Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury `T = 435. 5xx10^(-3) Nm^(-1)`

Options

A. `3.22 xx 10^(-6) J`
B. `4 xx 10^(-4) J`
C. `9 xx 10^(-3) J`
D. `6 xx 10^(-4) J`

Answer

A

Explanation

When two or more droplets collapse to form a bigger drop, then its surface area decreases and energy is released equal to S. `Delta A`.
Given, radii of mercury droplets,
`r_(1) = 0.1 cm = 1 xx 10^(-3) m`
and `r_(2) = 0.2 cm xx 2 xx 10^(-3) m`
Surface tension (S) `= 435.5 xx 10^(-3) Nm^(-1)`
Let the radius of the big drop formed by collapsing be R.
`:.` Volume of big drop = Volume of small droplets
`(4)/(3)pi R^(3) = (4)/(3)pi r_(1)^(3) + (4)/(3) pi r_(2)^(3)`
`rArr R^(3) = r_(1)^(3) + r_(2)^(3) = (0.1)^(3) + (0.2)^(3)`
`= 0.001 + 0.008 = 0.009`
`rArr R = 0.21 cm = 2.1 xx 10^(-3) m`
`:.` Change in surface area,
`Delta A = 4pi R^(2) - (4pi r_(1)^(2) + 4pi r_(2)^(2))`
`= 4pi [R^92) - (r_(1)^(2) + r_(2)^(2))]`
`:.` Energy released `= S.Delta A`
`= -S xx 4pi [R^(2) - (r_(1)^(2) + r_(2)^(2))]`
`= -435.5 xx 10^(-3) xx 4 xx 3.14 [(2.1 xx 10^(-3))^(2) - (1 xx 10^(-6) + 4 xx 10^(-6))]`
`= 435.5 xx 4 xx 3.14(4.41 - 5) xx 10^(-6) xx 10^(-3) = +3.22 xx 10^(-6) J`
Therefore, `3.22 xx 10^(-6) J` energy will be released.