Question

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
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 1

Correct Answer - C
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)`