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

Answer

A

Explanation

`W=T(A_2-A_1)`
`A_2 =4piR^2=3((4pi)/3)R^3/R=(3V)/V`
`A_1=n xx4pir^2 =V/(4/3pir^3)=(3V)/r [ as V= n(4)/3pir^3]`
`W=T[(3V)/R-(3V)/r]`
`=3TV [1/R-1/r] ` or `-3TV [1/R-1/r]`
Here (-ve) shows that energy releases.