A water drop whose radius is 0.0015 , mm is f | vedclass

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(A) According to Stokes' law,the terminal velocity $v$ of a spherical drop falling through a viscous medium,when buoyancy is neglected,is given by the

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$2.72 	imes 10^{-4} \, m/s$

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(A) According to Stokes' law,the terminal velocity $v$ of a spherical drop falling through a viscous medium,when buoyancy is neglected,is given by the formula:$v = \frac{2}{9} \frac{r^2 ho g}{\eta}$Given values:Radius $r = 0.0015 \, mm = 1.5 	imes 10^{-6} \, m$Density of water $ho = 1.0 	imes 10^3 \, kg/m^3$Acceleration due to gravity $g = 9.8 \, m/s^2$Coefficient of viscosity $\eta = 1.8 	imes 10^{-5} \, kg/(m \cdot s)$Substituting these values into the formula:$v = \frac{2}{9} 	imes \frac{(1.5 	imes 10^{-6})^2 	imes (1.0 	imes 10^3) 	imes 9.8}{1.8 	imes 10^{-5}}$$v = \frac{2}{9} 	imes \frac{2.25 	imes 10^{-12} 	imes 10^3 	imes 9.8}{1.8 	imes 10^{-5}}$$v = \frac{2}{9} 	imes \frac{2.205 	imes 10^{-8}}{1.8 	imes 10^{-5}}$$v = \frac{2}{9} 	imes 1.225 	imes 10^{-3}$$v \approx 2.72 	imes 10^{-4} \, m/s$

$A$ water drop whose radius is $0.0015 \, mm$ is falling through the air. If the coefficient of viscosity of air is $1.8 \times 10^{-5} \, kg/(m \cdot s)$,then assuming the buoyancy force is negligible,the terminal velocity of the drop will be:

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