An oil drop of diameter `4 xx 10^(-6)` m falls through air. iF the densities of the oil and air are 900 kg/m^(3)` and 1.293 kg/m^(3)` respectively and the coefficient of viscosity of air is `2.0 xx 10^(-5)` Ns//m^(2)`, then the terminal velocity of oil drop will be

Options

A. `0.2 xx 10^(-4)` m/s
B. `2xx 10^(-4)` m/s
C. `4 xx 10^(-4)` m/s
D. `8 xx 10^(-4)` m/s

Answer

To find the terminal velocity of the oil drop falling through air, we can use Stokes' law, which is applicable for small spherical objects moving through a viscous medium. The terminal velocity \( v_t \) can be calculated using the formula:

\[ v_t = \frac{2}{9} \frac{r^2 (ρ_o - ρ_a) g}{η} \]

Where:
- \( r \) = radius of the drop
- \( ρ_o \) = density of the oil
- \( ρ_a \) = density of the air
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( η \) = coefficient of viscosity of air

### Step 1: Calculate the radius of the oil drop
Given the diameter \( d = 4 \times 10^{-6} \, \text{m} \), we can find the radius \( r \):

\[ r = \frac{d}{2} = \frac{4 \times 10^{-6}}{2} = 2 \times 10^{-6} \, \text{m} \]

### Step 2: Identify the densities and viscosity
From the problem statement:
- Density of oil \( ρ_o = 900 \, \text{kg/m}^3 \)
- Density of air \( ρ_a = 1.293 \, \text{kg/m}^3 \)
- Coefficient of viscosity of air \( η = 2.0 \times 10^{-5} \, \text{Ns/m}^2 \)

### Step 3: Plug the values into the terminal velocity formula
Now we can substitute the values into the terminal velocity formula:

\[ v_t = \frac{2}{9} \frac{(2 \times 10^{-6})^2 (900 - 1.293) \cdot 9.81}{2.0 \times 10^{-5}} \]

### Step 4: Calculate the values step by step
1. Calculate \( (2 \times 10^{-6})^2 \):
\[ (2 \times 10^{-6})^2 = 4 \times 10^{-12} \, \text{m}^2 \]

2. Calculate \( 900 - 1.293 \):
\[ 900 - 1.293 = 898.707 \, \text{kg/m}^3 \]

3. Calculate the numerator:
\[ 4 \times 10^{-12} \cdot 898.707 \cdot 9.81 = 4 \times 10^{-12} \cdot 8817.98287 \approx 3.527 \times 10^{-8} \]

4. Now substitute back into the formula:
\[ v_t = \frac{2}{9} \cdot \frac{3.527 \times 10^{-8}}{2.0 \times 10^{-5}} \]

5. Calculate \( \frac{3.527 \times 10^{-8}}{2.0 \times 10^{-5}} \):
\[ \frac{3.527 \times 10^{-8}}{2.0 \times 10^{-5}} = 1.7635 \times 10^{-3} \]

6. Finally, calculate \( v_t \):
\[ v_t = \frac{2}{9} \cdot 1.7635 \times 10^{-3} \approx 3.927 \times 10^{-4} \, \text{m/s} \]

### Final Answer
The terminal velocity of the oil drop is approximately \( 3.93 \times 10^{-4} \, \text{m/s} \).