Energy needed in breaking a drop of radius `R` into `n` drops of radii `r` is given by

Options

A. `4piT(pir^2-R^2)`
B. `(4)/(3)pi(r^3n-R^2)`
C. `4piT(R^2-nr^2)`
D. `4piT(nr^2+R^2)`

Answer

A

Explanation

Energy needed `=` Increment in surface energy
`=("surface energy of" `n` small drops)`-`("surface energy of one big drop")`
`=n4pir^2T-4piR^2T=4piT(nr^2-R^2)`