Two small drops of mercury each of radius r form a single large drop. The ratio of surface energy before and after this change is

Options

A. `2 : 2^(2//3)`
B. `2^(2//3) : 1`
C. `2 : 1`
D. `1 :2`

Answer

A

Explanation

Suppose R be the radius of bigger drop, Then by equating volumes, we have
`2(4//3pir^(3))=(4//3)piR^(3)" or "R=(2)^(1//3)r`
Now, surface energy `prop` surface area
`therefore" "U_(1)/U_(2)=A_(1)/A_(2)=(2r^(2))/R^(2)=2/2^(2//3)`