Q 11-09-111JEE MainJEE Main 2022 (28 Jul, Shift 1)Medium
The diameter of an air bubble which was initially $2$ mm, rises steadily through a solution of density $1750\ \text{kg m}^{-3}$ at the rate of $0.35\ \text{cm s}^{-1}$. Coefficient of viscosity of the solution is ______ (Assume mass of the bubble to be negligible) (Answer in Poise to the nearest integer)
Numerical value type. Enter your answer.
Answer: 11
With negligible mass, at steady speed the buoyant force balances the viscous drag ($r = 1$ mm):
$$\frac43\pi r^3\rho g = 6\pi\eta rv \Rightarrow \eta = \frac{2r^2\rho g}{9v}$$
$$\eta = \frac{2(10^{-3})^2(1750)(9.8)}{9(3.5\times10^{-3})} \approx 1.09\ \text{Pa s} \approx 11\ \text{poise}$$
(1 Pa s = 10 poise.)
Solution by Sreeraj P, M.Sc Physics