In a Millikan's oil drop experiment what is the terminal speed of a drop of radius `2.0 xx 10^(-5) m`, and density `1.2 xx 10^(3) kg m^(-3) ?`. Take the viscosity of air at the temperature of the experiment to be `1.8 xx 10^(-5) Ns m^(-2) ` How much is the viscous force on the drop at that speed ? Neglect buoyancy of the drop in air.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write down the given data
- Radius of the drop, \( r = 2.0 \times 10^{-5} \, \text{m} \)
- Density of the drop, \( \rho_d = 1.2 \times 10^{3} \, \text{kg/m}^3 \)
- Viscosity of air, \( \eta = 1.8 \times 10^{-5} \, \text{Ns/m}^2 \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Calculate the terminal velocity using the formula
The formula for terminal velocity \( v_t \) of a spherical drop in a viscous medium is given by:

\[ v_t = \frac{2}{9} \cdot \frac{r^2 \cdot g \cdot (\rho_d - \rho_l)}{\eta} \]

Since we are neglecting the buoyancy, we can assume the density of the liquid \( \rho_l = 0 \).

Substituting the values:

\[ v_t = \frac{2}{9} \cdot \frac{(2.0 \times 10^{-5})^2 \cdot 9.8 \cdot (1.2 \times 10^{3} - 0)}{1.8 \times 10^{-5}} \]

Calculating \( (2.0 \times 10^{-5})^2 \):

\[ (2.0 \times 10^{-5})^2 = 4.0 \times 10^{-10} \, \text{m}^2 \]

Now substituting back into the equation:

\[ v_t = \frac{2}{9} \cdot \frac{4.0 \times 10^{-10} \cdot 9.8 \cdot 1.2 \times 10^{3}}{1.8 \times 10^{-5}} \]

Calculating the numerator:

\[ 4.0 \times 10^{-10} \cdot 9.8 \cdot 1.2 \times 10^{3} = 4.0 \times 9.8 \times 1.2 \times 10^{-7} = 47.04 \times 10^{-7} \]

Now substituting this back:

\[ v_t = \frac{2}{9} \cdot \frac{47.04 \times 10^{-7}}{1.8 \times 10^{-5}} = \frac{2}{9} \cdot 2.6133 \times 10^{-2} \]

Calculating \( v_t \):

\[ v_t \approx 5.8 \times 10^{-2} \, \text{m/s} \]

### Step 3: Calculate the viscous force using the formula
The formula for the viscous force \( F_v \) on the drop is given by:

\[ F_v = 6 \pi \eta r v_t \]

Substituting the values:

\[ F_v = 6 \pi (1.8 \times 10^{-5}) (2.0 \times 10^{-5}) (5.8 \times 10^{-2}) \]

Calculating \( 6 \pi \):

\[ 6 \pi \approx 18.8496 \]

Now substituting back:

\[ F_v = 18.8496 \cdot (1.8 \times 10^{-5}) \cdot (2.0 \times 10^{-5}) \cdot (5.8 \times 10^{-2}) \]

Calculating the force:

\[ F_v \approx 3.93 \times 10^{-10} \, \text{N} \]

### Final Answers
- Terminal velocity \( v_t \approx 5.8 \times 10^{-2} \, \text{m/s} \)
- Viscous force \( F_v \approx 3.93 \times 10^{-10} \, \text{N} \)