A mercury drop of radius 1 cm is broken into `10^6` droplets of equal size. The work done is `(T=35xx10^-2(N)/(m)`)

Options

A. `4.35 xx 10^(-2) J`
B. `4.35 xx 10^(-3) J`
C. `4.35 xx 10^(-6) J`
D. `4.35 xx 10^(-8) J`

Answer

A

Explanation

If r is the radius of smaller droplet and R is the radius of bigger drop, then according to question,
`(4)/(3) pi R^(3) = 10^(6) xx (4)/(3) pi r^(3) rArr r = (R)/(100) = 0.01 R`
`0.01 xx 10^(-2) m = 10^(-4) m`
`:.` Work done = Surface tension `xx` increase in area
`= 35 xx 10^(-2) xx [(10^(6) xx 4pi xx (10^(-4))^(2) - 4pi xx (10^(-3))^(2)]`
`= 4.35 xx 10^(-2) J`