InterviewSolution
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InterviewSolution
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A liquid drop of radius R breaks into N smaller droplets of radii r, If liquid has density `rho,` specific heat s and surface tension, T, than the drop in temeperature is givenA. `(NT)/(rhos)((1)/(R)-(1)/(r))`B. `(3NT)/(rhos)((R)/(r)-1)`C. `(3)/(4)(NT)/(rhos)((1)/(R)-(1)/(r))`D. `(3NT)/(rhos)((1)/(R)-(1)/(r))` |
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Answer» Volume will remain same, hence we can write `(4)/(3)piR^(3)=Nxx(4)/(3)pir^(3)rArrN(R^(3))/(r^(3))` Increase in surface energy due to breaking of drop inot N driplets. `" " `DeltaUtheta` `ms Deltatheta=DeltaU` `Deltatheta=(DeltaU)/(ms)=(4piT(R^(2)-Nr^(2)))/(((4)/(3)piR^(3)rho)s)` `rArrDeltatheta=(3T)/(rhos)((1)/(R)-(r^(2))/(R^(3))N)=(3T)/(rhos)((1)/(R)-(1)/r)" " ( :. N=(R^(3))/(r^(3)))` |
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