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(B) Let the radius of the large drop be $R$ and the radius of each small droplet be $r$.The surface energy of the large drop is $E = T \cdot 4\pi R^2$

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$8E$

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(B) Let the radius of the large drop be $R$ and the radius of each small droplet be $r$.The surface energy of the large drop is $E = T \cdot 4\pi R^2$, where $T$ is the surface tension.Since the volume remains conserved, the volume of the large drop equals the sum of the volumes of $512$ small droplets:$\frac{4}{3}\pi R^3 = 512 \cdot \frac{4}{3}\pi r^3$$R^3 = 512 r^3 \implies R = 8r$ or $r = \frac{R}{8}$.The final surface energy $E_{final}$ is the sum of the surface energies of $512$ droplets:$E_{final} = 512 \cdot (T \cdot 4\pi r^2)$Substitute $r = \frac{R}{8}$ into the equation:$E_{final} = 512 \cdot T \cdot 4\pi \left(\frac{R}{8}\right)^2$$E_{final} = 512 \cdot T \cdot 4\pi \cdot \frac{R^2}{64}$$E_{final} = \frac{512}{64} \cdot (T \cdot 4\pi R^2)$$E_{final} = 8E$.

$A$ liquid drop having surface energy $E$ is sprayed into $512$ droplets of the same size. Then the final surface energy is

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