Several spherical drops of a liquid of radius `r` coalesce to form a single drop of radius `R`. If `T` is surface tension and `V` is volume under consideration, then the release of energy is

Options

A. `3VS ((1)/(r)+(1)/(R))`
B. `3VS ((1)/(r)-(1)/(R))`
C. `VS ((1)/(r)-(1)/(R))`
D. `VS ((1)/(r^(2))+(1)/(R)^(2))`

Answer

B

Explanation

Energy released `=S xx` increase in surface area
Let n be the number of drops of radius r that coalesce to form single drop.
So, energy released `=S xx` (increase in surface area)
change in surface area `=4pi (nr^(2)-R^(2))` ...(i)
`:' n 4/3 pi r^(3)=4/3 pi R^(3) implies nr^(3) =R^(3)`
or `nr^(2)=(R^(3)/r)` ...(ii)
On putting value of Eq. (ii) in Eq. (i), we get
Change in surface area `=4pi R^(3) (1/r-1/R)`
energy released `=(4/3 pi R^(3))3S(1/r-1/R)=3VS (1/r-1/R)`