A rain drop of radius 0.3 mm falls through air with a terminal viscosity of `1 m s^(-1)`. The viscosity of air is `18 xx 10^(-5)` poise. The viscous force on the rain drop is

Options

A. `1.018 xx 10^(-2)` dyne
B. `2.018 xx 10^(-2)` dyne
C. `3.018 xx 10^(-2)` dyne
D. `4.018 xx 10^(-2)` dyne

Explanation

To calculate the viscous force acting on a raindrop falling through air, we can use Stokes' law, which is given by the formula:

\[ F = 6 \eta \pi r v \]

where:
- \( F \) is the viscous force,
- \( \eta \) is the viscosity of the fluid (air in this case),
- \( r \) is the radius of the raindrop,
- \( v \) is the terminal velocity of the raindrop.

### Step 1: Convert the radius to meters
The radius of the raindrop is given as \( 0.3 \, \text{mm} \). We need to convert this to meters.

\[ r = 0.3 \, \text{mm} = 0.3 \times 10^{-3} \, \text{m} = 3.0 \times 10^{-4} \, \text{m} \]

### Step 2: Identify the values
- Viscosity of air, \( \eta = 18 \times 10^{-5} \, \text{poise} \)
- Terminal velocity, \( v = 1 \, \text{m/s} \)

### Step 3: Convert viscosity to SI units
1 poise = 0.1 Pa·s, so:

\[ \eta = 18 \times 10^{-5} \, \text{poise} = 18 \times 10^{-5} \times 0.1 \, \text{Pa·s} = 1.8 \times 10^{-6} \, \text{Pa·s} \]

### Step 4: Substitute the values into Stokes' law
Now we can substitute the values into the formula:

\[ F = 6 \eta \pi r v \]

Substituting the known values:

\[ F = 6 \times (1.8 \times 10^{-6}) \times \pi \times (3.0 \times 10^{-4}) \times (1) \]

### Step 5: Calculate the force
Calculating the numerical value:

\[ F = 6 \times 1.8 \times 10^{-6} \times 3.14 \times 3.0 \times 10^{-4} \]

Calculating step by step:

1. \( 6 \times 1.8 = 10.8 \)
2. \( 10.8 \times 3.14 = 33.912 \)
3. \( 33.912 \times 3.0 = 101.736 \)
4. Now multiply by \( 10^{-6} \times 10^{-4} = 10^{-10} \):

\[ F = 101.736 \times 10^{-10} \, \text{N} = 1.01736 \times 10^{-2} \, \text{N} \]

### Step 6: Convert to dyne
Since \( 1 \, \text{N} = 10^5 \, \text{dyne} \):

\[ F = 1.01736 \times 10^{-2} \, \text{N} \times 10^5 \, \text{dyne/N} = 1.01736 \times 10^{3} \, \text{dyne} \approx 1.018 \times 10^{2} \, \text{dyne} \]

### Final Answer
The viscous force on the raindrop is approximately:

\[ F \approx 1.018 \times 10^{-2} \, \text{dyne} \]