© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Physical Situation
A spherical ball, of radius $r$ and density $\rho$, falls freely in air from a height $h$. Just before entering the water surface, it attains a certain velocity (due to free fall). Inside the water, the ball slows down due to the viscosity of water until it finally reaches terminal velocity.
Step 2: Express the Velocity of the Ball in Air
Since we ignore air viscosity, the ball experiences free fall over the distance $h$. The velocity $v$ just before entering water is given by energy conservation (or the equations of motion under constant acceleration):
$$v = \sqrt{2gh}.$$
Step 3: Express the Terminal Velocity in Water (Stokes' Law)
The terminal velocity $v_{t}$ of a spherical object falling through a viscous fluid (water in this case) can be found using Stokes' law:
$$v_{t} = \frac{2}{9}\,\frac{r^2(\rho - \rho_{w})\,g}{\eta},$$
where:
$r$ is the radius of the sphere,
$\rho$ is the density of the sphere,
$\rho_{w}$ is the density of water,
$\eta$ is the coefficient of viscosity of water,
$g$ is the acceleration due to gravity.
Step 4: Match the Velocities
Given that the terminal velocity in water is the same as the velocity of the ball just before entering the water, we set
$$v_{t} = \sqrt{2gh}.$$
So,
$$\frac{2}{9}\,\frac{r^2(\rho - \rho_{w})\,g}{\eta} = \sqrt{2gh}.$$
Step 5: Solve for the Height $h$
Square both sides:
$$\left(\frac{2}{9}\,\frac{r^2(\rho - \rho_{w})\,g}{\eta}\right)^{2} = 2gh.$$
Hence,
$$\frac{4}{81}\,\frac{r^4(\rho - \rho_{w})^2\,g^2}{\eta^2} = 2gh.$$
Rearranging to solve for $h$:
$$h = \frac{\frac{4}{81}\,\frac{r^4(\rho - \rho_{w})^2\,g^2}{\eta^2}}{2g}.$$
We see that $h$ depends on $r^4$ (and other constants). Therefore,
$$h \propto r^4.$$
Step 6: Conclusion
The distance $h$ through which the sphere falls in air (so that its velocity just equals the terminal velocity in water) is proportional to $r^4.$ Thus, the correct answer is
r4
