Two mercury drops each of radius r merge to form a bigger drop. Calculate the surface energy released.

Explanation

Surface asrea of one drop before meging `=4pir^2`
Total surface area of bogh the drops `=8pir^2`
Hence the surface enrgy before merging `=8pir^2S`.
When the drops merge the volume of the bigger drop
=2xx4/3R^3=8/3pir^3`
If the radius of this new drop is R.
`4/3pir^3=8/3pir^3`
or` R=o2^(1/3)r`
`or 4piR^2=4xx2^(2/3)xxpir^2S`
The released surface energy `=8pir^2S-4xx2^(2/3)pir^2S`
`=1.65pir^2S`