Physics - Fluid Question with Solution | TestHub

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

Equating the volumes before and after coalescing:

$\frac{4}{3}\pi R^3=2\times \frac{4}{3}\pi r^3$

$R=2^{1/3}r$

The total surface energy of the two small drops before coalescing is:

$E_i=2\times (4\pi r^{2}T)=8\pi r^{2}T$

The total surface energy of the big drop after coalescing is:

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

The ratio of the total surface energies after and before the change is:

$\frac{E_f}{E_i}=\frac{4\pi \times 2^{2/3}{r}^{2}T}{8\pi r^{2}T}=\frac{2^{2/3}}{2}=2^{-1/3}$

The ratio is $2^{-1/3}:1$.

Answer: $2^{-1/3}:1$