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(B) The volume of the original drop is equal to the sum of the volumes of the $n$ smaller droplets.Let $r$ be the radius of each small droplet.$n \cdo

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$4 \pi R^2(n^{1/3}-1) T$

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(B) The volume of the original drop is equal to the sum of the volumes of the $n$ smaller droplets.Let $r$ be the radius of each small droplet.$n \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$$r^3 = \frac{R^3}{n} \implies r = \frac{R}{n^{1/3}}$Work done $W$ is equal to the change in surface area multiplied by the surface tension $T$.$W = (A_{	ext{final}} - A_{	ext{initial}}) T$$W = (n \cdot 4 \pi r^2 - 4 \pi R^2) T$Substitute $r = R \cdot n^{-1/3}$:$W = (n \cdot 4 \pi (R \cdot n^{-1/3})^2 - 4 \pi R^2) T$$W = (4 \pi R^2 \cdot n \cdot n^{-2/3} - 4 \pi R^2) T$$W = 4 \pi R^2 (n^{1/3} - 1) T$

$A$ liquid drop of radius $R$ is broken into $n$ identical small droplets. The work done is $[T = \text{surface tension of the liquid}]$

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