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(B) Let the radius of the initial large drop be $R$ and the radius of each small droplet be $r$.Since the volume remains constant,the volume of the la

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$9 E$

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(B) Let the radius of the initial large drop be $R$ and the radius of each small droplet be $r$.Since the volume remains constant,the volume of the large drop equals the sum of the volumes of the $729$ small droplets:$\frac{4}{3} \pi R^3 = 729 	imes \frac{4}{3} \pi r^3$$R^3 = 729 r^3$$R = 9r$ or $r = \frac{R}{9}$.The initial surface energy is $E = T 	imes A_{initial} = T 	imes 4 \pi R^2$,where $T$ is the surface tension.The final surface energy $E'$ is the sum of the surface energies of all $729$ droplets:$E' = 729 	imes (T 	imes 4 \pi r^2)$Substitute $r = \frac{R}{9}$ into the equation:$E' = 729 	imes T 	imes 4 \pi \left(\frac{R}{9}ight)^2$$E' = 729 	imes T 	imes 4 \pi 	imes \frac{R^2}{81}$$E' = 9 	imes (T 	imes 4 \pi R^2)$$E' = 9 E$.

$A$ liquid drop having surface energy $E$ is spread into $729$ droplets of same size. The final surface energy of the droplets is

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