A spherical solid ball of volume V is made of a material of density rho_1. It is falling through a l

Q: Why is the viscous force given as $kv^2$ and not $6\pi\eta rv$ like in Stokes' Law?

Stokes' Law ($6\pi\eta rv$) is only valid for very small objects moving at very low velocities (laminar flow). For larger objects or higher speeds, the fluid flow becomes turbulent, and the drag force becomes proportional to the square of the velocity ($v^2$). The problem explicitly states to use this $v^2$ dependency.

Q: Does the buoyant force depend on the density of the ball?

No, buoyancy only depends on the volume of the fluid displaced and the density of the fluid itself ($\rho_2$). The density of the ball ($\rho_1$) only affects the gravitational force.

Q: What would happen if the liquid's density $\rho_2$ was greater than the ball's density $\rho_1$?

If $\rho_2 > \rho_1$, the buoyant force would be stronger than gravity. The ball would accelerate upwards instead of falling, and it would eventually reach an upward terminal velocity.

Q: Why does the acceleration become zero at terminal velocity?

As the ball falls faster, the viscous drag force ($kv^2$) increases. Eventually, this upward drag plus the constant upward buoyancy perfectly equals the downward pull of gravity. With zero net force, Newton's First Law tells us the acceleration must be zero.

Q: Is terminal velocity reached instantly as soon as the ball is dropped?

No, it takes time. The ball starts with zero velocity and accelerates. As velocity increases, acceleration decreases until it asymptotically approaches zero at the terminal velocity.

Question

A spherical solid ball of volume $V$ is made of a material of density $ho_1$. It is falling through a liquid of density $ho_2 (ho_2 < ho_1)$. [Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$, i.e., $F_{	ext{viscous}} = -kv^2 (k > 0)$]. The terminal speed of the ball is

Options:

(A) $\sqrt{\frac{Vg(ho_1-ho_2)}{k}}$

(B) $\frac{Vgho_1}{k}$

(C) $\sqrt{\frac{Vgho_1}{k}}$

(D) $\frac{Vg(ho_1-ho_2)}{k}$

Sigma Percentile · Accepted Answer

Have you ever wondered why a raindrop doesn't hit you with the speed of a bullet? Or why a pebble dropped in a lake eventually settles into a steady, constant speed as it sinks? This phenomenon is governed by the beautiful interplay of forces in a fluid, leading to what we call **terminal velocity**.

In this problem, we are looking at a solid spherical ball falling through a liquid. Let's break down the physics step-by-step.

### Analyzing the Setup

Imagine the ball submerged in the liquid. There is a constant battle of forces happening here.

First, we have the relentless pull of gravity downwards. The gravitational force, or weight, is given by the mass of the ball times the acceleration due to gravity. Since mass is volume times density, we can write this as:

$$F_g = Vho_1g$$

But the liquid doesn't just let the ball fall freely. It fights back! The first upward force is **Buoyancy**. According to Archimedes' principle, the buoyant force is equal to the weight of the displaced fluid. Since the ball is fully submerged, it displaces a volume $V$ of the liquid. Therefore, the buoyant force is:

$$F_B = Vho_2g$$

### The Viscous Drag

As the ball moves, it experiences a second upward force: the viscous drag. Usually, for very small objects moving slowly, we use Stokes' Law ($F = 6\pi\eta rv$). However, for larger objects or higher speeds, the drag force becomes proportional to the square of the velocity. The problem explicitly gives us this relationship:

$$F_{	ext{viscous}} = kv^2$$

Notice the negative sign in the problem statement ($F_{	ext{viscous}} = -kv^2$). This simply indicates that the force opposes the direction of motion. Since the ball is falling down, the viscous force acts upwards.

### The Master Equation: Dynamic Equilibrium

When the ball is first dropped, gravity is stronger than the upward forces, so it accelerates downwards. But as its speed $v$ increases, the viscous drag $kv^2$ also increases.

Eventually, the ball reaches a speed where the upward forces perfectly balance the downward force. At this exact moment, the net force becomes zero, and acceleration stops. The ball continues to fall, but at a constant, maximum speed. This is the **terminal velocity**, $v_T$.

Let's write down the force balance equation at terminal velocity:

$$F_{	ext{up}} = F_{	ext{down}}$$

$$F_B + F_{	ext{viscous}} = F_g$$

Substituting our expressions into this equation, we get:

$$Vho_2g + kv_T^2 = Vho_1g$$

### Final Calculation

Now, it's just a matter of simple algebra to isolate $v_T$. Let's move the buoyant force term to the right side:

$$kv_T^2 = Vho_1g - Vho_2g$$

We can factor out the common terms $V$ and $g$ on the right side:

$$kv_T^2 = Vg(ho_1 - ho_2)$$

Finally, divide by $k$ and take the square root to solve for the terminal velocity:

$$v_T = \sqrt{\frac{Vg(ho_1 - ho_2)}{k}}$$

And there we have it! This elegant expression shows exactly how the terminal velocity depends on the volume, the difference in densities, and the drag coefficient.

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Visualized Solution

$F r ee B o d y D ia g r am$

$F or ces$

$T er mina l V e l oc i t y C o n d i t i o n$

$F or ce B a l an ce E q u a t i o n$

$I so l a t in g V i sco u s F or ce$

$T er mina l V e l oc i t y$

The Sigma Insight: Viscosity and Stokes' Law

Analyzing the Setup

The Viscous Drag

The Master Equation

Final Calculation

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Visualized Solution

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