Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A ball of mass $m$ and radius $r$ is released in viscous liquid. The value of its terminal velocity is proportional to

Mechanical Properties of Fluids

Solution:

Terminal velocity $(v)=\frac{2}{9} \frac{r^{2}(\rho-\sigma) g}{\eta}$
$v=\frac{2 r^{2} \rho\left(1-\frac{\sigma}{\rho}\right) g}{9 \eta}=\frac{2 r^{3} \rho\left(1-\frac{\sigma}{\rho}\right) g}{9 r \eta}$
$v \propto \frac{m}{r}$
$\left(m=\frac{4}{3} \pi r^{3} \rho\right)$
$\left(m \propto r^{3} \rho\right)$