An oil drop falls through air with a terminal velocity of `(5xx10^(-4))/(sec)` viscosity of air is `18xx10^(-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. `2.5 xx 10^(-5) m`
B. `2 xx 10^(-4)`
C. `3 xx 10^(-6)m`
D. `4 xx 10^(-4)m`

Answer

To solve the problem of an oil drop falling through air with a given terminal velocity, we can use the formula for terminal velocity in a viscous medium. The formula is given by:

\[ V_t = \frac{2r^2 (ρ - ρ_{air}) g}{9η} \]

Where:
- \( V_t \) is the terminal velocity,
- \( r \) is the radius of the drop,
- \( ρ \) is the density of the oil,
- \( ρ_{air} \) is the density of air (which we neglect),
- \( g \) is the acceleration due to gravity (approximately \( 10 \, m/s^2 \)),
- \( η \) is the viscosity of air.

Given:
- Terminal velocity \( V_t = 5 \times 10^{-4} \, m/s \)
- Viscosity of air \( η = 18 \times 10^{-5} \, N \cdot s/m^2 \)
- Density of oil \( ρ = 900 \, kg/m^3 \)
- Density of air is neglected.

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

We can rearrange the terminal velocity formula to solve for \( r^2 \):

\[ r^2 = \frac{9η V_t}{2(ρ - ρ_{air}) g} \]

Since we neglect the density of air, we can simplify this to:

\[ r^2 = \frac{9η V_t}{2ρ g} \]

### Step 2: Substituting the known values

Now we substitute the known values into the equation:

- \( η = 18 \times 10^{-5} \, N \cdot s/m^2 \)
- \( V_t = 5 \times 10^{-4} \, m/s \)
- \( ρ = 900 \, kg/m^3 \)
- \( g = 10 \, m/s^2 \)

\[ r^2 = \frac{9 \times (18 \times 10^{-5}) \times (5 \times 10^{-4})}{2 \times 900 \times 10} \]

### Step 3: Calculating the numerator and denominator

Calculating the numerator:

\[ 9 \times (18 \times 10^{-5}) \times (5 \times 10^{-4}) = 9 \times 18 \times 5 \times 10^{-9} = 810 \times 10^{-9} = 8.1 \times 10^{-7} \]

Calculating the denominator:

\[ 2 \times 900 \times 10 = 18000 \]

### Step 4: Putting it all together

Now we can calculate \( r^2 \):

\[ r^2 = \frac{8.1 \times 10^{-7}}{18000} \]

Calculating:

\[ r^2 = 4.5 \times 10^{-11} \, m^2 \]

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

Now we take the square root to find \( r \):

\[ r = \sqrt{4.5 \times 10^{-11}} = 6.7 \times 10^{-6} \, m \]

### Conclusion

The radius of the oil drop is approximately \( 6.7 \times 10^{-6} \, m \).