The work done in splitting a drop of water of 1 mm radius into 10 6 droplets is [S.T. of water 72×10−3J/m2

. In splitting, the volume remains unchanged.∴ 43πR3=43πr3×n or R=n1/3r  or  r=R/(n)1/3   work done W=   surface tension x change in area =4πr2n−4πR2 =4πn×R2n2/3−R2=4πR2n1/3−1 ∴ W=4πR2n1/3−1T Hero n=106,R=1mm=10−3m and T=72×10−3J/m2  Substitutingthesevalues,weget W=8⋅95×10−5 joule