InterviewSolution
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InterviewSolution
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A large number of droplets, each of radius a, coalesce to form a bigger drop of radius `b`. Assume that the energy released in the process is converted into the kinetic energy of the drop. The velocity of the drop is `sigma =` surface tension, `rho =` density)A. `[(sigma)/(rho)(1/a-1/b)]^(1/2)`B. `[(2sigma)/(rho)(1/a-1/b)]^(1/2)`C. `[(3sigma)/(rho)(1/a-1/b)]^(1/2)`D. `[(6sigma)/(rho)(1/a-1/b)]^(1/2)` |
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Answer» Correct Answer - D Energy realeased `=[nxx4pia^(2)=-4pib^(2)]sigma` Now `nxx4/3pir^(3)=4/3pib^(3)` or `n=(b^(3))/(a^(3))` Therefore energy released is `=[(b^(3))/(a^(3))=4pia^(2)-4pib^(2)]sigma=4pib^(2)[b/a-1]sigma` Now `1/2(4/3pib^(3))rhov^(2)=4pib^(2)[b/a-1]sigma` or `v=[(6sigma)/(rho)(1/a-1/b)]^(1/2)` |
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