A spherical liquid drop splits in to 729 identical spherical drops. If $E$ is the surface energy of the…

The volume of liquid remains constant when the spherical drop splits into 729 identical smaller drops. The radius $R$ of the original drop and radius $r$ of each small drop satisfy $R^3 = 729r^3$, so $R = 9r$.

The surface energy of the original drop is $E = \sigma \cdot 4\pi R^2$, where $\sigma$ is the surface tension. The total surface energy of the 729 smaller drops is $U = \sigma \cdot 729 \cdot 4\pi r^2$.

The ratio becomes

$\frac{E}{U} = \frac{R^2}{729r^2}$

Substituting $R = 9r$ gives

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

Since $\frac{E}{U} = \frac{1}{x}$, we conclude that $x = 9$.