Two small drops of mercury each of radius r coalesce to form a single large drop of radius R. The ratio of the total surface energies before and after the change is ……

Options

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

Answer

b

Explanation

When n small drops coalesce to form a large drop then ,
Total volume of n drops = volume of large drop. Each small drop has radius r , then
`n(4)/(3)pir^(3)=(4)/(3)piR^(3)`
`thereforeR=n^((1)/(3))r` [Here n=2]
`thereforeR=2^((1)/(3))r`
(The ratio of surface energies of small drops and large drop)
`(E_(i))/E_(f)=(2(4pir^(2)T))/(4piR^(2)T)`
`=(2r^(2))/(R^(2))=(2r^(2))/(R^((2)/(3)r^(2)))`
`=(2)/(2^((2)/(3)))=(2^((1)/(3)))/(1)`