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A drop of water of radius 0.0015 mm is f...

A drop of water of radius `0.0015 mm` is falling in air. If the coefficient of viscosity of air is `1.8 xx 10^(-5)kg//ms`, what will be the terminal velocity of the drop? Density of water `= 1.0 xx 10^(3) kg//m^(3)` and `g = 9.8 N//kg`. Density of air can be neglected.

Text Solution

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By stoke's law, the terminal velocity of a water drop of radius `r` is given by
`v = (2)/(9) (r^(2)(rho - sigma)g)/(eta)`
where `rho` is the density of water, `sigma` is the density of air and `eta` the coefficient of viscity of air. Here `sigma` is negligible and `r = 0.0015 mm = 1.5 xx 10^(-3) mm = 1.5 xx 10^(-6) m`. Substituting the values :
`v = (2)/(9) xx ((1.5 xx 10^(-6))^(2) xx (1.0 xx 10^(3)) xx 9.8)/(1.8 xx 10^(-5))`
`= 2.72 xx 10^(-4) m//s`
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