Surface area of big liquid drop of radius R, A1=4πR2 Surface area of small liquid drop of radius r, a2=4πr2 Total surface area of n small liquid drops of radius R ‌A2=na2=n⋅4πr2 ‌‌ Required energy ‌=T×∆A ‌‌=T(A2−A1)=T(4πr2n−4πR2)