If T is the surface tension of a liquid, the energy needed to break a liquid drop of radius R into 64 drops is

Options

A. `6pi R^(2)S`
B. `4pi R^(2)S`
C. `12 pi R^(2)S`
D. `8 pi R^(2)S`

Answer

C

Explanation

Let r be the radius of each small drop. Then Volume of 64 small drops= Volume of big drop
or `64 xx (4)/(3) pi r^(3) = (4)/(3) pi R^(3) or r = (R )/(4)`…(i)
Surface area of big drop `= 4pi R^(2)`
Surface area of 64 small drops `= 64 xx 4pi r^(2)`
`:.` Increase in surface area `=64 xx 4pi r^(2) - 4pi R^(2) = 4pi [64r^(2) - R^(2)]`
`=4pi [4R^(2) - R^(2)] = 12pi R^(2)` (Using (i))
Energy needed = surface tension `xx` increase in surface area `=S xx 12pi R^(2) = 12pi R^(2)S`