n drops of a liquid, each with surface energy E. joining to form a single drop
(a). Some energy will be released in the process
(b). Some energy will be absorbed in the process
(c). The energy released or absorbed will be `E(n-n^(2//3))`
(d). the energy released or absorbed will be `nE(2^(2//3)-1)`

Answer

A, C

Explanation

`E=T4pir^(2)implies(4)/(3)piR^(3)=nxx(4pi)/(3)r^(3)impliesn=(R^(3))/(r^(3))impliesR=n^(1//3)r`
Surface energy of big drop `E=T4piR^(2)=T4pin^(2//3)r^(2)=En^(2//3)`
Energy released `=nE-E'=nE-n^(2//3)E=E(n-n^(2//3))`