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(C) According to Stokes' Law,the terminal velocity $v_T$ of a spherical body falling through a viscous fluid is given by the formula:$v_T = \frac{2}{9

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$4v$

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(C) According to Stokes' Law,the terminal velocity $v_T$ of a spherical body falling through a viscous fluid is given by the formula:$v_T = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta}$Where $r$ is the radius of the sphere,$\rho$ is the density of the sphere,$\sigma$ is the density of the fluid,$g$ is the acceleration due to gravity,and $\eta$ is the coefficient of viscosity.From this relation,we can see that $v_T \propto r^2$.Given the initial radius is $r$ and the initial speed is $v$,we have $v \propto r^2$.When the radius is changed to $2r$,the new terminal velocity $v'$ will be:$v' \propto (2r)^2 = 4r^2$Therefore,$v' = 4v$.

$A$ copper ball of radius $r$ travels with a uniform speed $v$ in a viscous fluid. If the ball is changed with another ball of radius $2r$,then the new uniform speed will be

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