A drop of water of radius 0.0015 mm is falling in air .If the cofficient of viscosity of air is `2.0 xx 10^(-5) kg m^(-1)s^(-1)` ,the terminal velocity of the drop will be
(The density of water = `10^(3) kg m^(-3)` and g = `10 m s^(-2)` )

Options

A. 1. `1.0 xx 10^(-4) m s^(-1)`
B. 2. `2.0 xx 10^(-4) m s^(-1)`
C. 3. `2.5 xx 10^(-4) m s^(-1)`
D. 4. `5.0 xx 10^(-4) m s^(-1)`

Explanation

To find the terminal velocity of a water drop falling in air, we can use the formula for terminal velocity \( V_T \):

\[ V_T = \frac{2}{9} \cdot r^2 \cdot \frac{( \rho - \sigma ) \cdot g}{\eta} \]

Where:
- \( r \) = radius of the drop
- \( \rho \) = density of the drop (water)
- \( \sigma \) = density of the fluid (air)
- \( g \) = acceleration due to gravity
- \( \eta \) = coefficient of viscosity of the fluid (air)

### Step 1: Convert the radius from millimeters to meters
Given:
- Radius \( r = 0.0015 \, \text{mm} \)

Convert to meters:
\[ r = 0.0015 \, \text{mm} = 0.0015 \times 10^{-3} \, \text{m} = 1.5 \times 10^{-6} \, \text{m} \]

### Step 2: Identify the values for density and viscosity
Given:
- Density of water \( \rho = 10^3 \, \text{kg/m}^3 \)
- Density of air \( \sigma \approx 0 \, \text{kg/m}^3 \) (for simplicity, we can consider it negligible)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)
- Coefficient of viscosity of air \( \eta = 2.0 \times 10^{-5} \, \text{kg/(m \cdot s)} \)

### Step 3: Substitute the values into the terminal velocity formula
Substituting the values into the formula:
\[ V_T = \frac{2}{9} \cdot (1.5 \times 10^{-6})^2 \cdot \frac{(10^3 - 0) \cdot 10}{2.0 \times 10^{-5}} \]

### Step 4: Calculate \( (1.5 \times 10^{-6})^2 \)
\[ (1.5 \times 10^{-6})^2 = 2.25 \times 10^{-12} \, \text{m}^2 \]

### Step 5: Substitute and simplify
Now substituting back:
\[ V_T = \frac{2}{9} \cdot 2.25 \times 10^{-12} \cdot \frac{(10^3) \cdot 10}{2.0 \times 10^{-5}} \]
\[ = \frac{2}{9} \cdot 2.25 \times 10^{-12} \cdot \frac{10^4}{2.0 \times 10^{-5}} \]
\[ = \frac{2}{9} \cdot 2.25 \times 10^{-12} \cdot 5 \times 10^{8} \]

### Step 6: Calculate the final value
\[ = \frac{2 \cdot 2.25 \cdot 5}{9} \times 10^{-12 + 8} \]
\[ = \frac{22.5}{9} \times 10^{-4} \]
\[ = 2.5 \times 10^{-4} \, \text{m/s} \]

### Final Answer
The terminal velocity of the drop is:
\[ V_T = 2.5 \times 10^{-4} \, \text{m/s} \]