What will be the approximate terminal velocit | vedclass

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(B) The formula for terminal velocity $v$ is given by $v = \frac{2}{9} r^{2} \frac{(\rho - \sigma)}{\eta} g$. Since we neglect the buoyancy of air, we

Vedclass · Accepted Answer

$98$

Vedclass · Answer

(B) The formula for terminal velocity $v$ is given by $v = \frac{2}{9} r^{2} \frac{(ho - \sigma)}{\eta} g$. Since we neglect the buoyancy of air, we take $\sigma \approx 0$. The formula simplifies to $v = \frac{2}{9} \frac{ho}{\eta} r^{2} g$. Given: diameter $d = 1.8 	imes 10^{-3} \ m$, so radius $r = 0.9 	imes 10^{-3} \ m$. Density $ho = 10^{3} \ kg \ m^{-3}$, viscosity $\eta = 1.8 	imes 10^{-5} \ N \ s \ m^{-2}$, and $g = 9.8 \ m \ s^{-2}$. Substituting the values: $v = \frac{2}{9} 	imes \frac{10^{3}}{1.8 	imes 10^{-5}} 	imes (0.9 	imes 10^{-3})^{2} 	imes 9.8$. $v = \frac{2}{9} 	imes \frac{10^{3}}{1.8 	imes 10^{-5}} 	imes (0.81 	imes 10^{-6}) 	imes 9.8$. $v = \frac{2}{9} 	imes \frac{10^{3} 	imes 0.81 	imes 10^{-6} 	imes 9.8}{1.8 	imes 10^{-5}} = \frac{2}{9} 	imes \frac{0.81 	imes 10^{-3} 	imes 9.8}{1.8 	imes 10^{-5}} = \frac{2}{9} 	imes 0.45 	imes 10^{2} 	imes 9.8 = 0.1 	imes 100 	imes 9.8 = 98 \ m \ s^{-1}$.

What will be the approximate terminal velocity of a rain drop of diameter $1.8 \times 10^{-3} \ m$, when density of rain water $\approx 10^{3} \ kg \ m^{-3}$ and the coefficient of viscosity of air $\approx 1.8 \times 10^{-5} \ N \ s \ m^{-2}$ (in $m \ s^{-1}$)? (Neglect buoyancy of air)

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