Question 179 of 192

The diameter of an air bubble which was initially $2mm$, rises steadily through a solution of density $1750 \mathrm{~kg} / \mathrm{m}^3$ at the rate of $0.35 \mathrm{~cm} / \mathrm{s}$. Coefficient of viscosity of the solution is ______ (Assume mass of the bubble to be negligible) (Answer in Poise to the nearest integer)

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Answer: 11

As the bubble is rising steadily the net force acting on it will be zero.

(Because of density of air is negligible, so the value of $mg$ can be neglected)
So, the force of buoyancy $B$ is equal to the weight of the displaced liquid acting upward is equal to the viscous force acting downward, i.e., $B=F$.
$\Rightarrow \dfrac{4\pi }{3}{R}^{3}\rho g=6\pi \eta Rv$
Putting $R=1mm={10}^{-3}m$, $\rho =1.75\times {10}^{3}kg{m}^{-3}$, $g=10m{s}^{-2}$ and $v=0.35\times {10}^{-2}m{s}^{-1}$
$\eta =\dfrac{10}{9}=1\cdot 11$
In SI unit, $\eta =11$ poise $\left(CGS\right)$.