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(C) 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

$1:9$

Vedclass · Answer

(C) 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 $r$ is the radius of the sphere, $\rho$ 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 the spheres are made of the same material and fall through the same liquid, the parameters $\rho, \sigma, g,$ and $\eta$ are constant.Therefore, the terminal velocity is proportional to the square of the radius: $v \propto r^2$.For radii $R$ and $3R$, the ratio of terminal velocities is: $\frac{v_1}{v_2} = \frac{R^2}{(3R)^2} = \frac{R^2}{9R^2} = \frac{1}{9}$.Thus, the ratio is $1:9$.

Two spheres of the same material, but of radii $R$ and $3R$ are allowed to fall vertically downwards through a liquid of density $\rho$. The ratio of their terminal velocities is

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