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)

Two mercury droplets of radii 0.1 cm . And 0.2 cm . Collapse into one

1.	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)
Answer» Solution : `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.