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

Step-by-step solution

Total surface energy before coalesce E₁=2 (4 R^2 ) T But 4 3 R ^3 2= 4 3 R ¹³ R^ =(2)^ 1 / 3 R Total surface energy after coalescing aligned & E ₂=4 R^ 2 ~T =4 (2)^ 2 / 3 R^2 ~T & E ₁ E ₂ = 2 (4 R^2 ) T 4 (2)^ 2 / 3 R ^2 ~T =2^ 1- 2 3 =2^ 1 / 3 aligned