The energy of the free surface of a liquid drop is `5pi` times the surface tesion of the liqid. Find the diameter of the drop in C.G.S system .

Explanation

Given that `E = 5 piT`
Surface energy, `E = T xx dA "……." (i)`
`dA = 4 pir^(2)`
(where, r is the radius of the liquid drop)
Substituting in equation (i), we get
`E = T xx 4pir^(2)`
`5 pi T = T xx 4pir^(2) ( :' E = 5 piT)`
`:. r^(2) = 5/4`
`:. r = (sqrt(5))/(2)`
Diameter `d = 2r = 2 xx (sqrt(5))/(2)`
`:. d = sqrt(5) = 2.23 cm`