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

Answer

C

Explanation

Let there be n small drop each of radius r, which when coalesce from a big drop of radius R. Then
`V=nxx(4)/(3)pi r^(3) =(4)/(3)pi R^(3)` or `n=(R^(3))/(r^(3))`
Energy released = surface tension x decrease in surface area
`=T[4pi r^(2)xxn-4pi R^(2)]`
`=T[4pi r^(2)xx(R^(3))/(r^(3)) - (4pi R^(3))/(R)]`
`=3T[(4)/(3)pi R^(3)xx(1)/(r) - (4)/(3)pi R^(3)xx(1)/(R)]`
`=3T[Vxx(1)/(r)-Vxx(1)/(R)] =3VT[(1)/(r) - (1)/(R)]`