A liquid drop of diameter D breaks up into 27 tiny drops. Find the resulting change in energy. Take surface tension of the liquid as `sigma`

Explanation

Radius of larger drop = `(D)/(2)`
Radius of each small drop= r
`27xx(4)/(3)pir^(3)=(4)/(3)pi((D)/(2))^(2) implies r=(D)/(6)`
Initial surface area of large drop `4pi((D)/(2))^(2)=piD^(2)`
Final surface area of 27 small drop
`=27xx4pir^(2)=27xx4pi(D^(2))/(36)=3piD^(2)`
`:.` Change in energy = Increase in area `xx rho`
`=2piD^(2) rho`