What is the change in surface energy, when a mercury drop of radius R splits up into $1000$ droplets of radius r? T is the surface tension.

Hint: Surface tension examines to reduce the surface area of fluid. For growing surface area, the work has to be performed against the surface tension, and it is collected in the surface molecules in the sort of potential energy. Surface energy is described as the work executed per unit area to raise the accessible surface area under isothermal conditions.

Complete step by step answer:
Let T be the surface tension.
When a mercury drop of radius R splits up into 1000 droplets of radius r.
Initially, it was a big drop.
So, initial surface area of a single drop $= 4 \pi R^{2}$
 Final surface after splitting $=1000 \times 4 \pi r^{2}$
Initial volume $= \dfrac{4}{3} \pi R^{3}$
Final volume $=1000 \times \dfrac{4}{3} \pi r^{3}$
Initial volume is equal to the final volume.
Initial Volume $=$ final volume.
$\dfrac{4}{3} \pi R^{3} = 1000 \times \dfrac{4}{3} \pi r^{3}$
$\implies R^{3} = 1000 r^{3}$
$\implies R = 10 r$
$r = \dfrac{R}{10}$
Change in surface area $= $ Final surface area $-$ Initial surface area
Change in surface area $= 1000 \times 4 \pi r^{2} - 4 \pi R^{2}$
$= 1000 \times 4 \pi \dfrac{R^{2}}{100} - 4 \pi R^{2}$
$10 \times 4 \pi R^{2} – 4 \pi r^{2}$
$= 36 \pi R^{2}$
Change in surface energy is equal to the product of the surface tension and change in surface area.
Change in surface energy $=T \times 36 \pi R^{2}$
Hence, change of surface energy is $ 36 \pi R^{2}T$.

Note: Surface free energy estimates the excess power present at the material's surface compared to its bulk. It can be utilized to define wetting and adhesion among materials but is not often applied quantitatively. Surface tension is the aim of liquid covers at rest to contract into the smallest surface area possible.