Question

A spherical ball of density $\rho$ and radius $0.003\text{m}$ is dropped into a tube containing a viscous fluid up to the $0\text{cm}$ mark as shown in the figure. Viscosity of the fluid $=1.2160{\text{N.s m}}^{-2}$ and its density ${\rho }_L=\frac{\rho }{2}=1260{\text{kg.m}}^{-3}$. Assume the ball reaches a terminal speed by the $10\text{cm}$ mark. Find the time taken rounded off to the nearest integer (in $\text{sec}$) by the ball to traverse the distance between the $10\text{cm}$ and $20\text{cm}$ mark. Take $g=10{\text{ms}}^{-2}$

Answer 1

We know that terminal velocity $v_T=\frac{2r^2({\rho }_{\text{body}}-{\rho }_{\text{fluid}})g}{9\eta }$
Here, $r=3\times {10}^{-3}\text{m}$
Putting the values, we get,
$v_T=\frac{2r^2(\rho -{\rho }_L)g}{9\eta }...\left(i\right)$
${\rho }_L=\frac{\rho }{2}$
$\Rightarrow \rho =2{\rho }_L=2520{\text{kg.m}}^{-3}$
and $\eta =1.2160{\text{N.s m}}^{-2}$
On substituting in Eq.$\left(i\right)$,
$v_T\simeq 2\times {10}^{-2}\text{m/s}$

Once terminal velocity is attained, body will move with constant velocity $v_T$ thereafter.
Distance travelled by the body,
$d=20-10=10\text{cm}=0.1\text{m}$
Time taken to traverse distance:
$t=\frac{d}{v_T}=5\text{sec}$