A solid sphere of radius R acquires a terminal velocity v1 when falling (due to gravity) through a v

Q: Why does the terminal velocity depend on the square of the radius?

The driving force (gravity minus buoyancy) scales with the volume of the sphere ($r^3$), while the resisting viscous force (Stokes' drag) scales with the radius ($r$). When you equate them to find the terminal velocity, the ratio $r^3/r$ leaves you with an $r^2$ dependence.

Q: Why is the volume conserved when the sphere breaks?

The total amount of matter (mass) remains the same. Since the material hasn't changed, its density is constant. Therefore, the total volume before and after breaking must also remain perfectly constant.

Q: Does the density of the fluid affect the ratio of the velocities?

No. Because both the large sphere and the small spheres fall through the exact same fluid, the density terms (and the viscosity term) cancel out completely when you take the ratio of their velocities.

Q: What if the sphere broke into 8 pieces instead of 27?

If it broke into 8 pieces, the volume equation would be $R^3 = 8r^3$, which means $r = R/2$. The velocity ratio would then be $R^2 / (R/2)^2 = 4$.

Q: Why do we ignore the initial velocity of the spheres?

Terminal velocity is the steady-state speed achieved after a long time, where the net force becomes zero. It is completely independent of whatever initial velocity the sphere started with.

Question

A solid sphere of radius $R$ acquires a terminal velocity $v_1$ when falling (due to gravity) through a viscous fluid having a coefficient of viscosity $\eta$. The sphere is broken into 27 identical solid spheres. If each of these spheres acquires a terminal velocity, $v_2$ when falling through the same fluid, the ratio $(v_1 / v_2)$ equals

Options:

(A) 9

(B) $\frac{1}{27}$

(C) $\frac{1}{9}$

(D) 27

Sigma Percentile · Accepted Answer

### The Physics of Falling Spheres: Terminal Velocity and Volume Conservation

Imagine a large solid sphere falling gracefully through a thick, viscous fluid like honey or glycerin. Initially, it accelerates due to gravity, but soon, the viscous drag force balances the net downward force, and the sphere reaches a constant speed. This steady speed is known as the **terminal velocity**.

### The Master Equation

The terminal velocity $v_T$ of a spherical body falling through a viscous fluid is given by Stokes' Law:

$$v_T = \frac{2}{9} \frac{r^2}{\eta} (ho_0 - ho_f) g$$

Where:
- $r$ is the radius of the sphere.
- $\eta$ is the coefficient of viscosity of the fluid.
- $ho_0$ is the density of the sphere.
- $ho_f$ is the density of the fluid.
- $g$ is the acceleration due to gravity.

Notice a beautiful proportionality here. If the material of the sphere and the fluid remain the same, all terms except the radius become constant. This leads us to a powerful conclusion:

$$v_T \propto r^2$$

The terminal velocity is directly proportional to the square of the radius. A larger sphere falls significantly faster than a smaller one of the same material!

### Conservation of Volume

In our problem, the large sphere of radius $R$ breaks into 27 identical smaller spheres, each of radius $r$. Even though the shape has fragmented, the total amount of matter—and therefore the total volume—must remain perfectly conserved.

Let's set up the volume conservation equation:

$$	ext{Volume of large sphere} = 27 	imes 	ext{Volume of one small sphere}$$

$$\frac{4}{3} \pi R^3 = 27 	imes \frac{4}{3} \pi r^3$$

By canceling out the common factor of $\frac{4}{3} \pi$ on both sides, we get a simple relationship between the radii:

$$R^3 = 27 r^3$$

Taking the cube root of both sides, we find:

$$R = 3r \quad 	ext{or} \quad r = \frac{R}{3}$$

This means each of the 27 small spheres has exactly one-third the radius of the original large sphere.

### Final Calculation

We are asked to find the ratio of the initial terminal velocity $v_1$ to the new terminal velocity $v_2$. Using our proportionality rule ($v_T \propto r^2$), we can write:

$$\frac{v_1}{v_2} = \frac{R^2}{r^2}$$

Now, substitute the value of $r$ we just found:

$$\frac{v_1}{v_2} = \frac{R^2}{(R/3)^2}$$

$$\frac{v_1}{v_2} = \frac{R^2}{R^2 / 9}$$

The $R^2$ terms cancel out beautifully, and the denominator's fraction flips, giving us:

$$\frac{v_1}{v_2} = 9$$

The original large sphere falls 9 times faster than the smaller fragmented spheres. The elegance of scaling laws in physics makes this complex-sounding problem incredibly simple to solve!

Select Answer:

Visualized Solution

Visualizing the Setup

Terminal Velocity Formula

Breaking into Smaller Spheres

Conservation of Volume

Finding the New Radius

Ratio of Terminal Velocities

Final Calculation

The Sigma Insight: Viscosity and Stokes' Law

The Physics of Falling Spheres

The Master Equation

Conservation of Volume

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Setup

Terminal Velocity Formula

Breaking into Smaller Spheres

Conservation of Volume

Finding the New Radius

Ratio of Terminal Velocities

Final Calculation

The Sigma Insight: Viscosity and Stokes' Law

The Physics of Falling Spheres

The Master Equation

Conservation of Volume

Final Calculation

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