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

Step-by-step solution

Radius of one drop of mercury is R . The volume of one drop = 4 3 R^3 Total volume of the two drops, V=2 4 3 R^3= 8 3 R^3 Let the radius of the large drop formed be R^ . The volume of the large drop is also V . 4 3 R^ 3 = 8 3 R^3 R^ 3 =2 R^3 R^ =2^ 1 / 3 R . Now the surface area of the two drops is S₁=2 4 R^2=8 R^2 and the surface area of the resultant drop is S₂=4 R^ 2 =4 2^ 2 / 3 R^2 Let T be the surface tension of mercury. Therefore the surface energy of the two drops before coalescing is U₁=S₁ T=8 R^2 T and the