Physics - Fluid Question with Solution | TestHub

Let $T$ be the surface tension of water.

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

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

The energy loss during the formation of the big drop is given by:
$\Delta E=E_i-E$

Given that the energy loss is $3E$:
$E_i-E=3E$

$E_i=4E$

Substituting the expressions for $E_i$ and $E$:
$n\times 4\pi r^{2}T=4\times (4\pi R^{2}T)$

$n{r}^2=4R^2$

$n=\frac{4R^2}{r^2}$

Answer: $\frac{4R^2}{r^2}$