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

Step-by-step solution

Let R be the radius of the coalesced drop. aligned & 4 3 R^3=8 4 3 r^3 & R^3=8 r^3 & R=2 r aligned Surface Energy E = T .dA aligned & E₁=8 T d A=8 T 4 r^2 & E₂=T d A=T 4 R^2 aligned Where E₁ and E₂ are the surface energies before and after coalescing. Dividing equation (i) by (ii), aligned & E ₁ E ₂ = 8 r ^2 R ^2 & but R =2 r & E ₁ E ₂ = 8 r ^2 4 r ^2 = 2 1 & aligned