Two small drop of mencury , each of radius R coulesece 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

Volume of dropes after cosdesescing remain same let radius of one drop of noymaly by a zero drop is spherical is slope its volume is `(4)/(3) pi r^(2)`
Total volume of two drop ` = 2 xx (4)/(3) pi v^(2) = (8)/(3) pi v^(2)`
Radius of lerge drop formed be R
Volume of `2` drops = Volume of lerge drop
`(8)/(3) pi v^(2) = (4)/(3) pi R^(2)`
`rArr R^(2) = 2v^(2) = 2^(1/3)r`
The surface area of the two drop is
`S_(1) = 2 xx 4 pi r^(2) = 8 pi r^(2)`
Surface area of resultant drop is
`S_(2) = 4 pi r^(2) = 4 pi 2^(3)R^(2)`
surface energy of two drop before codesescing is
`W_(1) = T, S_(1) = 8 pi r^(2) T`
end surface energy after contescing is
`W_(1) = S_(2) T = 3^(3) xx 4 pi R^(2) T`
`rArr (W_(1))/(W_(2)) = (2)/(2^(2/3)) = 2^(1)/(3)`