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

Answer

C

Explanation

Let radius of one drop of mercury be r since drop is spherical in shape, its volume is `(4)/(3)pir^3`
Total volume of two drops`=2xx(4)/(3)pir^3=(8)/(3)pir^3`
Radius of large drops formed by `R`. Volume of 2 drops`=`volume of a large drop
`(8)/(3)pir^3=(4)/(3)piR^3`
`impliesR^3=2r^3=2^((1)/(3))r`
The surface area of the two drops is
`S_1=2xx4pir^2=8pir^2`
Surface area of resultant drop is
`S_2=4pir^2=4pi2((2)/(3))R^2`
Surface energy of two drops before coalescing is
`W_1=T`,`S_1=8pir^2T`
and surface energy after coalescing is
`W_2=S_2T=2^((2)/(3))xx4piR^2T`
`implies(W_1)/(W_2)=(2)/(2^((2)/(3)))=2^((1)/(3))`