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 = 3VT (1/r+1/R) is absorbed
(c) Energy = 3VT (1/r-1/R) is released
(d) Energy is neither released nor absorbed
Here, if the surface are a changes, it will change the surface energy as well. If the surface area decreases, it means that energy is released and vfce-versa.
Change in surface energy ΔAxT ...(i)
Let we have 'n' number of drops initially.
So, ΔA=4πR2-n(4πr2) ...(ii)
Volume is constant
So, n4/3πR3-n(4πr2) ...(iii)
From Eqs. (ii) and (iii)
ΔA=3/R x 4π/3 x R3-3/r(n4π/r3)
=3/R x V-3/r x V
ΔA=3V(1/R-1/r)= -ve value
As K > r, so ΔA is negative It means surface area is decreased, so energy must be released.
Energy released=ΔA x T=-3VT(1/r-1/R)
Above expression shows the magnitude of energy released.