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

Radius of one drop of mercury is R.
`therefore` the volume of the drop `=4/3 piR^3`
`therefore` total volume of the two drops,
`V=2xx4/3piR^3=8/3piR^3`
Let the radius of the large drop formed be R'.
The volume of the large drop is also V.
`therefore 4/3piR'^3=8/3piR^3 rArr R'^3=2R^3 rArr R' = 2^(1//3)R.`
Now the surface area of the two drops is
`S_1= 2xx4piR^2=8piR^2`
and the surface area of the resultant drop is
`S_2=4piR'^2=4pi2^(2//3)R^2`
Let T be the surface tension of mercury. therefore the surface energy of the two drops before coalescing is
`U_1 = S_1 T= 8 pi R^2T`
and the surface energy after coalescing,
`U_2=S_2T=2^(2//3) xx 4piR^2T`
`therefore U_1/U_2=(8piR^2T)/(2^(2//3)xx4piR^2T)=(2)/(2^(2//3)) = 2^(1//3)`.