Physics - Fluid Question with Solution | TestHub

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

Since volume remains constant during the process:

$\frac{4}{3}\pi R^3=n\times \frac{4}{3}\pi r^3$

$R=n^{1/3}r$

The surface energy of one small drop is $E=4\pi r^{2}T$, where $T$ is the surface tension.

Initial total surface energy of $n$ drops:

$E_i=n\times (4\pi r^{2}T)=nE$

Final surface energy of the single large drop:

$E_f=4\pi R^{2}T=4\pi (n^{1/3}r{)}^{2}T=n^{2/3}(4\pi r^{2}T)=n^{2/3}E$

Since $E_i>E_f$, the surface energy decreases, which means energy is released.

Energy released $=E_i-E_f=nE-n^{2/3}E=E(n-n^{2/3})$

Answer: energy released will be $\left[E\left(n-n^{2/3}\right)\right]$