A spherical ball of density rho and radius 0. | vedclass

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(D) The formula for terminal velocity $v_T$ is given by Stokes' Law: $v_T = \frac{2}{9} \frac{r^2(\rho - \rho_L)g}{\eta}$.Given: $r = 0.003 \ m$,$\rho

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$5 \ s$

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(D) The formula for terminal velocity $v_T$ is given by Stokes' Law: $v_T = \frac{2}{9} \frac{r^2(ho - ho_L)g}{\eta}$.Given: $r = 0.003 \ m$,$ho_L = 1260 \ kg \cdot m^{-3}$,$ho = 2 	imes ho_L = 2520 \ kg \cdot m^{-3}$,$\eta = 1.260 \ N \cdot s \cdot m^{-2}$,and $g = 10 \ m \cdot s^{-2}$.Substituting the values:$v_T = \frac{2}{9} 	imes \frac{(0.003)^2 	imes (2520 - 1260) 	imes 10}{1.260}$$v_T = \frac{2}{9} 	imes \frac{9 	imes 10^{-6} 	imes 1260 	imes 10}{1.260}$$v_T = 2 	imes 10^{-6} 	imes 1000 	imes 10 = 0.02 \ m \cdot s^{-1}$.The distance $d$ between the $10 \ cm$ and $20 \ cm$ marks is $10 \ cm = 0.1 \ m$.Since the ball has reached terminal velocity,the time $t$ taken is:$t = \frac{d}{v_T} = \frac{0.1 \ m}{0.02 \ m \cdot s^{-1}} = 5 \ s$.

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