If a ball of steel (density rho = 7.8 ; g cdo | vedclass

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(B) The terminal velocity $v$ of a sphere falling through a viscous fluid is given by the formula:$v = \frac{2r^2(\rho - \rho_0)g}{9\eta}$where $\rho$

Vedclass · Accepted Answer

$6.25 	imes 10^{-4} \; cm \cdot s^{-1}$

Vedclass · Answer

(B) The terminal velocity $v$ of a sphere falling through a viscous fluid is given by the formula:$v = \frac{2r^2(ho - ho_0)g}{9\eta}$where $ho$ is the density of the sphere,$ho_0$ is the density of the fluid,and $\eta$ is the coefficient of viscosity.From this,we see that $v \propto \frac{(ho - ho_0)}{\eta}$.Let $v_1$ and $\eta_w$ be the terminal velocity and viscosity in water,and $v_2$ and $\eta_g$ be the terminal velocity and viscosity in glycerine.$\frac{v_2}{v_1} = \frac{(ho - ho_g)}{\eta_g} 	imes \frac{\eta_w}{(ho - ho_w)}$Given: $ho = 7.8 \; g \cdot cm^{-3}$,$ho_w = 1.0 \; g \cdot cm^{-3}$,$ho_g = 1.2 \; g \cdot cm^{-3}$,$v_1 = 10 \; cm \cdot s^{-1}$,$\eta_w = 8.5 	imes 10^{-4} \; Pa \cdot s$,$\eta_g = 13.2 \; Pa \cdot s$.$\frac{v_2}{10} = \frac{(7.8 - 1.2)}{13.2} 	imes \frac{8.5 	imes 10^{-4}}{(7.8 - 1.0)}$$\frac{v_2}{10} = \frac{6.6}{13.2} 	imes \frac{8.5 	imes 10^{-4}}{6.8} = 0.5 	imes 1.25 	imes 10^{-4} = 0.625 	imes 10^{-4}$$v_2 = 10 	imes 0.625 	imes 10^{-4} = 6.25 	imes 10^{-4} \; cm \cdot s^{-1}$.

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