A mercury drop of radius 1.0 cm is sprayed into `10^(6)` droplets of equal sizes. The energy expended in this process is (Given, surface tension of mercury is `32 xx 10^(-2) Nm^(-1)`)

Answer

`3.98`

Explanation

Let r be the radius of one small droplet, R = radius of mercury drop
Now, ` (4)/(3) pi R ^3 = 10 ^6 xx (4)/(3) pi r ^ 3 , r = (R )/(100)= (1)/(100 ) cm = 10 ^(-4) m `
Area of mercury drop ` A _i = 4pi R^2`, Area of small droplets ` A_f = 10 ^6 xx 4pi r ^ 2 `
Change in area, ` Delta A = A_f - A_ i = 4pi xx 99 xx 10 ^(-4) m ^(2) `
Increase in surface energy ` = S Delta A = 32 xx 10 ^(-2) xx 4pi xx 99 xx 10 ^(-4) J = 3.98 xx 10 ^(-2) J `
The increase in surface energy is on the expense of internal energy expended ` = 3.98 xx 10 ^(-2) J `