Two small drop of mercury, each of radius R coalesce in from a simple 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

If R is the radius of large drop, then
`(4)/(3)pi R^(3) =2((4)/(3)pi r^(3))` or `R=2^(1//3)r`
`("surface energy of 2 small drops")/("Surface energy of the large drop") =(2[Sxx4pi r^(2)])/(Sxx4pi R^(2))`
`=2[(r)/(R)]^(2)`
`=2(2^(-1//3))^(2)` `[.:(r)/(R)=2^(-1//3)]`
`=2^(1//3) : 1`