Two small drops of mercury, each of radius r , coalesce to form a single large drop of radius R . The ratio of the total surface energies before and after the change is

Step-by-step solution

As, radius of bigger drop, R=n^ 1 / 3 r=2^ 1 / 3 r aligned & R²=2^ 2 / 3 r² r² R² or 2^ -2 / 3 & aligned Initial surface energy Final surface energy &= 2 (4 r² T ) (4 R² T ) =2 ( r² R² ) &=2 2^ -2 / 3 =2^ 1 / 3 or 2^ 1 / 3 : 1 aligned aligned