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 = 3 VT `((1)/(r )+(1)/( R))` is absorbed
C. energy `=3V T ((1)/(R )-(1)/(r ))` is released
D. energy is neither released nor absorbed

Answer

C

Explanation

Here, if the surface area changes, it will change the surface energy as well. If the surface area decreases, it means that energyu is released and vice versa.
Change in surface energy `Deltaa xx T…..(i)`
Let we have n number of drops initially
`So, DeltaA=4piR^(2)-n(4pir^(2))...(ii)`
Volume is constant
`So, n(4)/(3)pir^(3)=(4)/(3)piR^(3)=V...(iii)`
From (ii) and (iii) we get
`DeltaA=(3)/(R)(4pi)/(3)xxR^(3)-(3)/(r)(n(4pi)/(3r^(3)))=(3)/(R)xxV-(3)/(r)V`
`Rightarrow DeltaA=3V ((1)/(R)-(1)/(r))="negative value"`
As, `R gt r, "so" DeltaA` is negative. It means that surface areais decreased, so energy must be released.
`"Energy released"=DeltaAxxT=-3VT((1)/(r)-(1)/R)`