A spherical water drop of radius R is split into 8 equal droplets. If T is the surface tension of water, what is the work done in the process?

Explanation

For a water drop, W = `4piR^2 xx T`
`4/3 piR^3 = 4/3 pir^3 xx 8`
`therefore` r = r/2`
change in area = `4pir^2 xx 8 - 4piR^2`
dA = `4pi (8r^2 - R^2) = 4piR^2`
`therefore` change in energy = T dA = `4piR^2T`