Question

An air bubble of diameter 2 mm rises steadily through a solution of density 1750 kg m^-3 at the rate of 0.35 cm s^-1. Calculate the coefficient of viscosity of the solution. The density of air is negligible.

• A. 11 poise
• B. 13.6 poise
• C. 2.6 poise
• D. 8 poise

Answer 1

The correct option is A

11 poise

The force of buoyancy B is equal to the weight of the displaced liquid. Thus,
$B=\frac{4}{3}\pi r^3\sigma g$
This force is upward. The viscous force acting downward is $F=6\pi \eta rv$
The weight of the air bubble may be neglected as the density of air is small. For uniform velocity F = B
or,$6\pi \eta rv=\frac{4}{3}\pi r^3\sigma g$
or,$\eta =\frac{2r^2\sigma g}{9v}$
or,=$\frac{2\times (1\times {10}^{-3}m{)}^2\times (1750kg{m}^{-3}\left)\right(9.8m{s}^{-2})}{9\times (0.35\times {10}^{-2}m{s}^{-1})}$
$\approx$ 11 poise
This appears to be a highly viscous liquid.