(c) Surface area of the liquid drop A = 4Ï€R2 Let E be the surface energy of liquid drop. When the drop splits into 512 droplets, the surface area becomes A2 = 512 x 4Ï€r2 [r = radius of smaller drop] Comparing the volumes of biggerand all smaller droplets, we get i.e.
4
3
πR3=512×
4
3
πr3 ⇒ r=
R
8
Total area of smaller droplets is A1=512×4×π×(