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

Explanation

Let the radius of the big drop be R and that of small drop be `r = d//2, R = D//2`
Volume of one big drop = Volume of two drops
`( 4)/(3) pi R^(3) = 2 xx ( 4)/( 3)pi r^(3) , (D^(3))/( 8) = 2 xx ( d^(3))/( 8) ` or ` D =2 ^(1//3) d `
Change in surface area `=2 xx 4pi "" ( d^(2))/( 4) - 4pi xx ( 2^(2//3) d^(2))/( 4)`
Energy evolved `=2 pi d^(2) [1- 2^(-1//3)] T` Joule