A big water drop is formed by the combination of ' $n$ ' small water droplets of equal radii. The ratio of the surface energy of ' $n$ ' droplets to the surface energy of the big drop is

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Aicharya · Accepted Answer

Let R be the radius of bigger drop and r be the radius of single small water drop. Volume of big drop $=n($ Volume of small drop $)$ $\begin{aligned} \therefore \quad & \frac{4}{3} \pi R^3=n \times \frac{4}{3} \pi r^3 \\ & \Rightarrow R^3=n r^3 \\ & R=n^{\frac{1}{3}} r \end{aligned}$
Surface energy of $n$ drops $\left(E_n\right)=n \times 4 \pi r^2 \times T$ Surface energy of big drop $(E)=4 \pi R^2 T$ $\therefore \quad \frac{E_n}{E}=\frac{n r^2}{R^2}=\frac{n r^2}{\left(n^{\frac{1}{3}} r\right)^2}=\frac{n r^2}{n^{\frac{2}{3}} r^2}=n^{\frac{1}{3}}=\sqrt[3]{n}: 1$

A big water drop is formed by the combination of ' $n$ ' small water droplets of equal radii. The ratio of…

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