Question

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

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

Answer 1

The Correct Option is B

To solve this problem, we need to understand the dynamics of a spherical ball falling through air and then entering water, adhering to Stokes' Law principles. Let's go through the steps:

1. First, the spherical ball of radius $r$ and density $\rho$ falls through air. Since the viscosity of air is ignored, there's no significant resistance while the ball is falling in air. Thus, the ball accelerates under the influence of gravity.
2. Upon entering the water, the ball reaches a terminal velocity. According to Stokes' Law, the terminal velocity $v_t$ in a fluid (here, water) is given by the formula:

v_t = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta}

• Here, $\rho$ is the density of the sphere, $\sigma$ is the density of the water, $g$ is the acceleration due to gravity, and $\eta$ is the viscosity of the water.

3. The condition in the problem states that the terminal velocity $v_t$ in the water is equal to the velocity of the ball just before hitting the water surface.
4. Using the principle of conservation of energy, the velocity of the ball just before entering the water due to falling through height $h$ can be given by:

v = \sqrt{2gh}

5. Since $v_t = v$, equating the two expressions for velocity gives us:

\frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta} = \sqrt{2gh}

6. Squaring both sides:

\left( \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta} \right)^2 = 2gh

7. Solving for $h$ gives:

h \propto r^4

8. This shows that the height $h$ is proportional to the fourth power of the radius, $r^4$.

Hence, the correct option is $r^4$.