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. the radius of is `4.18xx10^(6)m`
B. The radius of drop is `2.14xx10^(-6)m`
C. The terminal velocity of a drop of half of this radius is `1.25xx10^(-4)(m)/(sec)`.
D. The terminal velocity of a drop of half of this radius is `2.5xx10^(-4)(m)/(sec)`.

Explanation

To solve the problem, we will use the formula for terminal velocity of a sphere falling through a viscous fluid, which is given by:

\[ V_t = \frac{2r^2(\rho_{oil} - \rho_{air})g}{9\eta} \]

Where:
- \( V_t \) is the terminal velocity,
- \( r \) is the radius of the drop,
- \( \rho_{oil} \) is the density of the oil,
- \( \rho_{air} \) is the density of the air (which we will neglect),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, m/s^2 \)),
- \( \eta \) is the viscosity of the fluid (air in this case).

Given:
- \( V_t = 5 \times 10^{-4} \, m/s \)
- \( \eta = 1.8 \times 10^{-5} \, N \cdot s/m^2 \)
- \( \rho_{oil} = 900 \, kg/m^3 \)
- \( \rho_{air} \) is neglected.

### Step 1: Rearranging the formula to find \( r^2 \)

Since we are neglecting the density of air, we can simplify the equation:

\[ V_t = \frac{2r^2 \cdot \rho_{oil} \cdot g}{9\eta} \]

Rearranging for \( r^2 \):

\[ r^2 = \frac{9\eta V_t}{2\rho_{oil}g} \]

### Step 2: Substituting the known values

Now we substitute the known values into the equation:

- \( V_t = 5 \times 10^{-4} \, m/s \)
- \( \eta = 1.8 \times 10^{-5} \, N \cdot s/m^2 \)
- \( \rho_{oil} = 900 \, kg/m^3 \)
- \( g \approx 9.81 \, m/s^2 \)

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

### Step 3: Calculating the numerator and denominator

Calculating the numerator:

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

Calculating the denominator:

\[ 2 \cdot 900 \cdot 9.81 = 17898 \]

### Step 4: Final calculation for \( r^2 \)

Now substituting back into the equation:

\[ r^2 = \frac{8.1 \times 10^{-8}}{17898} \]

Calculating:

\[ r^2 \approx 4.53 \times 10^{-12} \, m^2 \]

### Step 5: Finding \( r \)

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

\[ r = \sqrt{4.53 \times 10^{-12}} \approx 2.13 \times 10^{-6} \, m \]

### Final Answer

The radius of the oil drop is approximately \( 2.13 \times 10^{-6} \, m \).