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

Options

A. `1:2`
B. `2:1`
C. `1:2^(1//3)`
D. `2^(1//3):1`

Answer

D

Explanation

`2[4/3 piR^(3)]d = 4/3 pir^(3)d`
` :. r^(3) 2r^(3) " " :. r = 2^(1//3) R`
` :' E = T xxA`
` :. (E_(1))/(E_(2)) =(2(4piR^(2))T)/(4pir^(2)T)=(2R^(2))/(r^(2))`
` =(2R^(2))/(2^(1//3)R^(2))=2^(1//3):1 `