InterviewSolution
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InterviewSolution
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Two equal drops of water falling through air with a steady velocity `5 cm//s`. If the drops combire to from a single drop, what will be new terminal velocity? |
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Answer» Here, `v=5 cms^(-1)` .Let r be the radius of each drop. Terminal velocity of each drop is `v=(2pi r^(2)(rho-sigma)g)/(9 eta)=5` ……(i) Let R be the radius of by drop formed when two small drop coalese together. Then `(4)/(3) piR^(3) =2xx(4)/(3) pi r^(3)` or `R=2^(1//3) r` Terminal velocity of big drop is `v=(2R^(2)(rho- sigma)g)/(9 eta)` .....(ii) Dividing (ii) by (i), we get `(V)/(5) =(R^(2))/(r^(2))` or `V=5xx(R^(2))/(r^(2))` `V=5xx((2^(1//3)r)^(2))/(r^(2))=5xx2^(2//3)=5xx1.5874` `=7.937 cms^(-1)` |
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