1000 small water drops of equal size combine to form a big drop. The ratio of final surface energy to the total initial surface energy is

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Aicharya · Accepted Answer

Radii of the bigger and smaller drops are related as, $\mathrm{R} \propto \mathrm{n}^{\frac{1}{3}} \mathrm{r}$ Now, surface energy $E=T d A=T 	imes 4 \pi R^2$ $\begin{array}{ll} 	herefore \quad & \frac{\mathrm{E}_2}{\mathrm{E}_1}=\frac{4 \pi \mathrm{R}^2 	imes \mathrm{T}}{\mathrm{n} 	imes 4 \pi \mathrm{r}^2 \mathrm{~T}} \ 	herefore \quad \frac{\mathrm{E}_2}{\mathrm{E}_1} & =\frac{\mathrm{R}^2}{\mathrm{nr}^2}=\frac{\mathrm{n}^{\frac{2}{3}} 	imes \mathrm{r}^2}{\mathrm{nr}^2} \ & \frac{\mathrm{E}_2}{\mathrm{E}_1}=\frac{1}{\mathrm{n}^{\frac{1}{3}}}=\frac{1}{1000^{\frac{1}{3}}} \ 	herefore \quad \frac{\mathrm{E}_2}{\mathrm{E}_1} & =1: 10 \end{array}$

1000 small water drops of equal size combine to form a big drop. The ratio of final surface energy to the…

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