Question

An air bubble of radius $5 \times {10^{ - 4}}m$ rises in a liquid of viscosity $0.1Pas\left( {g = 10m{s^{ - 2}}} \right)$ and density $900kg{m^{ - 3}}$ . The terminal velocity of the bubble is:
\left( A \right)0.005m{s^{ - 1}} \\\
\left( B \right)0.01m{s^{ - 1}} \\\
\left( C \right)0.5m{s^{ - 1}} \\\
\left( D \right)0.4m{s^{ - 1}} \\\

Answer 1

Hint : In order to solve this question, we are going to first write the formula for the terminal velocity, then, by putting the values for the tension, radius, density and viscosity, the terminal velocity value is determined. It is also to be noted that the terminal velocity does not exceed the maximum value.
For an air bubble with tension, $T$ , radius $r$ , viscosity of the bubble, acceleration due to gravity, $g$ and the density $\sigma$ , the terminal velocity of the bubble is given by the formula:
${v_T} = \dfrac{{2{r^2}\left( {T - \sigma } \right)g}}{{9\eta }}$

Complete Step By Step Answer:
Let us first of all start by using the formula given below for the terminal velocity,
i.e.
${v_T} = \dfrac{{2{r^2}\left( {T - \sigma } \right)g}}{{9\eta }}$
It is given in the question that the radius of the bubble is
$r = 5 \times {10^{ - 4}}m$ ,
The viscosity of the liquid is
$\eta = 0.1Pas$
And the density is
$\sigma = 900kg{m^{ - 3}}$
Now, putting these above values in the formula, we can find the terminal velocity as
{v_T} = \dfrac{{2{{\left( {5 \times {{10}^{ - 4}}} \right)}^2} \times 900 \times 10}}{{9 \times 0.1}} \\\ \Rightarrow {v_T} = 0.005m{s^{ - 1}} \\\
Thus, the option $\left( A \right)0.005m{s^{ - 1}}$ is the correct answer.

Note :
Terminal velocity is the maximum velocity attainable by an object as it falls through a fluid (air is the most common example). ... As the speed of an object increases, so does the drag force acting on it, which also depends on the substance it is passing through.
It is determined that the terminal rise velocity of a single air bubble in an otherwise water saturated porous medium cannot exceed $18.5cm{s^{ - 1}}$ .