A spherical liquid drop of radius R is divided into eight equal droplets. If the surface tension is 7, then the work done in this process will be

Radius of the larger drop = RSuppose radius of the droplets = rSince volume will remain constant,43πR3=8×43πr2 as no. of droplets =8∴ r=R3813=R2Therefore, work done = increase in surface area x Surface tension=8×4πR22−4πR2×T=8πR2−4πR2×T=4πR2T