In an experiment to verify Stoke's law, a small spherical ball of radius r and density rho falls und

Q: Why do we ignore the viscosity of air?

Air has a very low viscosity compared to water. Ignoring it simplifies the problem by allowing us to treat the fall in air as a perfect free fall, where the only force acting is gravity.

Q: Why does the ball reach terminal velocity in water?

As the ball enters the water, it experiences a viscous drag force that increases with its velocity. When this drag force plus the buoyant force exactly balances the gravitational force, the net force becomes zero, and the ball moves with a constant terminal velocity.

Q: What is the buoyant force acting on the ball?

The buoyant force is the upward force exerted by the water, equal to the weight of the water displaced by the ball. It is given by $\frac{4}{3}\pi r^3 \rho_l g$, where $\rho_l$ is the density of water.

Q: Why is the terminal velocity proportional to $r^2$?

The driving force (weight minus buoyancy) is proportional to the volume, which scales as $r^3$. The resisting viscous drag force is proportional to $r \cdot v$. Equating them gives $r^3 \propto r \cdot v$, which simplifies to $v \propto r^2$.

Q: What if the ball's density was less than water?

If the ball's density was less than water ($\rho < \rho_l$), the buoyant force would be greater than its weight. The ball would not sink; instead, it would experience a net upward force and eventually float on the surface.

Question

In an experiment to verify Stoke's 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)

Options:

(A) $r^4$

(B) $r$

(C) $r^3$

(D) $r^2$

Sigma Percentile · Accepted Answer

### The Perfect Dive: When Free Fall Meets Terminal Velocity

Imagine standing at the edge of a diving board, looking down at the water below. When you jump, you accelerate through the air, gaining speed until the moment you hit the water. But once you're submerged, the water pushes back, eventually slowing you down to a steady, constant speed. This is exactly what our little spherical ball is experiencing in this classic physics problem!

### The Free Fall in Air

Our journey begins in the air. The problem explicitly tells us to ignore the viscosity of air. This is a huge relief because it means we can treat the ball's descent as a perfect free fall under gravity.

Starting from rest, the ball falls through a height $h$. By the time it reaches the water surface, it has converted its gravitational potential energy into kinetic energy. Using the fundamental equations of kinematics, the velocity $v$ of the ball just before it kisses the water surface is given by:

$$v = \sqrt{2gh}$$

This is our first critical piece of the puzzle. It tells us exactly how fast the ball is moving when it transitions from air to water.

### The Terminal Velocity in Water

Now, the environment changes drastically. As the ball plunges into the water, it is no longer in free fall. It now faces two new forces: an upward buoyant force and a viscous drag force that opposes its motion.

According to **Stokes' Law**, as the ball speeds up, the viscous drag increases until the net force on the ball becomes exactly zero. At this point, the ball stops accelerating and moves with a constant speed known as the **terminal velocity**, $v_T$. The formula for terminal velocity is a beautiful balance of forces:

$$v_T = \frac{2r^2g(\rho - \rho_l)}{9\eta}$$

Here, $r$ is the radius of the ball, $\rho$ is its density, $\rho_l$ is the density of the water, and $\eta$ is the viscosity of the water.

### The Master Equation

The problem gives us a fascinating condition: the velocity of the ball just before entering the water is exactly equal to its terminal velocity inside the water. This means the ball doesn't need to speed up or slow down once it's submerged; it's already moving at the perfect speed!

Let's equate our two expressions:

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

### Final Calculation

Our goal is to find how the height $h$ depends on the radius $r$. To isolate $h$, we need to get rid of that square root. Let's square both sides of the equation:

$$2gh = \frac{4r^4g^2(\rho - \rho_l)^2}{81\eta^2}$$

Now, we simply divide by $2g$ to solve for $h$:

$$h = \frac{2r^4g(\rho - \rho_l)^2}{81\eta^2}$$

Look closely at this final expression. The terms $g$, $\rho$, $\rho_l$, and $\eta$ are all constants for a given setup. The only variable that $h$ depends on is the radius $r$.

We can clearly see that $h$ is directly proportional to $r^4$:

$$h \propto r^4$$

And there we have it! A beautiful synthesis of kinematics and fluid dynamics leading us straight to the correct answer.

Select Answer:

Visualized Solution

The Physical Setup

Velocity after Free Fall

Terminal Velocity

Applying the Given Condition

Squaring Both Sides

Proportionality of $h$

The Way Forward

The Sigma Insight: Viscosity and Stokes' Law

The Perfect Dive

The Free Fall in Air

The Terminal Velocity in Water

The Master Equation

Final Calculation

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