Question

An air bubble of radius ${10}^{-2}m$ is rising up at a steady rate of $2\times {10}^{-3}m/s$ through a liquid of density $1.5\times {10}^{3}kg/m^3,$ the coefficient of viscosity neglecting the density of. air, will be $(g=10m/s^2)$

• A. 23.2 units
• B. 83.5 units
• C. 334 units
• D. 167 units

Answer 1

Answer: (D)

Since air bubble is moving up with a constant velocity, there is no acceleration in it.
Let bubble of radius $r$ and density $\rho$ is falling in a liquid whose density is $\sigma$ and coefficient of viscosity $\eta$.

It attains a terminal velocity due to two forces effective force acting downward
$=V(\rho -\sigma )g=\frac{4}{3}\pi r^3(\rho -\sigma )g$
Viscous force acting upward $=6\pi \eta rv$.
Since, ball is moving up with constant velocity $v$, there is no acceleration in it, the net force acting on it must be zero.
$\therefore 6\pi \eta rv=\frac{4}{3}\pi r^3(\rho -\sigma )g$
$\eta =\frac{2}{9}\frac{r^2(\rho -\sigma )}{v}g$
Given, $v=-2\times 10^{-3}m/s,r=10^{-2}m$
$\rho =0,\sigma =1.5\times 10^{3}kg/m^3,g=10m/s^2$
$\therefore \eta =\frac{2\times {\left(10^{-2}\right)}^2\times \left(-1.5\times 10^3\right)\times 10}{9\times \left(-2\times 10^{-3}\right)}$
$=\frac{3}{18\times 10^{-3}}=\frac{1}{6}\times 10^3$
$=0.167\times 10^3=167$ units.