Question

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

• A. $8\pi R^{2}T$
• B. $16\pi R^{2}T$
• C. $24\pi R^{2}T$
• D. $36\pi R^{2}T$

Answer 1

The correct option is D $36\pi R^{2}T$

Let Surface Tension be T.

Initial surface = $4\Pi R^2$

Final Surface area = $1000\times 4\Pi r^2$

& Initial Volume = final volume

$\frac{4}{3}\Pi R^3=1000\times \frac{4}{3}\Pi r^3$

R = 10r

So,

Final surface area = $1000\times 4\Pi \times \frac{R^2}{100}=40\Pi R^2$

So, Workdone = Change in Surface energy

= $T\Delta A=(40-4)\Pi R^{2}T$

$ChangeinSurfaceenergy=36\Pi r^{2}T$.