A drop of water of radius 0.0015 mm is falling in air .If the coefficient 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. `1.0 xx 10^(-4) m s^(-1)`
B. `2.0 xx 10^(-4) m s^(-1)`
C. `2.5 xx 10^(-4) m s^(-1)`
D. `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 derived from Stokes' law:

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

where:
- \( v_t \) is the terminal velocity,
- \( r \) is the radius of the drop,
- \( \rho \) is the density of the fluid (water in this case),
- \( g \) is the acceleration due to gravity,
- \( \eta \) is the coefficient of viscosity of the fluid (air in this case).

### Step 1: Convert the radius from mm to meters
The radius given is \( 0.0015 \) mm. To convert this 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 the other variables
- Density of water, \( \rho = 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Coefficient of viscosity of air, \( \eta = 2.0 \times 10^{-5} \, \text{kg/m/s} \)

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

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

### Step 5: Substitute and calculate
Now substituting this back into the equation:
\[ v_t = \frac{2}{9} \cdot \frac{(2.25 \times 10^{-12}) \cdot (10^3) \cdot (10)}{2.0 \times 10^{-5}} \]
\[ = \frac{2}{9} \cdot \frac{(2.25 \times 10^{-12}) \cdot (10^4)}{2.0 \times 10^{-5}} \]
\[ = \frac{2}{9} \cdot \frac{2.25 \times 10^{-8}}{2.0 \times 10^{-5}} \]
\[ = \frac{2}{9} \cdot 1.125 \times 10^{-3} \]
\[ = \frac{2.25 \times 10^{-3}}{9} \]
\[ = 2.5 \times 10^{-4} \, \text{m/s} \]

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