A spherical liquid drop of radius `R` is divided into eight equal droplets. If the surface tension is `T`, then the work done in this process will be

Options

A. `2piR^2T`
B. `3piR^2T`
C. `4piR^2T`
D. `2piRT^2`

Answer

C

Explanation

Radius of the larger drop `=R`
Suppose radius of the droplets `=r`
Since, volume will be remain constant,
`(4)/(3)piR^3=8xx(4)/(3)pir^2`
[`because` No. of droplets`=8pixx(0.01)xx75`
work done`=`(increase in surface area)`xx`Surface tension
`=[84pi((R )/(2))^2-4piR^2]xxT`
`=(8piR^2-4piR^2)xxT=4piR^2T`