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Question

(A) The terminal velocity $v_t$ of a sphere of radius $r$ and density $\rho$ falling through a viscous liquid of density $\sigma$ and viscosity $\eta$

Vedclass · Accepted Answer

$\frac{m - m'}{r}$

Vedclass · Answer

(A) The terminal velocity $v_t$ of a sphere of radius $r$ and density $\rho$ falling through a viscous liquid of density $\sigma$ and viscosity $\eta$ is given by Stokes' Law:$v_t = \frac{2}{9} r^2 g \frac{(\rho - \sigma)}{\eta}$The mass of the ball is $m = \frac{4}{3} \pi r^3 \rho$,so $\rho = \frac{3m}{4 \pi r^3}$.The mass of the displaced liquid is $m' = \frac{4}{3} \pi r^3 \sigma$,so $\sigma = \frac{3m'}{4 \pi r^3}$.Substituting these into the terminal velocity formula:$v_t = \frac{2}{9} r^2 g \frac{1}{\eta} \left( \frac{3m}{4 \pi r^3} - \frac{3m'}{4 \pi r^3} \right)$$v_t = \frac{2}{9} r^2 g \frac{1}{\eta} \left( \frac{3(m - m')}{4 \pi r^3} \right)$$v_t = \frac{6 g (m - m')}{36 \pi \eta r} = \frac{g (m - m')}{6 \pi \eta r}$Since $g$,$\pi$,and $\eta$ are constants,we have $v_t \propto \frac{m - m'}{r}$.

$A$ ball of mass $m$ and radius $r$ is gently released in a viscous liquid. The mass of the liquid displaced by it is $m'$ such that $m > m'$. The terminal velocity is proportional to

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