Neglecting the density of air, the terminal velocity obtained by a raindrop of radius `0.3 mm` falling through the air of viscosity `1.8 xx10^(-5) N//m^(2)` will be

Options

A. `10.9m//s`
B. `8.3m//s`
C. `9.2m//s`
D. `7.6m//s`

Explanation

To find the terminal velocity of a raindrop falling through the air, we can use the formula for terminal velocity, which is given by:

\[ v_t = \frac{2}{9} \frac{r^2 (\rho_{rain} - \rho_{air}) g}{\eta} \]

Where:
- \( v_t \) = terminal velocity
- \( r \) = radius of the raindrop
- \( \rho_{rain} \) = density of the raindrop (water)
- \( \rho_{air} \) = density of air (which we will neglect)
- \( g \) = acceleration due to gravity
- \( \eta \) = coefficient of viscosity of air

### Step 1: Identify the values
- Radius of the raindrop, \( r = 0.3 \, \text{mm} = 0.3 \times 10^{-3} \, \text{m} \)
- Density of water, \( \rho_{rain} = 1000 \, \text{kg/m}^3 \)
- Density of air, \( \rho_{air} = 0 \, \text{(neglected)} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)
- Coefficient of viscosity, \( \eta = 1.8 \times 10^{-5} \, \text{N/m}^2 \)

### Step 2: Substitute the values into the formula
Since we are neglecting the density of air, the equation simplifies to:

\[ v_t = \frac{2}{9} \frac{r^2 \cdot \rho_{rain} \cdot g}{\eta} \]

Substituting the values:

\[ v_t = \frac{2}{9} \cdot \frac{(0.3 \times 10^{-3})^2 \cdot 1000 \cdot 9.8}{1.8 \times 10^{-5}} \]

### Step 3: Calculate \( r^2 \)
Calculating \( r^2 \):

\[ (0.3 \times 10^{-3})^2 = 0.09 \times 10^{-6} = 9 \times 10^{-8} \, \text{m}^2 \]

### Step 4: Substitute \( r^2 \) back into the equation
Now substituting \( r^2 \) back:

\[ v_t = \frac{2}{9} \cdot \frac{9 \times 10^{-8} \cdot 1000 \cdot 9.8}{1.8 \times 10^{-5}} \]

### Step 5: Calculate the numerator
Calculating the numerator:

\[ 9 \times 10^{-8} \cdot 1000 \cdot 9.8 = 9 \times 9.8 \times 10^{-5} = 88.2 \times 10^{-5} \]

### Step 6: Calculate the entire expression
Now substituting back into the equation:

\[ v_t = \frac{2}{9} \cdot \frac{88.2 \times 10^{-5}}{1.8 \times 10^{-5}} \]

Calculating the fraction:

\[ \frac{88.2}{1.8} = 49 \]

So,

\[ v_t = \frac{2}{9} \cdot 49 \]

Calculating \( \frac{2}{9} \cdot 49 \):

\[ v_t = \frac{98}{9} \approx 10.89 \, \text{m/s} \]

### Final Answer
Thus, the terminal velocity of the raindrop is approximately:

\[ \boxed{10.9 \, \text{m/s}} \]