If `n` drops of a liquid, form a single drop, then

Options

A. some energy will be released in the process
B. some energy will be released in the process
C. the energy released or absorbed will be `E(n-n^((2)/(3)))`
D. the enrgy released or absorbed will be `nE(2^((2)/(3))-1)`

Answer

A, C

Explanation

Let S `=` surface energy per unit drop
`r=` radius of each small drop
`R` radius of a single drop
`nxx(4)/(3)pir^(3)=(4)/(3)piR^(3)` or `R=rn^((1)/(3))`
Initial surface energy `E_(i)=nxx4pir^(2)xxS=nE`
Final surface energy
`E_(f)=4piR^(2)S=4piR^(2)n^((2)/(3))S=n^((2)/(3))E`
Therefore, energy released
`E_(i)-E_(f)=(n-n^((2)/(3)))`