Two mercury drops (each of radius 'r') merge to form bigger drop. The surface energy of the bigger drop, if T is the surface tension is :

Options

A. `4pir^(2)T`
B. `2pir^(2)T`
C. `2^(8//3)pir^(2)T`
D. `2^(5//3)pir^(2)T`

Answer

D

Explanation

`2. 4/3pir^(3)=4/3piR^(3)`
`R=2^(1//3)r`
`S.E.=T4piR^(2)`
`=T4pi2^(2//3)r^(2)=T2^(5//3)pir^(2)`
Correct choice : (d).