Question

A mercury droplet of radius R and surface tension $\alpha$ is broken into 8 smaller droplets of equal size. The work done by the external energy is :

• A. $\frac{4}{3}\pi R^3\alpha$
• B. $\pi R^2\alpha$
• C. $8\pi R^2\alpha$
• D. $4\pi R^2\alpha$

Answer 1

The correct option is D $4\pi R^2\alpha$

Volume of bigger drop = Total volume of 8 droplets
(4/3)*pi* R3   =  8 (4/3)* pi* r3
=> r = R/2
Change in surface area in this process = dS = 4* pi * [8*r2 - R2] = 4* pi* R2
Hence work done = alpha.dS = alpha* 4*pi*R2