Consider two solid spheres P and Q each of density 8 gm cm^-3 and diameters 1 cm and 0.5 cm, respect

Q: Why did we use the diameter to find the radius? Can't we just use the diameter directly in the formula?

While the ratio of diameters is the same as the ratio of radii, the standard formula is derived using the radius. It is always safer to convert to radius to avoid any confusion, especially if other formulas in the problem depend on radius.

Q: Why is the viscosity ratio inverted in the final ratio equation?

Terminal velocity is inversely proportional to viscosity ($\eta$). Therefore, when we take the ratio $v_P / v_Q$, the viscosity term becomes $\eta_Q / \eta_P$.

Q: Do we need to convert the units from CGS (g/cm^3, poise, cm) to SI units?

No! Since we are calculating a ratio, as long as all quantities are in the same system of units (CGS in this case), the units will cancel out completely. Converting to SI would just be a waste of precious time.

Q: What if the density of the liquid was greater than the density of the sphere?

If the liquid density $\rho$ is greater than the sphere density $d$, the term $(d - \rho)$ becomes negative. This physically means the buoyant force is greater than the weight, and the sphere will rise upwards instead of falling, eventually reaching an upward terminal velocity.

Q: Does Stokes' Law hold true for any speed of the falling sphere?

No, Stokes' Law is only valid for streamline, low-speed flow (low Reynolds number). If the sphere falls too fast, turbulence occurs, and the drag force becomes proportional to the square of the velocity, making Stokes' Law inapplicable.

Question

Consider two solid spheres P and Q each of density $8	ext{ gm cm}^{-3}$ and diameters $1	ext{ cm}$ and $0.5	ext{ cm}$, respectively. Sphere P is dropped into a liquid of density $0.8	ext{ gm cm}^{-3}$ and viscosity $\eta = 3	ext{ poiseulles}$. Sphere Q is dropped into a liquid of density $1.6	ext{ gm cm}^{-3}$ and viscosity $\eta = 2	ext{ poiseulles}$. The ratio of the terminal velocities of P and Q is

Sigma Percentile · Accepted Answer

### Introduction: The Magic of Terminal Velocity

Imagine dropping a stone from a high cliff. It keeps accelerating, going faster and faster under the relentless pull of gravity.

But what happens when you drop a tiny sphere into a jar of honey? It quickly settles into a steady, constant speed.

This constant speed is what we call **terminal velocity**. It is a beautiful state of dynamic equilibrium where the forces of nature perfectly balance each other out.

### Deconstructing the Physics: The Three-Way Tug of War

When a solid sphere falls through a viscous fluid, it is caught in a three-way tug of war.

First, **gravity** pulls the sphere downwards with its weight:

$$W = mg = V d g$$

where $V$ is the volume of the sphere and $d$ is its density.

Second, **buoyancy** pushes the sphere upwards, equal to the weight of the displaced liquid:

$$F_b = V ho g$$

where $ho$ is the density of the liquid.

Third, **viscous drag** opposes the downward motion, acting upwards. According to Stokes' Law, this force is:

$$F_v = 6\pi\eta r v$$

where $\eta$ is the viscosity of the liquid, $r$ is the radius of the sphere, and $v$ is its velocity.

### The Master Equation: Stokes' Law to the Rescue

As the sphere accelerates, the viscous drag increases. Eventually, the upward forces perfectly balance the downward weight.

At this point, the net force becomes zero, and the sphere continues to fall at a constant **terminal velocity** $v_T$:

$$F_v + F_b = W$$

Substituting the expressions for each force:

$$6\pi\eta r v_T + \frac{4}{3}\pi r^3 ho g = \frac{4}{3}\pi r^3 d g$$

Solving for $v_T$, we get the master equation:

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

This elegant formula shows that terminal velocity is directly proportional to the square of the radius and the density difference, and inversely proportional to the viscosity.

### Parameter Extraction: Knowing Our Players

Let's look at the two spheres, $P$ and $Q$, and their respective environments.

For Sphere $P$:
- Diameter $D_P = 1	ext{ cm} \implies$ Radius $r_P = 0.5	ext{ cm}$
- Density of sphere $d = 8	ext{ g cm}^{-3}$
- Density of liquid $ho_P = 0.8	ext{ g cm}^{-3}$
- Viscosity of liquid $\eta_P = 3	ext{ poise}$

For Sphere $Q$:
- Diameter $D_Q = 0.5	ext{ cm} \implies$ Radius $r_Q = 0.25	ext{ cm}$
- Density of sphere $d = 8	ext{ g cm}^{-3}$
- Density of liquid $ho_Q = 1.6	ext{ g cm}^{-3}$
- Viscosity of liquid $\eta_Q = 2	ext{ poise}$

### Step-by-Step Calculation: The Power of Ratios

Instead of calculating the absolute terminal velocities, we can find their ratio directly. This is a highly efficient strategy for competitive exams like JEE.

Let's write the ratio of the terminal velocities:

$$\frac{v_P}{v_Q} = \left(\frac{r_P}{r_Q}ight)^2 	imes \left(\frac{\eta_Q}{\eta_P}ight) 	imes \left(\frac{d - ho_P}{d - ho_Q}ight)$$

Now, let's compute each component of this ratio step-by-step.

First, the radius ratio:

$$\frac{r_P}{r_Q} = \frac{0.5}{0.25} = 2 \implies \left(\frac{r_P}{r_Q}ight)^2 = 4$$

Second, the viscosity ratio:

$$\frac{\eta_Q}{\eta_P} = \frac{2}{3}$$

Third, the density difference ratio:

$$\frac{d - ho_P}{d - ho_Q} = \frac{8 - 0.8}{8 - 1.6} = \frac{7.2}{6.4} = \frac{9}{8}$$

Now, let's substitute these three components back into our ratio equation:

$$\frac{v_P}{v_Q} = 4 	imes \frac{2}{3} 	imes \frac{9}{8}$$

Multiplying these fractions together:

$$\frac{v_P}{v_Q} = \frac{8}{3} 	imes \frac{9}{8} = 3$$

### Conclusion: The Elegance of Cancellation

The ratio of the terminal velocities of $P$ and $Q$ is exactly **3**.

Notice how beautifully the complex units and constants cancelled out, leaving us with a clean integer.

This problem teaches us the importance of looking at ratios and scaling laws rather than getting bogged down in tedious absolute calculations.

Enter Numerical Value:

Visualized Solution

Visualizing the Falling Spheres

The Terminal Velocity Formula

Setting up the Ratio

Parameters for Sphere $P$

Parameters for Sphere $Q$

Calculating the Radius Ratio

Calculating the Viscosity Ratio

Calculating the Density Difference Ratio

Combining All Ratios

Calculating the Final Ratio

Exploring Variations

The Sigma Insight: Viscosity and Stokes' Law

Introduction

Deconstructing the Physics

The Master Equation

Parameter Extraction

Step-by-Step Calculation

Conclusion

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