Question 275: Properties of Matter

Question

An air bubble of diameter $$6 \mathrm{~mm}$$ rises steadily through a solution of density $$1750 \mathrm{~kg} / \mathrm{m}^{3}$$ at the rate of $$0.35 \mathrm{~cm} / \mathrm{s}$$. The co-efficient of viscosity of the solution (neglect density of air) is ___________ Pas (given, $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$ ).

Accepted Answer

The terminal velocity of a small spherical object moving under the action of gravity through a fluid medium is given by Stokes' Law, which is stated as:
$v = \frac{2}{9} \frac{r^2 g (ho_p - ho_f)}{\eta}$,
where:
$v$ is the velocity of the object (in this case, the air bubble),
$r$ is the radius of the object,
$g$ is the acceleration due to gravity,
$ho_p$ is the density of the object (negligible in this case, as it's an air bubble),
$ho_f$ is the density of the fluid (the solution), and
$\eta$ is the coefficient of viscosity of the fluid.
Since we are neglecting the density of the air bubble, the formula simplifies to:
$v = \frac{2}{9} \frac{r^2 g ho_f}{\eta}$.
Rearranging for $\eta$, we get:
$\eta = \frac{2}{9} \frac{r^2 g ho_f}{v}$.
Given that $r = \frac{6 \, 	ext{mm}}{2} = 3 \, 	ext{mm} = 3 	imes 10^{-3} \, 	ext{m}$, $g = 10 \, 	ext{ms}^{-2}$, $ho_f = 1750 \, 	ext{kg/m}^{3}$, and $v = 0.35 \, 	ext{cm/s} = 0.35 	imes 10^{-2} \, 	ext{m/s}$, we can substitute these values into the formula to find $\eta$:
$\eta = \frac{2}{9} \frac{(3 	imes 10^{-3})^2 	imes 10 	imes 1750}{0.35 	imes 10^{-2}} = 10 \, 	ext{Pas}$.
Therefore, the coefficient of viscosity of the solution is $10 \, 	ext{Pas}$.

Properties of Matter

Integer / Numeric Answer Type