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

Explanation

`2[4/3 pi R^3]d = 4/3 pi r^3d` `therefore r^3 = 2R^3` `therefore r = 2^(1/3)R`
`because E = T xx A` `therefore E_1/E_2 = (2(4piR)^2T)/(4pir^2T) = (2R^2)/r^2`
= `(2R^2)/(2^(2/3)R^2) = 2^(1/3):1`