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

`r_(1)=0.1cm=10^(-3)m`
`r_(2)=0.2cm=2xx10^(-3)m`
Surface tension `S=435.5xx10^(-3)N//m`
Suppose radius of larger drop is R.
Volume of large drop =Volume of two small drops
`therefore(4)/(3)piR^(3)=(4)/(3)pir_(1)^(3)+(4)/(3)pir_(2)^(3)`
`thereforeR^(3)=r_(1)^(3)+r_(2)^(3)`
`R^(3)=(0.1)^(3)+(0.2)^(3)`
`=(0.001)+(0.008)`
`=0.009`
`thereforeR=0.21cm=2.1xx10^(-3)cm`
Change in area,
`DeltaA=` (surface area of large drop)-(surface area of two small drops)
`DeltaA=4piR^(2)-(4pir_(1)^(2)+4pir_(2)^(2))`
`=4pi[R^(2)=(r_(1)^(2)+r_(2)^(2))]`
`therefore` The energy increased `DeltaU=W=SDeltaA`
`thereforeDeltaU=S(4pi)[R^(2)-(r_(1)^(2)-r_(2)^(2))]`
`DeltaU=435.5xx4xx3.14[(2.1xx10^(-32))-(1xx10^(-6)+4xx10^(-6))]`
`=435.5xx4xx3.14(4.41-5)xx10^(-6)xx10^(-3)`
`=-32.23xx10^(-7)`
Negative sign indicates energy is absorbed Hence , energy absorbed is `3.22xx10^(-6)J`. Hence same amount of energy increases.