An oil drop falls through air with a terminal velocity of `(5xx10^(-4))/(sec)` viscosity of air is `1.8xx10^(-5)(N-s)/(m^(2))` and density of oil is 900 kg `m^(3)` neglect density of air as compared to that of oil.

Options

A. radius of the drop is `6.20xx10^(-2) m`
B. radius of the drop is `2.14xx10^(-6) m`
C. terminal velocity of the drop at half of this radius is `1.25xx10^(-4) m//s`
D. terminal velocity of the drop at half of this radius is `2.5xx10^(-4) m//s`

Answer

To solve the problem, we need to find the radius of the oil drop falling through air at terminal velocity. We will use the formula for terminal velocity of a sphere falling through a viscous medium:

\[ V_t = \frac{2}{9} \cdot \frac{R^2 \cdot (P - \rho) \cdot g}{\eta} \]

Where:
- \( V_t \) = terminal velocity
- \( R \) = radius of the drop
- \( P \) = density of the oil
- \( \rho \) = density of the air (neglected in this case)
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, m/s^2 \))
- \( \eta \) = viscosity of the air

### Given Data:
- \( V_t = 5 \times 10^{-4} \, m/s \)
- \( P = 900 \, kg/m^3 \)
- \( \eta = 1.8 \times 10^{-5} \, N \cdot s/m^2 \)

### Step 1: Rearranging the formula to solve for \( R^2 \)

We can rearrange the formula for terminal velocity to isolate \( R^2 \):

\[ R^2 = \frac{9 \cdot V_t \cdot \eta}{2 \cdot (P - \rho) \cdot g} \]

Since we are neglecting the density of air (\( \rho \)), we can simplify it to:

\[ R^2 = \frac{9 \cdot V_t \cdot \eta}{2 \cdot P \cdot g} \]

### Step 2: Plugging in the values

Now, we substitute the known values into the equation:

\[ R^2 = \frac{9 \cdot (5 \times 10^{-4}) \cdot (1.8 \times 10^{-5})}{2 \cdot 900 \cdot 9.81} \]

### Step 3: Calculating the numerator

Calculating the numerator:

\[ 9 \cdot (5 \times 10^{-4}) \cdot (1.8 \times 10^{-5}) = 8.1 \times 10^{-8} \]

### Step 4: Calculating the denominator

Calculating the denominator:

\[ 2 \cdot 900 \cdot 9.81 = 17658 \]

### Step 5: Final calculation for \( R^2 \)

Now, we can calculate \( R^2 \):

\[ R^2 = \frac{8.1 \times 10^{-8}}{17658} \approx 4.59 \times 10^{-12} \]

### Step 6: Finding \( R \)

Taking the square root to find \( R \):

\[ R = \sqrt{4.59 \times 10^{-12}} \approx 2.14 \times 10^{-6} \, m \]

### Conclusion

The radius of the oil drop is approximately \( 2.14 \times 10^{-6} \, m \) or \( 2.14 \, \mu m \).