Question

An air bubble of diameter $4\text{mm}$ rises steadily through a solution of density $1800{\text{kg/m}}^3$ at the rate of $0.05\text{m/s}$. Neglecting density of air, the co-efficient of viscosity of the solution is
[Take $g=10{\text{m/s}}^2$]

• A. $0.64\text{poise}$
• B. $6.4\text{poise}$
• C. $0.32\text{poise}$
• D. $3.2\text{poise}$

Answer 1

The correct option is D $3.2\text{poise}$

Given, diameter of air bubble, $d=4\text{mm}$
Radius of air bubble, $r=2\text{mm}=2\times {10}^{-3}\text{m}$
Density of solution, $\rho =1800{\text{kg/m}}^3$
Speed of air bubble, $v=0.05\text{m/s}$
Co-efficient of viscosity of solution, $\eta =?$
As air bubble rises steadily, therefore net force on it should be zero. Also, density of the air is neglegile, so we can neglect force due to gravity.


$\sum F=0$
$\Rightarrow F_b-F_v=0$
$\Rightarrow \frac{4}{3}\pi r^3\rho g=6\pi \eta rv$
$\Rightarrow \eta =\frac{2r^2\rho g}{9v}$
$=\frac{2\times (2\times {10}^{-3}{)}^2\times 1800\times 10}{9\times 0.05}$
$\Rightarrow \eta =0.32\text{Pa-s}=3.2\text{poise}$