The speed of a ball of radius 2 ,cm in a visc | vedclass

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(C) According to Stoke's Law, the terminal velocity $V$ of a spherical body falling through a viscous liquid is given by $V = \frac{2}{9} \frac{r^2 g

Vedclass · Accepted Answer

$5$

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(C) According to Stoke's Law, the terminal velocity $V$ of a spherical body falling through a viscous liquid is given by $V = \frac{2}{9} \frac{r^2 g (\rho - \sigma)}{\eta}$.From this formula, it is clear that the terminal velocity is directly proportional to the square of the radius of the ball, i.e., $V \propto r^2$.Given:Radius $r_1 = 2 \,cm$, Velocity $V_1 = 20 \,cm / s$.Radius $r_2 = 1 \,cm$, Velocity $V_2 = ?$.Using the proportionality $V \propto r^2$:$\frac{V_2}{V_1} = \left(\frac{r_2}{r_1}\right)^2$$\frac{V_2}{20} = \left(\frac{1}{2}\right)^2$$\frac{V_2}{20} = \frac{1}{4}$$V_2 = \frac{20}{4} = 5 \,cm / s$.

The speed of a ball of radius $2 \,cm$ in a viscous liquid is $20 \,cm / s$. What will be the speed of a ball of radius $1 \,cm$ in the same liquid (in $\,cm / s$)?

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