Question

In an experiment to verify Stokes law, a small spherical ball of radius r and density $\sigma$ falls under gravity through a distance h in the air before entering a tank of water. If the terminal velocity of the ball inside water is the same as its velocity just before entering the water surface, then the value of h is proportional to (ignore viscosity of air)

• A. ${\mathrm{r}}^4$
• B. $\mathrm{r}$
• C. ${\mathrm{r}}^3$
• D. ${\mathrm{r}}^2$

Answer 1

The correct option is A

${\mathrm{r}}^4$

Explanation for correct option:

Option (a) ${\mathrm{r}}^4$

Step 1: Given data

Radius of the ball $=\mathrm{r}$

Density of the ball is $\sigma$.

Distance between the initial height of ball and water $=\mathrm{h}$

Step 2: Formula used

Terminal velocity: Terminal velocity is the maximum velocity (speed) attainable by an object as it falls through a fluid.

$V_{\mathrm{t}}=\frac{2{\mathrm{r}}^2(\mathrm{\rho }-\mathrm{\sigma })\mathrm{g}}{9\mathrm{\eta }}$
${\mathrm{V}}_{\mathrm{t}}=$ terminal velocity
$\mathrm{m}=$ mass of the falling object
$\mathrm{g}=$ acceleration due to gravity
$\mathrm{\rho }=$ density of the fluid through which the object is
$\mathrm{A}=$ projected area of the object

Step 3: Find value of $\mathrm{h}$

by the comparing the formula mentioned above and velocity according to the laws of motion, we get:

${\mathrm{V}}_{\mathrm{t}}=\sqrt{2\mathrm{gh}}$
$\left[(2/9){\mathrm{r}}^2\right]\left[\right(\mathrm{\rho }-\mathrm{\sigma })/\mathrm{\eta }]=\sqrt{2\mathrm{gh}}$
$$\begin{gathered} {\text{r}}^2\propto \sqrt{\mathrm{h}} \\ \Rightarrow {\mathrm{r}}^4\propto \mathrm{h} \end{gathered}$$
$\mathrm{h}\propto {\mathrm{r}}^4$

Hence option (a) is correct.