1.

Two mercury drops each of radius r merge to form a bigger drop. Calculate the surface energy released.A. `8pir^2sigma-4xx2^((2)/(3))pir^2sigma`B. zeroC. negativeD. `4pir^2sigma-8xx2^((1)/(3))r^2sigma`

Answer» Correct Answer - A
`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`


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