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(B) Let $R$ be the radius of the bigger drop and $r$ be the radius of a single small water drop.Since the volume remains conserved,the volume of the b

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

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(B) Let $R$ be the radius of the bigger drop and $r$ be the radius of a single small water drop.Since the volume remains conserved,the volume of the big drop equals the sum of the volumes of $n$ small drops:$\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3$$R^3 = n r^3 \Rightarrow R = n^{1/3} r$Surface energy of $n$ small drops is $E_n = n 	imes (4 \pi r^2 T)$,where $T$ is the surface tension.Surface energy of the big drop is $E = 4 \pi R^2 T$.The ratio of the surface energy of $n$ droplets to the surface energy of the big drop is:$\frac{E_n}{E} = \frac{n 	imes 4 \pi r^2 T}{4 \pi R^2 T} = \frac{n r^2}{R^2}$Substituting $R = n^{1/3} r$:$\frac{E_n}{E} = \frac{n r^2}{(n^{1/3} r)^2} = \frac{n r^2}{n^{2/3} r^2} = n^{1 - 2/3} = n^{1/3} = \sqrt[3]{n}$Thus,the ratio is $\sqrt[3]{n}: 1$.

$A$ 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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