Physics - Fluid Question with Solution | TestHub

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

Equating the initial and final volumes:

$\frac{4}{3}\pi R^3=729\times \frac{4}{3}\pi r^3$

$R^3=729r^3$

$R=9r$

The initial surface energy of $729$ droplets is:

$E_i=729\times 4\pi r^{2}T$

The final surface energy of the big drop is:

$E_f=4\pi R^{2}T=4\pi (9r{)}^{2}T=81\times 4\pi r^{2}T$

The ratio of final surface energy to the total initial surface energy is:

$\frac{E_f}{E_i}=\frac{81\times 4\pi r^{2}T}{729\times 4\pi r^{2}T}=\frac{81}{729}=\frac{1}{9}$

Answer: $1:9$