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

Radius of the larger drop = RSuppose, radius of the droplets = rSince, volume will be remain constant,     43πR3 = 8×43πr2[No. of droplets = 8]Work done = (Increase in surface area) ×Surface tension.= [8[4π(R2)2}-4πR2] ×T= (8πR2-4πR2)×T = 4πR2T