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=10\,m/{{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.
$\therefore 6 \pi \,\eta r v=\frac{4}{3} \pi r^{3}(\rho-\sigma) g$
$\eta=\frac{2}{9} \frac{r^{2}(\rho-\sigma)}{v} g$
Given, $ v=-2 \times 10^{-3} \,m / s , r=10^{-2} \,m$
$\rho=0, \sigma=1.5 \times 10^{3} \,kg / m ^{3}, g=10 \,m / 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.