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(D) The terminal velocity $v$ of a sphere falling through a viscous liquid is given by the formula: $v = \frac{2gr^2(\rho - \sigma)}{9\eta}$,where $g$

Vedclass · Accepted Answer

$3$

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(D) The terminal velocity $v$ of a sphere falling through a viscous liquid is given by the formula: $v = \frac{2gr^2(ho - \sigma)}{9\eta}$,where $g$ is the acceleration due to gravity,$r$ is the radius of the sphere,$ho$ is the density of the sphere,$\sigma$ is the density of the liquid,and $\eta$ is the coefficient of viscosity.The ratio of terminal velocities is: $\frac{v_P}{v_Q} = \frac{r_P^2(ho_P - \sigma_1) \eta_2}{r_Q^2(ho_Q - \sigma_2) \eta_1}$.Given values:$ho_P = ho_Q = 8 \ g \ cm^{-3}$$r_P = 0.5 \ cm$,$r_Q = 0.25 \ cm$$\sigma_1 = 0.8 \ g \ cm^{-3}$,$\sigma_2 = 1.6 \ g \ cm^{-3}$$\eta_1 = 3 \ 	ext{poiseuille}$,$\eta_2 = 2 \ 	ext{poiseuille}$Substituting these values into the ratio formula:$\frac{v_P}{v_Q} = \frac{(0.5)^2 	imes (8 - 0.8) 	imes 2}{(0.25)^2 	imes (8 - 1.6) 	imes 3}$$\frac{v_P}{v_Q} = \frac{0.25 	imes 7.2 	imes 2}{0.0625 	imes 6.4 	imes 3}$$\frac{v_P}{v_Q} = \frac{3.6}{1.2} = 3$.Thus,the ratio of the terminal velocities is $3$.

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