Compute the terminal velocity of a rain drop of radius 0.3 mm. Take coefficient of viscosity of air `1.83 X 10^(-5)` poise and density of air = `1.3 kg m^(-3)` Density of water - `10^(3) kg m^(-3) and take g= 9.8 ms^(-2)`

Explanation

To compute the terminal velocity of a raindrop with a radius of 0.3 mm, we will use Stokes' law, which relates the viscous force acting on a sphere moving through a fluid to its velocity. Here are the steps to solve the problem:

### Step 1: Understand the Forces Acting on the Raindrop
When a raindrop falls, it experiences three forces:
1. The gravitational force (weight) acting downward, \( W = mg \).
2. The viscous force acting upward, given by Stokes' law: \( F = 6 \pi \eta r v \).
3. The buoyant force, which can be neglected in this case since the density of air is much less than that of water.

### Step 2: Write the Expression for Weight
The weight of the raindrop can be expressed as:
\[ W = \rho_{water} \cdot V \cdot g \]
Where:
- \( \rho_{water} = 10^3 \, \text{kg/m}^3 \) (density of water)
- \( V = \frac{4}{3} \pi r^3 \) (volume of the sphere)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the volume:
\[ W = \rho_{water} \cdot \left(\frac{4}{3} \pi r^3\right) \cdot g \]

### Step 3: Write the Expression for Viscous Force
According to Stokes' law, the viscous force is given by:
\[ F = 6 \pi \eta r v \]
Where:
- \( \eta = 1.83 \times 10^{-5} \, \text{poise} = 1.83 \times 10^{-5} \times 0.1 \, \text{N s/m}^2 = 1.83 \times 10^{-6} \, \text{N s/m}^2 \) (converting poise to SI units)
- \( r = 0.3 \, \text{mm} = 0.3 \times 10^{-3} \, \text{m} \)

### Step 4: Set Up the Equation for Terminal Velocity
At terminal velocity, the viscous force equals the weight:
\[ 6 \pi \eta r v = \rho_{water} \cdot \left(\frac{4}{3} \pi r^3\right) \cdot g \]

### Step 5: Solve for Terminal Velocity \( v \)
Cancel out \( \pi \) from both sides:
\[ 6 \eta r v = \rho_{water} \cdot \left(\frac{4}{3} r^3\right) g \]
Rearranging gives:
\[ v = \frac{\rho_{water} \cdot \left(\frac{4}{3} r^3\right) g}{6 \eta r} \]
Simplifying further:
\[ v = \frac{2 \rho_{water} r^2 g}{9 \eta} \]

### Step 6: Substitute the Values
Substituting the known values:
- \( \rho_{water} = 10^3 \, \text{kg/m}^3 \)
- \( r = 0.3 \times 10^{-3} \, \text{m} \)
- \( g = 9.8 \, \text{m/s}^2 \)
- \( \eta = 1.83 \times 10^{-6} \, \text{N s/m}^2 \)

Calculating \( v \):
\[ v = \frac{2 \cdot 10^3 \cdot (0.3 \times 10^{-3})^2 \cdot 9.8}{9 \cdot 1.83 \times 10^{-6}} \]

Calculating the numerator:
\[ = 2 \cdot 10^3 \cdot 0.09 \times 10^{-6} \cdot 9.8 = 2 \cdot 9.8 \cdot 0.09 \times 10^{-3} = 1.764 \times 10^{-3} \]

Calculating the denominator:
\[ = 9 \cdot 1.83 \times 10^{-6} = 16.47 \times 10^{-6} \]

Finally:
\[ v = \frac{1.764 \times 10^{-3}}{16.47 \times 10^{-6}} \approx 107.0 \, \text{m/s} \]

### Step 7: Convert to cm/s
Converting to cm/s:
\[ v \approx 1.07 \, \text{cm/s} \]

### Final Answer
The terminal velocity of the raindrop is approximately **1.07 cm/s**.