Energy needed in breaking a drop of radius R into n drops of radius r, is where T is surface tension and P is atmospheric pressure.

Options

A. `(4pir^(2)n - 4piR^(2))T`
B. `((4pi)/(3) n r^(2) - (4)/(3)piR^(2))T`
C. `(4piR^(2) - 4pir^(2))n T `
D. `(4piR^(2) - n4pir^(2))P`

Answer

A

Explanation

`(4//3) pi R^(2) = (4//3) pi r^(3) xx n, r = (R//n)^(1//3)`
Energy needed ` = T (4pi r^(2)n-piR^(2))`