A raindrop with radius R = 0.2 mm falls from a cloud at a height h = 2000 m above the ground. Assume

Q: Why does the raindrop attain a constant velocity instead of accelerating continuously?

As the raindrop accelerates, the viscous drag force from the air increases proportionally with its velocity. Eventually, this upward drag force perfectly balances the downward gravitational pull, resulting in zero net force and a constant 'terminal' velocity.

Q: What if we didn't neglect buoyancy?

If we included buoyancy, the net downward force would be $W - F_B$. The terminal velocity formula would then use the effective density $(\rho_w - \rho_a)$ instead of just $\rho_w$. The result would be very slightly smaller.

Q: Why is the height ($h = 2000 \text{ m}$) given if we don't use it?

In many physics problems, extra information is provided to test your conceptual clarity. The height is given to set the physical context, but terminal velocity depends only on the properties of the drop and the fluid, not the height it falls from (assuming it falls long enough to reach it).

Q: What is Stokes' Law and when is it applicable?

Stokes' Law ($F_v = 6\pi\eta R v$) calculates the viscous drag on a small spherical object moving through a viscous fluid. It is strictly applicable only for laminar flow (low Reynolds number), which is a reasonable assumption for tiny raindrops.

Question

A raindrop with radius $R = 0.2 	ext{ mm}$ falls from a cloud at a height $h = 2000 	ext{ m}$ above the ground. Assume that, the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is (Take, density of water, $ho_w = 1000 	ext{ kg m}^{-3}$ and density of air, $ho_a = 1.2 	ext{ kg m}^{-3}$, $g = 10 	ext{ m/s}^2$, coefficient of viscosity of air, $\eta = 1.8 	imes 10^{-5} 	ext{ N-s m}^{-2}$)

Options:

(A) $250.6 	ext{ ms}^{-1}$

(B) $43.56 	ext{ ms}^{-1}$

(C) $4.94 	ext{ ms}^{-1}$

(D) $14.4 	ext{ ms}^{-1}$

Sigma Percentile · Accepted Answer

### The Setup: A Falling Raindrop

Imagine you are standing on the ground, looking up at a cloud $2000 	ext{ m}$ above you. A tiny raindrop, perfectly spherical with a radius of $0.2 	ext{ mm}$, begins its descent.

At first glance, you might think it will keep accelerating due to gravity until it hits the ground with a massive, destructive speed. But nature has a built-in speed limit for falling objects, thanks to the fluid they fall through—in this case, the air.

### The Battle of Forces

As the raindrop falls, it doesn't just keep accelerating forever. Why? Because it's moving through a fluid! This means it experiences a viscous drag force.

There are two primary forces at play here. First, the **weight** of the drop, $W = mg$, pulling it relentlessly downwards. Second, the **viscous drag force**, $F_v$, pushing upwards, trying to slow it down.

*Note: The problem explicitly tells us to neglect buoyancy. We'll discuss why this is a safe assumption at the end!*

### Reaching Terminal Velocity

As the drop speeds up, the viscous drag force increases. Eventually, a magical moment occurs: the upward drag force perfectly balances the downward weight.

When this happens, the net force on the raindrop becomes exactly zero. According to Newton's First Law, an object with zero net force stops accelerating and moves with a constant velocity. We call this the **terminal velocity**, $v_T$.

### The Master Equation

Let's bring in the math. According to **Stokes' Law**, the viscous drag force on a small spherical object moving through a fluid is given by:

$$F_v = 6\pi\eta R v_T$$

The weight of the drop can be expressed in terms of its volume and density:

$$W = mg = \left(\frac{4}{3}\pi R^3ight) ho_w g$$

Equating the two forces ($F_v = W$), we get:

$$6\pi\eta R v_T = \left(\frac{4}{3}\pi R^3ight) ho_w g$$

By rearranging the terms and canceling out $\pi$ and one $R$, we arrive at the beautiful, standard formula for terminal velocity:

$$v_T = \frac{2R^2 ho_w g}{9\eta}$$

### The Final Calculation

Now, we just need to plug in the numbers. But wait, watch out for the units! We must convert the radius from millimeters to meters: $R = 0.2 	ext{ mm} = 2 	imes 10^{-4} 	ext{ m}$.

Substituting the given values:

$$v_T = \frac{2 	imes (2 	imes 10^{-4})^2 	imes 1000 	imes 10}{9 	imes 1.8 	imes 10^{-5}}$$

$$v_T = \frac{2 	imes 4 	imes 10^{-8} 	imes 10^4}{16.2 	imes 10^{-5}}$$

$$v_T = \frac{8 	imes 10^{-4}}{16.2 	imes 10^{-5}} = \frac{80}{16.2} \approx 4.94 	ext{ m/s}$$

### A Quick Note on Buoyancy

Why were we allowed to neglect buoyancy? The buoyant force depends on the density of the displaced fluid (air, $ho_a = 1.2 	ext{ kg/m}^3$). Since the density of water ($ho_w = 1000 	ext{ kg/m}^3$) is nearly a thousand times greater, the buoyant force is incredibly tiny compared to the weight. Neglecting it makes our calculation simpler without sacrificing much accuracy!

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Forces on the Raindrop$

$Terminal Velocity Condition$

$Expanding the Terms$

$Formula for v_{T}$

$Substituting Values$

$Final Calculation$

$Why Neglect Buoyancy?$

The Sigma Insight: Viscosity and Stokes' Law

The Setup

The Battle of Forces

Reaching Terminal Velocity

The Master Equation

The Final Calculation

A Quick Note on Buoyancy

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Forces on the Raindrop

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Terminal Velocity Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expanding the Terms

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Formula for vT​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Why Neglect Buoyancy?