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)`

Explanation

Let `V_1 and V_2` be the volume of the two droplets and V be that of the resulting drop
So, `V_(1)+V_(2)=V`
`or 4/3 pi r_(1)^(3)+4/3 pi r_(2)^(3)=4/3 pi r^(3)`
`or r^(3)=r_(1)^(3)+r_(2)^(3)`
=0.001 +0.008
`r^(3)=0.009 cm^(3)`
or `r=0.21 cm =2.1 xx 10^(-3)m`
Amount of energy released =change in surface energy due to change in surface area.
`or triangleU=4pi [r^(2)-(r_(1)^(2)+r^(2)]T`
`=4 xx (3.14) (2.1 xx 10^(-3))^(2)-(1 xx 10^(-3)+2 xx 10^(-3))^(2)] 435.5 xx 10^(-3)`
`=-32 xx 10^(-7)J`