Question

A spherical ball of density $\rho$ and radius $0.003m$ is dropped into a tube containing a viscous fluid. If the viscosity of the fluid $=1.260Ns/m^2$ and its density is ${\rho }_L=\frac{\rho }{2}=1260kg/m^3$, then find the terminal speed (in cm/s) of the ball. ($g$= Acceleration due to gravity $=10m/s^2$ )

Answer 1

We know that
$v_T=\frac{2}{9}\frac{r^2(\sigma -\rho )g}{\eta }$
Given, radius of the sphere,
$r=0.003m$
Viscosity of the liquid,
$\eta =1.260Ns/m^2$
Density of the liquid,
${\rho }_L=\frac{\rho }{2}=1260kg/m^3$
terminal speed of the sphere,
$v_T=\frac{2}{9}\frac{r^2(\sigma -\rho )g}{\eta }=\frac{2}{9}\frac{r^2(\rho -\rho /2)g}{\eta }$
$v_T=\frac{2}{9}\frac{\left({0.003}^2\right)\times (2\times 1260-1260)\times 10}{1.260}$
$v_T=0.02m/s=2cm/s$

Final answer: $\left(2\right)$