A solid steel ball of diameter 3.6 mm acquir | vedclass

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(D) The formula for terminal velocity $v_T$ is given by Stokes' Law: $v_T = \frac{2}{9} \frac{(\rho_s - \rho_o) r^2 g}{\eta}$,where $\rho_s$ is the de

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$1.99$

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(D) The formula for terminal velocity $v_T$ is given by Stokes' Law: $v_T = \frac{2}{9} \frac{(ho_s - ho_o) r^2 g}{\eta}$,where $ho_s$ is the density of steel,$ho_o$ is the density of oil,$r$ is the radius of the ball,$g$ is the acceleration due to gravity,and $\eta$ is the coefficient of viscosity.Given:Diameter $d = 3.6 \ mm = 3.6 	imes 10^{-3} \ m$,so radius $r = 1.8 	imes 10^{-3} \ m$.$v_T = 2.45 	imes 10^{-2} \ m/s$.$ho_s = 7825 \ kg/m^3$.$ho_o = 925 \ kg/m^3$.$g = 9.8 \ m/s^2$.Rearranging for $\eta$: $\eta = \frac{2}{9} \frac{(ho_s - ho_o) r^2 g}{v_T}$.Substituting the values:$\eta = \frac{2}{9} 	imes \frac{(7825 - 925) 	imes (1.8 	imes 10^{-3})^2 	imes 9.8}{2.45 	imes 10^{-2}}$.$\eta = \frac{2}{9} 	imes \frac{6900 	imes 3.24 	imes 10^{-6} 	imes 9.8}{2.45 	imes 10^{-2}}$.$\eta = \frac{2}{9} 	imes \frac{21902.4 	imes 10^{-6}}{2.45 	imes 10^{-2}} \approx 1.99 \ Pa \cdot s$.

$A$ solid steel ball of diameter $3.6 \ mm$ acquires a terminal velocity of $2.45 \times 10^{-2} \ m/s$ while falling under gravity through an oil of density $925 \ kg/m^3$. Take the density of steel as $7825 \ kg/m^3$ and $g$ as $9.8 \ m/s^2$. The viscosity of the oil in $SI$ units is:

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