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 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

Let `S=` surface tension `=` surface energy per unit area.
`r=` radius of each small drop
`R=` radius of single drop
`nxx4/3pir^(3)=4/3piR^(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)-Ef=E(n-n^(2//3))`