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 find the viscous force on a raindrop falling through air, we can use Stokes' law, which states that the viscous force \( F \) acting on a sphere moving through a viscous fluid is given by the formula:

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

where:
- \( F \) is the viscous force,
- \( \eta \) is the viscosity of the fluid,
- \( r \) is the radius of the sphere (raindrop),
- \( v \) is the velocity of the sphere.

### Step-by-Step Solution:

1. **Convert the radius of the raindrop to centimeters:**
- Given radius \( r = 0.3 \, \text{mm} = 0.03 \, \text{cm} \).

2. **Convert the terminal velocity to centimeters per second:**
- Given terminal velocity \( v = 1 \, \text{m/s} = 100 \, \text{cm/s} \).

3. **Identify the viscosity of air:**
- Given viscosity \( \eta = 18 \times 10^{-5} \, \text{poise} \).

4. **Substitute the values into Stokes' law:**
- Using the formula \( F = 6 \pi \eta r v \):
\[ F = 6 \times 3.14 \times (18 \times 10^{-5}) \times (0.03) \times (100) \]

5. **Calculate the viscous force:**
- First, calculate \( 6 \times 3.14 \):
\[ 6 \times 3.14 = 18.84 \]
- Now calculate \( 18.84 \times (18 \times 10^{-5}) \):
\[ 18.84 \times 18 = 338.88 \quad \text{and} \quad 338.88 \times 10^{-5} = 3.3888 \times 10^{-3} \]
- Now multiply by \( 0.03 \):
\[ 3.3888 \times 10^{-3} \times 0.03 = 1.01664 \times 10^{-4} \]
- Finally, multiply by \( 100 \):
\[ 1.01664 \times 10^{-4} \times 100 = 1.01664 \times 10^{-2} \, \text{dyne} \]

6. **Final result:**
- The viscous force on the raindrop is approximately:
\[ F \approx 1.018 \times 10^{-2} \, \text{dyne} \]

### Conclusion:
The viscous force on the raindrop is \( 1.018 \times 10^{-2} \, \text{dyne} \).