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A spherical drop of radius $R$ is divided into eight equal droplets. If surface tension is $\mathrm{T}$, then the work done in this process is
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$4 \pi R^{2} T$
Volume of 8 small droplets $=$ Volume of 1 big drop $\Rightarrow \frac{4}{3} \pi r^{3} \times 8=\frac{4}{3} \pi \mathrm{R}^{3} \Rightarrow r=\frac{\mathrm{R}}{(8)^{1 / 3}}$
Work done $=($ Change in area $) \times$ surface tension $=\left(4 \pi r^{2} n-4 \pi \mathrm{R}^{2}\right) \mathrm{T}$
$\left\{4 \pi\left(\frac{\mathrm{R}^{2}}{(8)^{2 / 3}}\right) \times 8-4 \pi \mathrm{R}^{2}\right\} \mathrm{T}=4 \pi \mathrm{R}^{2} \mathrm{~T}$
Work done $=($ Change in area $) \times$ surface tension $=\left(4 \pi r^{2} n-4 \pi \mathrm{R}^{2}\right) \mathrm{T}$
$\left\{4 \pi\left(\frac{\mathrm{R}^{2}}{(8)^{2 / 3}}\right) \times 8-4 \pi \mathrm{R}^{2}\right\} \mathrm{T}=4 \pi \mathrm{R}^{2} \mathrm{~T}$
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