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) = 2 xx (4)/(3) piR^(3)`
where R is radius of bigger drop and r is radius of each smaller drop
`therefore R = 2^(1//3)r`
Now, surface energy per unit surface area is the surface tension. SO,
Surface energy, `W = TtriangleA`
or `W = 4piR^(2)T`
Therefore, surface energy of bigger drop
`W_(1) = 4pi(2^(t//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`