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)piR^(3)=2xx(4)/(4)pir^(3)`
where, surface energy per unit surface are is the surface tension, So,
or `W=4piR^(2)T`
Therefore, surface energy of bigger drop
`W_(1)=4pi(2^(1//3)r)^(2)T`
`=(2^(2//3))4pir^(2)T`
Surface energy of smaller drop
`W_(2)=4pir^(2)T`
Hence, required ratio
`(W_(1))/(W_(2))=2^(2//3):1`