Question

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)`)
A. `3.98 xx 10^(-4) J`
B. `8.46 xx 10^(-4) J`
C. `3.98 xx 10^(-2) J`
D. `8.46 xx 10^(-2) J`

Answer 1

Correct Answer - C
Let r be the radius of one droplet.
Now, `(4)/(3)pi R^(3) = 10^(6) xx (4)/(3)pi r^(3)`
`rArr r = (R)/(100) = (1)/(100) cm = 10^(-4) m`
`A_(i) = 4pi R^(2)` and `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)`
Increases in surface energy
`= S Delta A = 32 xx 10^(-2) xx 4pi xx 99 xx 10^(-4) J = 3.98 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, so expended `= 3.98 xx 10^(-2) J`