Physics - Fluid Question with Solution | TestHub

Let $r$ be the radius of each small droplet and $R$ be the radius of the big drop.

By conservation of volume:
$n\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$
$R=n^{1/3}r$

Surface energy of $n$ small droplets is $E_1=n\times 4\pi r^{2}T$
Surface energy of the big drop is $E_2=4\pi R^{2}T$

The required ratio is:
$\frac{E_1}{E_2}=\frac{n\times 4\pi r^{2}T}{4\pi R^{2}T}=\frac{n{r}^2}{R^2}$

Substituting $R=n^{1/3}r$:
$\frac{E_1}{E_2}=\frac{n{r}^2}{(n^{1/3}r{)}^2}=\frac{n}{n^{2/3}}=n^{1/3}$

The ratio is $\sqrt[3]{n}:1$.

Answer: $\sqrt[3]{\mathrm{n}}:1$