The surface energy of a liquid drop is U. It | vedclass

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(A) Let $R$ be the radius of the large drop and $r$ be the radius of each small droplet.Since the volume remains constant,we have $\frac{4}{3} \pi R^3

Vedclass · Accepted Answer

$8$

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(A) Let $R$ be the radius of the large drop and $r$ be the radius of each small droplet.Since the volume remains constant,we have $\frac{4}{3} \pi R^3 = 512 	imes \frac{4}{3} \pi r^3$.This simplifies to $R^3 = 512 r^3$,which gives $R = 8r$ or $r = \frac{R}{8}$.The initial surface energy is $U = 4 \pi R^2 T$,where $T$ is the surface tension.The new surface energy $U'$ is the sum of the surface energies of $512$ droplets: $U' = 512 	imes (4 \pi r^2 T)$.Substituting $r = \frac{R}{8}$ into the equation: $U' = 512 	imes 4 \pi (\frac{R}{8})^2 T$.$U' = 512 	imes 4 \pi \frac{R^2}{64} T = 8 	imes (4 \pi R^2 T) = 8 U$.

The surface energy of a liquid drop is $U$. It splits up into $512$ equal droplets. The surface energy becomes (in $U$)

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