A certain number of spherical drops of a liquid of radius `r` coalesce to form a single drop of radius `R` and volume `V`. If `T` is the surface tension of the liquid, then

Options

A. Energy `=4VT(1/r-1/R)` is released
B. Energy `=3VT(1/r+1/R)` is absorbed
C. Energy `=3VT(1/r-1/R)` is released
D. Energy is neither released nor absorbed.

Answer

C

Explanation

Let n drops having radius r be combined to form one bigger drop having radius R.
Volume would be same, `n 4/3 pi r^(3)=4/3 pi R^(3)=V `
`n=R^(3)/r^(3)`
Surface energy `E=T triangle A=T[4piR^(2)-pi4piR^(2)]`
`=4piT [R^(2)-nr^(2)]=4piT [R^(2)-R^(3)/r^(3) r^(2)]`
`=4piR^(3)T [1/R-1/r]`
`R gt r therefore E=3 (4/3 piR^(3)) T(1/R-1/r)`
`E=3VT (1/R-1/r)`= negative
`E=3VT (1/R-1/r) ="negative"`
`E=3VT (1/r-1/R)` is released