Physics - Fluid Question with Solution | TestHub

The surface area of the liquid drop is$A=4\pi R^2$

Its surface energy is$E$

When the drop splits in$512$droplets, the surface area of each droplet is$4\pi r^2$

$\therefore$Total surface area${\mathit{\text{A}}}_2$$=512\times 4\pi r^2$

The volume of the bigger drop is$\frac{4}{3}\pi R^3$and the volume of small droplets is$512\times \frac{4}{3}\pi r^3$

$\therefore \frac{4}{3}\pi R^3=512\times \frac{4}{3}\pi r^3\Rightarrow r=\frac{R}{8}$

$\therefore A_2=512\times 4\pi r^2=512\times 4\mathrm{\pi }{\left(\frac{R}{8}\right)}^2=8A_1$

Surface energy$E=A.T$($T$is surface tension and$A$is area)

$\therefore \frac{E_{\mathrm{n}}}{E_0}=\frac{A_2\cdot \mathrm{T}}{A_1\cdot \mathrm{T}}=\frac{8\cdot A_1}{A_1}=8$

$\therefore \mathrm{ }{E}_{\mathrm{n}}=8E$