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^(-3)kg//m^(3)`, 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 Stokes' 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` the coefficient of viscosity of air. Here, `sigma` is negligible and `r=0.0015mm =1.5xx10^(-3)mm=1.5xx10^(-6)m`. Substituting the values `:` `v=(2)/(9)xx((1.5xx10^(-6))^(2)xx(1.0xx10^(3))xx9.8)/(1.8xx10^(-5))=2.72xx10^(-4)m//s`
A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is 1.8xx10^(-8)kgm^(-1)s^(-1) what will be the terminal velocity of the drop. Density of air can be neglected.
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