Two drops of equal radius coalesce to form a bigger drop. What is ratio of surface energy of bigger drop to smaller one?

Step-by-step solution

Volume remains constant after coalescing. Thus, 4 3 R³=2 4 3 r³ where R is radius of bigger drop and r is radius of each smaller drop. R=2^ 1 / 3 r Now, surface energy per unit surface area is the surface tension. So, Surface energy, W=T A Or W=4 R² T Therefore, surface energy of bigger drop aligned W₁ &=4 (2^ 1 / 3 r )² T &= (2^ 2 / 3 ) 4 r² T aligned Surface energy of smaller drop W₂=4 r² T Hence, required ratio W₁ W₂ =2^ 2 / 3 : 1