What is the terminal velocity of a rain drop | vedclass

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(A) Terminal velocity $(v_t)$ is given by the formula derived from Stokes' Law: $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$.Given: Radius $r = 0.02 \ m

Vedclass · Accepted Answer

$4.9$

Vedclass · Answer

(A) Terminal velocity $(v_t)$ is given by the formula derived from Stokes' Law: $v_t = \frac{2r^2(ho - \sigma)g}{9\eta}$.Given: Radius $r = 0.02 \ mm = 2 	imes 10^{-5} \ m$,viscosity $\eta = 1.8 	imes 10^{-5} \ N \ s \ m^{-2}$,density of water $ho = 1000 \ kg \ m^{-3}$,density of air $\sigma \approx 0$,and $g = 10 \ m \ s^{-2}$.Substituting the values: $v_t = \frac{2 	imes (2 	imes 10^{-5})^2 	imes 1000 	imes 10}{9 	imes 1.8 	imes 10^{-5}}$.$v_t = \frac{2 	imes 4 	imes 10^{-10} 	imes 10^4}{16.2 	imes 10^{-5}} = \frac{8 	imes 10^{-6}}{16.2 	imes 10^{-5}} = \frac{80}{16.2} 	imes 10^{-2} \approx 4.938 	imes 10^{-2} \ m \ s^{-1}$.Converting to $cm \ s^{-1}$: $v_t \approx 4.938 	imes 10^{-2} 	imes 100 \ cm \ s^{-1} = 4.938 \ cm \ s^{-1}$.Rounding to the nearest option,we get $4.9 \ cm \ s^{-1}$.

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