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Two mercury drops (each of radius $\mathrm{r}$ ) merge to form a bigger drop. The surface energy of the bigger drop, if $\mathrm{T}$ is the surface tension, is
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Verified Answer
The correct answer is:
$2^{8 / 3} \pi \mathrm{r}^{2} \mathrm{~T}$
Let $\mathrm{R}$ be the radius of the bigger drop, then Volume of bigger drop $=2 \times$ volume of small drop
$$
\frac{4}{3} \pi R^{3}=2 \times \frac{4}{3} \pi r^{3} \Rightarrow \mathrm{R}=2^{1 / 3} r
$$
Surface energy of bigger drop,
$$
\mathrm{E}=4 \pi R^{2} T=4 \times 2^{2 / 3} \pi r^{2} T=2^{8 / 3} \pi r^{2} T
$$
$$
\frac{4}{3} \pi R^{3}=2 \times \frac{4}{3} \pi r^{3} \Rightarrow \mathrm{R}=2^{1 / 3} r
$$
Surface energy of bigger drop,
$$
\mathrm{E}=4 \pi R^{2} T=4 \times 2^{2 / 3} \pi r^{2} T=2^{8 / 3} \pi r^{2} T
$$
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