Physics - Fluid Question with Solution | TestHub

Let $R$ be the radius of the original drop and $r$ be the radius of each of the $729$ small drops.

By conservation of volume:
$\frac{4}{3}\pi R^3=729\times \frac{4}{3}\pi r^3$

$R^3=729r^3$

$R=9r$

The surface energy of the original drop is:
$E=4\pi R^{2}T$

The total surface energy of the $729$ small drops is:
$U=729\times 4\pi r^{2}T$

The ratio of the surface energies is:
$\frac{E}{U}=\frac{4\pi R^{2}T}{729\times 4\pi r^{2}T}=\frac{R^2}{729r^2}$

Substituting $R=9r$:
$\frac{E}{U}=\frac{(9r{)}^2}{729r^2}=\frac{81r^2}{729r^2}=\frac{1}{9}$

Given $\frac{E}{U}=\frac{1}{x}$, we get $x=9$.

Answer: 9