Two mercury droplets of radii 0.2 cm and 0.4 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury `440 xx 10^(-3) Nm^(-1)` and `g=10ms^(-2)`.

Explanation

Here, `r_(1) = 0.2 c, = 0.2 xx 10^(-2)`
` = 2 xx 10^(-3) m`,
`r_(2) = 0.4 cm = 0.4 xx 10^(-2) m = 4 xx 10^(-3) m`
`S= 440 xx 10^(-3)Nm^(-1)`
Let R be the radius of big drop formed.
Now volume of big drop = volume of two small drops
`:. 4/3 pi R^(3) = 4/3 pi r_(1)^(3) + 4/3 pi r_(2)^(3)`
or, `R^(3) = r_(1)^(3) + r_(2)^(3) or R=[r_(1)^(3)+r_(2)^(3)]^(1//3)`
or `R=[(2xx10^(-3))^(3) + (4 xx 10^(-3))^(3)]^(1//3)`
`=(72 xx10^(-9))^(1//3) = 4.16 xx 10^(-3)m`
Decrease in surface area
`Delta A = 4 pi (r_(1)^(2)+r_(2)^(2))-4 pi R^(2)`
`= 4 pi [(2 xx 10^(-3))^(2)+(4 xx 10^(-3))^(2) - (4.16 xx 10^(-3))^(2)]`
`= 4 pi [ 20 xx 10^(-6)-17.30 xx 10^(-6)]`
`= 4 xx 22/7 xx 2.70 xx 10^(-6) = 33.94 xx 10^(-6)^(2)`
Energy released = `S xx Delta A`
`=(440 xx 10^(-3)) xx 933.94 xx 10^(-6)`
`=14.935 xx 10^(-6)J`.