness of soap, both can be assumed to have same radius} Now final surface area=4π(2r)2 =16πr2 Final surface energy = 2S(16πr2) {same reason} =32πr2S Vf−Vi=32πr2S−8πr2S =24πr2S So work done in the process = 24πr2S

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