InterviewSolution
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InterviewSolution
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A drop of water of radius 0.0015 mm is falling in air .If the cofficient of viscosity of air is `2.0 xx 10^(-5) kg m^(-1)s^(-1)` ,the terminal velocity of the drop will be (The density of water = `10^(3) kg m^(-3)` and g = `10 m s^(-2)` )A. `1.0 xx 10^(-4) m s^(-1)`B. `2.0 xx 10^(-4) m s^(-1)`C. `2.5 xx 10^(-4) m s^(-1)`D. `5.0 xx 10^(-4) m s^(-1)` |
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Answer» Correct Answer - C Here, r=0.0015 mm `=0.0015xx10^(-3) m` `eta=2.0xx10^(-5) kg m^(-1) s^(-1)` `rho=1.0xx10^(3) kg m^(-3)` `g=10 m s^(-2)` Neglecting the density of air, the terminal velocity of the water drop is `v_(T)=(2)/(9)(r^(2)rhog)/(eta)=(2xx(0.0015xx10^(-3))^(2)xx1.0xx10^(3)xx10)/(9xx2.0xx10^(-5))` `=2.5xx10^(-4) m s^(-1)` |
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