Two drops of equal radius coalesce to form a bigger drop. What is ratio of surface energy of bigger drop to smaller one?

Options

A. `2^(1//2) : 1`
B. `1:1`
C. `2^(2//3) :1`
D. None of these

Answer

C

Explanation

Volume remains constant after coalescing.
Thus, `(4)/(3)pi R^(3) = 2 xx (4)/(3)pi r^(3)`
where, R is radius of bigger drop and r is radius of each smaller drop.
`:. R = 2^(1//3) r`
Surface energy per unit surface area is the surface tension. So, surface energy `E = Delta DA = 4pi R^(2)S`
for bigger drop, `E_(1) = 4pi (2^(1//3)r)^(2) S = (2^(1//3)) 4pi r^(2)S`
For smaller drop, `E_(2) = 4pi r^(2)S`
Hence, required ratio, `(S_(1))/(S_(2)) = 2^(2//3) : 1`