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(D) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.Since the volume remains constant,the volume of the big drop equals

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$\sqrt[3]{n}: 1$

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(D) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.Since the volume remains constant,the volume of the big drop equals the sum of the volumes of $n$ small drops:$\frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3$$R^3 = n r^3 \implies R = n^{1/3} r$Surface energy $E$ is given by $E = T \cdot A$,where $T$ is surface tension and $A$ is the surface area.Surface energy of $n$ small drops $(E_2)$ = $n 	imes (4 \pi r^2 T) = 4 n \pi r^2 T$Surface energy of the big drop $(E_1)$ = $4 \pi R^2 T$Ratio $\frac{E_2}{E_1} = \frac{4 n \pi r^2 T}{4 \pi R^2 T} = \frac{n r^2}{R^2}$Substituting $R = n^{1/3} r$:$\frac{E_2}{E_1} = \frac{n r^2}{(n^{1/3} r)^2} = \frac{n r^2}{n^{2/3} r^2} = n^{1 - 2/3} = n^{1/3}$Thus,the ratio is $\sqrt[3]{n}: 1$.

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

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