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

Options

A. `8pir^2sigma-4xx2^((2)/(3))pir^2sigma`
B. zero
C. negative
D. `4pir^2sigma-8xx2^((1)/(3))r^2sigma`

Answer

A

Explanation

`2xx(4)/(3)pir^2=(4)/(3)piR^3` or `R=2^((1)/(3))r`
Final surface energy`=4piR^2sigma=4pi2^((2)/(3))r^2sigma`
Initial surface energy`=2xx4pir^2sigma`
Energy released`=[8pir^2-4xx2^((2)/(3))pir^2]sigma`