If sigma is the surface tension,the work done | vedclass

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(B) The volume of the liquid remains conserved during the process.Let $R$ be the radius of the big drop and $r$ be the radius of each of the $n$ small

Vedclass · Accepted Answer

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

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(B) The volume of the liquid remains conserved during the process.Let $R$ be the radius of the big drop and $r$ be the radius of each of the $n$ small drops.Volume of big drop = $n 	imes$ Volume of small drop$\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3$$R^3 = n r^3 \implies r = R n^{-1/3}$.Work done $(W)$ is equal to the change in surface energy:$W = \sigma 	imes (	ext{Final Surface Area} - 	ext{Initial Surface Area})$$W = \sigma 	imes (n 	imes 4\pi r^2 - 4\pi R^2)$Substitute $r = R n^{-1/3}$:$W = 4\pi \sigma (n 	imes (R n^{-1/3})^2 - R^2)$$W = 4\pi \sigma (n 	imes R^2 n^{-2/3} - R^2)$$W = 4\pi R^2 \sigma (n^{1-2/3} - 1)$$W = 4\pi R^2 (n^{1/3} - 1) \sigma$.

If $\sigma$ is the surface tension,the work done in breaking a big drop of radius $R$ into $n$ drops of equal radius is:

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