Question

A spherical solid ball of volume $V$ is made of a material of density $\rho_1$ It is falling through a liquid of density $\rho_2(\rho_2 < \rho_1)$. Assume that the liquid applies a viscous force on the ball that is $F = krv$ where $r$ is the radius of the ball and $v$ is the speed and $k$ is the constant of proportionality. The terminal speed of the ball is

• A. $\frac{4\pi r^{2} g\left(\rho_{1} -\rho_{2}\right)}{3k}$
• B. $\frac{\pi r^{2} g\left(\rho_{1} -\rho_{2}\right)}{3k}$
• C. $\frac{\pi r^{2} g\left(\rho_{1} -\rho_{2}\right)}{k}$
• D. $\frac{\pi r^{2} g\left(\rho_{1} -\rho_{2}\right)}{4k}$

Answer 1

Answer: (A)

$\frac{4\pi r^{2} g\left(\rho_{1} -\rho_{2}\right)}{3k}$

Answer 2

When the falling ball attains the terminal velocity $\upsilon_T$, then the acceleration of the ball is zero. In this situation $mg = F_{B} +F_{V}$ $V\rho_{1}g = V\rho_{2}g +kr\upsilon_{T}$ $\upsilon_{T} = \frac{V\left(\rho_{1}-\rho_{2}\right)g}{kr}$ $\upsilon_{T} =\frac{\frac{4}{3}\pi r^{3}\left(\rho_{1}-\rho_{2}\right)g }{kr}$ $= \frac{4\pi^{2}\left(\rho_{1} -\rho_{2}\right)g}{3k}$