A raindrop with radius R=0.2 , mm falls from | vedclass

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(D) At terminal speed,the net force on the raindrop is zero.$Mg = F_{v} = 6 \pi \eta R v$Here,$M$ is the mass of the raindrop,$\eta$ is the coefficien

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$4.94$

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(D) At terminal speed,the net force on the raindrop is zero.$Mg = F_{v} = 6 \pi \eta R v$Here,$M$ is the mass of the raindrop,$\eta$ is the coefficient of viscosity,$R$ is the radius,and $v$ is the terminal velocity.Substituting $M = ho_{w} \cdot \frac{4}{3} \pi R^3$:$ho_{w} \cdot \frac{4}{3} \pi R^3 g = 6 \pi \eta R v$Solving for $v$:$v = \frac{2 ho_{w} R^2 g}{9 \eta}$Given values: $ho_{w} = 1000 \, kg/m^3$,$R = 0.2 	imes 10^{-3} \, m$,$g = 10 \, m/s^2$,$\eta = 1.8 	imes 10^{-5} \, Ns/m^2$.$v = \frac{2 	imes 1000 	imes (0.2 	imes 10^{-3})^2 	imes 10}{9 	imes 1.8 	imes 10^{-5}}$$v = \frac{20000 	imes 0.04 	imes 10^{-6}}{16.2 	imes 10^{-5}}$$v = \frac{800 	imes 10^{-6}}{16.2 	imes 10^{-5}} = \frac{80}{16.2} \approx 4.94 \, m/s$

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