A big water drop is formed by the combination of 'n' small water drops of equal radii. The ratio of the surface energy of 'n' drops to the surface energy of big drop is

Options

A. `n^(2) : 1`
B. `n : 1`
C. `sqrt(n) : 1`
D. `""^(3)sqrt(n) : 1`

Answer

D

Explanation

Volume of big drop = n (Volume of small drop)
`rArr 4/3 piR^(3) = n. (4)/(3) pir^(3)`
Where, R = radius of big water drop.
and r = radius of small water drop.
`rArr R^(3) = nr^(3)`
`rarr R = n^(1//3) .r"….."(i)`
Surface energy of n drops,
`E_(2) = n xx 4pir^(2) xx T`
Surface energy of big drop,
`E_(1) = 4piR^(2)T`
`:. (E_(2))/(E_(1)) = (nr^(2))/(R^(2)) = (nr^(2))/((n^(1//3).r)^(2))`
`= (nr^(2))/(n^(2//3).r^(2)) = n^(1//3)` [ `:'` from Eq. `(i)`]
`E_(2) : E_(1) = .^(3)sqrt(n) : 1`