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

Vedclass · Accepted Answer

$\sqrt{\frac{2}{3}}$

Vedclass · Answer

(B) The terminal velocity $v$ of a sphere falling through a viscous liquid is given by the formula: $v = \frac{2r^2(ho - \sigma)g}{9\eta}$ where $r$ is the radius of the sphere,$ho$ is the density of the sphere,$\sigma$ is the density of the liquid,$g$ is the acceleration due to gravity,and $\eta$ is the coefficient of viscosity. Since both spheres fall with the same uniform speed,their terminal velocities are equal: $v_1 = v_2$ $\frac{2r_1^2(ho_1 - \sigma)g}{9\eta} = \frac{2r_2^2(ho_2 - \sigma)g}{9\eta}$ $r_1^2(ho_1 - \sigma) = r_2^2(ho_2 - \sigma)$ $\frac{r_1^2}{r_2^2} = \frac{ho_2 - \sigma}{ho_1 - \sigma}$ $\frac{r_1}{r_2} = \sqrt{\frac{ho_2 - \sigma}{ho_1 - \sigma}}$ Given: $ho_1 = 11.5 	imes 10^3 \ kg/m^3$,$ho_2 = 8.5 	imes 10^3 \ kg/m^3$,and $\sigma = 2.5 	imes 10^3 \ kg/m^3$. Substituting these values: $\frac{r_1}{r_2} = \sqrt{\frac{8.5 	imes 10^3 - 2.5 	imes 10^3}{11.5 	imes 10^3 - 2.5 	imes 10^3}} = \sqrt{\frac{6.0 	imes 10^3}{9.0 	imes 10^3}} = \sqrt{\frac{6}{9}} = \sqrt{\frac{2}{3}}$

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