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

Here, radius of the drop, `R=1 cm = 1 xx 10^(-2)m`
Let r be the radius of each droplet. Then Volume of `10^(6)` droplets= Volume of the drop
or `10^(6) xx (4)/(3) pi r^(3) =(4)/(3) pi R^(3) or r = (10^(-2))/(10^(2)) = 10^(-4)m`
Increase in surface area `= 10^(6) xx 4pi r^(2) - 4pi R^(2`
`=10^(6) xx 4 xx pi xx 10^(-8) - 4pi xx 10^(-4)`
`= 4 xx 9.9 xx pi xx 10^(3) m^(2)`
`:.` Work done = surface tension `xx` increase in surface area `=35 xx 10^(-2) xx 4 xx 9.9 xx pi xx 10^(-3) = 4.35 xx 10^(-2)J`