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Question

A spherical liquid drop of radius $$R$$ is divided into eight equal droplets. If the surface tension is $$T$$, then the work done in this process will be


A
2πR2T
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B
3πR2T
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C
4πR2T
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D
2πRT2
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Solution

The correct option is C $$4\pi { R }^{ 2 }T$$
Let us say $$r$$ is the radius of each eight of the bubble. Since volume of water remains the same, we have
$$\frac { 4 }{ 3 } \pi R^{ 3 }=8 \times \frac { 4 }{ 3 } \pi r^{ 3 }\\ \Rightarrow r=\frac { R }{ 2 } $$
So the work done in this process will be = change in the surface energy i.e.
$$\left( 8\times 4\pi r^{ 2 }\times T \right) -\left( 4\pi R^{ 2 }\times T \right) \\ =\left( 8\times 4\pi \left( \frac { R }{ 2 }  \right) ^{ 2 }\times T \right) -\left( 4\pi R^{ 2 }\times T \right) \quad \quad \quad \because \quad r=\frac { R }{ 2 } \\ =8\pi R^{ 2 }T-4\pi R^{ 2 }T=4\pi R^{ 2 }T$$

Physics

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