A water drop of radius 1 mm is broken into `10^(6)` identical drops, surface tension of water is 72 dynes/cm find the energy spent in this process.

Explanation

As volume of water remains constant, so
`(4)/(3)piR^(3)=n(4)/(3)pir^(3)" "implies" "r=(R)/(n^(1//3))`
Increase in surface area
`DeltaA=n(4pir^(2))-4piR^(2)=4pi(n^(1//3)-1)R^(2))`
`=4pi(100-1)10^(-6)`
`:.` Energy spent =
`TDeltaA=4pixx99xx10^(-6)xx72xx10^(-3)`
`=89.5xx10^(-6)J`