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(B) The terminal velocity $v_T$ of a spherical body falling through a viscous fluid is given by the formula: $v_T = \frac{2}{9} \frac{r^2 g (\rho - \s

Vedclass · Accepted Answer

$5$

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(B) The terminal velocity $v_T$ of a spherical body falling through a viscous fluid is given by the formula: $v_T = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta}$.Here,$r$ is the radius of the ball,$\rho$ is the density of the material of the ball,$\sigma$ is the density of the fluid,$\eta$ is the coefficient of viscosity,and $g$ is the acceleration due to gravity.Since the material of the ball and the fluid are the same,$v_T \propto r^2$.Given: $r_1 = 6 \ mm$,$v_{T1} = 20 \ cm/s$,and $r_2 = 3 \ mm$.Using the proportionality: $\frac{v_{T2}}{v_{T1}} = (\frac{r_2}{r_1})^2$.Substituting the values: $\frac{v_{T2}}{20} = (\frac{3}{6})^2 = (\frac{1}{2})^2 = \frac{1}{4}$.Therefore,$v_{T2} = \frac{20}{4} = 5 \ cm/s$.

The terminal velocity of a metallic ball of radius $6 \ mm$ in a viscous fluid is $20 \ cm/s$. The terminal velocity of another ball of same material and having radius $3 \ mm$ in the same fluid will be . . . . . . $cm/s$.

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