A water drop of radius `1mum` falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is `1.8xx10^(-5)Nsm^(-2)` and its density is negligible as compared to that of water `10^(6)gm^(-3)`. Terminal velocity of the water drop is :
(Take acceleration due to gravity `=10ms^(-2)`)

Options

A. `145.4xx10^(-6)ms^(-1)`
B. `118.0xx10^(-6)ms^(-1)`
C. `132.6xx10^(-6)ms^(-1)`
D. `123.4xx10^(-6)ms^(-1)`

Explanation

To find the terminal velocity of a water drop falling through air, we can use the principles of fluid dynamics, specifically Stokes' law. The terminal velocity can be derived from the balance of forces acting on the drop.

### Step-by-Step Solution:

1. **Identify the forces acting on the water drop:**
- The weight of the water drop acting downwards: \( W = mg \)
- The viscous drag force acting upwards: \( F_d = 6 \pi \eta r v_t \)

2. **Write the expression for the weight of the water drop:**
- The mass \( m \) of the water drop can be expressed as:
\[ m = \rho V = \rho \left( \frac{4}{3} \pi r^3 \right) \]
- Therefore, the weight becomes:
\[ W = mg = \rho \left( \frac{4}{3} \pi r^3 \right) g \]

3. **Set the forces equal at terminal velocity:**
- At terminal velocity, the weight of the drop equals the viscous drag force:
\[ \rho \left( \frac{4}{3} \pi r^3 \right) g = 6 \pi \eta r v_t \]

4. **Rearranging the equation to solve for terminal velocity \( v_t \):**
- Cancel \( \pi \) from both sides:
\[ \rho \left( \frac{4}{3} r^3 g \right) = 6 \eta r v_t \]
- Dividing both sides by \( 6 \eta r \):
\[ v_t = \frac{2}{9} \frac{\rho g r^2}{\eta} \]

5. **Substituting the known values:**
- Given:
- \( r = 1 \times 10^{-6} \, \text{m} \)
- \( \rho = 10^6 \, \text{g/m}^3 = 10^3 \, \text{kg/m}^3 \)
- \( g = 10 \, \text{m/s}^2 \)
- \( \eta = 1.8 \times 10^{-5} \, \text{N s/m}^2 \)

- Plugging in the values:
\[ v_t = \frac{2}{9} \cdot \frac{(10^3 \, \text{kg/m}^3)(10 \, \text{m/s}^2)(1 \times 10^{-6} \, \text{m})^2}{1.8 \times 10^{-5} \, \text{N s/m}^2} \]

6. **Calculating the terminal velocity:**
- Calculate \( r^2 \):
\[ r^2 = (1 \times 10^{-6})^2 = 1 \times 10^{-12} \, \text{m}^2 \]
- Substitute \( r^2 \) into the equation:
\[ v_t = \frac{2}{9} \cdot \frac{(10^3)(10)(1 \times 10^{-12})}{1.8 \times 10^{-5}} \]
- Simplifying:
\[ v_t = \frac{2}{9} \cdot \frac{10^4 \times 10^{-12}}{1.8 \times 10^{-5}} = \frac{2 \times 10^{-8}}{1.8 \times 9} = \frac{2 \times 10^{-8}}{16.2} \approx 1.23 \times 10^{-9} \, \text{m/s} \]

7. **Final result:**
- The terminal velocity of the water drop is:
\[ v_t \approx 1.23 \times 10^{-6} \, \text{m/s} \]