The work done to break a spherical drop of ra | vedclass

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(C) The volume of the liquid remains constant during the process.Let $R$ be the radius of the large drop and $r$ be the radius of each of the $n$ smal

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$n^{1/3} - 1$

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(C) The volume of the liquid remains constant during the process.Let $R$ be the radius of the large drop and $r$ be the radius of each of the $n$ small drops.Volume of large drop = $n 	imes$ Volume of small drop$\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3$$R^3 = n r^3 \implies r = \frac{R}{n^{1/3}}$The work done $W$ is equal to the increase in surface energy:$W = S 	imes \Delta A = S 	imes (A_{final} - A_{initial})$$W = S 	imes (n 	imes 4 \pi r^2 - 4 \pi R^2)$Substitute $r = R n^{-1/3}$:$W = S 	imes 4 \pi (n 	imes (R n^{-1/3})^2 - R^2)$$W = S 	imes 4 \pi R^2 (n 	imes n^{-2/3} - 1)$$W = 4 \pi R^2 S (n^{1/3} - 1)$Since $4 \pi R^2 S$ is constant,the work done is proportional to $(n^{1/3} - 1)$.

The work done to break a spherical drop of radius $R$ into $n$ drops of equal size is proportional to .............

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