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

Answer

B

Explanation

`W_S/W_B=(2xx4piTr^2)/(4piTR^2)=(2r^2)/((n^(1//3)r)^2)=(2r^2)/(2^(1//3)r^2)=1/(2^(-1//3))`
`implies (2^(1//3))/(1)=2^(1//3):1`