The terminal velocity of a copper ball of radius `2mm` falling through a tank of oil at `20^(@)C` is `6.5cm//s`. Find the viscosity of the oil at `20^(@)C`. Density of oil is `1.5xx10^(3)Kg//m^(3)`, density of copper is `8.9xx10^(3)Kg//m^(3)`.

Options

A. `9.9xx10^(-1)kgm^(-1)s^(-1)`
B. `9.0xx10^(-1)kgm^(-1)s^(-1)`
C. `8.0xx10^(-1)kgm^(-1)s^(-1)`
D. `8.5xx10^(-1)kgm^(-1)s^(-1)`

Explanation

To find the viscosity of the oil at 20°C, we can use the formula for the coefficient of viscosity (η) for a spherical object falling through a fluid, which is given by:

\[ \eta = \frac{2}{9} r^2 g \frac{(\rho - \sigma)}{v_t} \]

Where:
- \( r \) = radius of the sphere (in meters)
- \( g \) = acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \))
- \( \rho \) = density of the sphere (copper ball)
- \( \sigma \) = density of the fluid (oil)
- \( v_t \) = terminal velocity (in meters per second)

### Step 1: Convert the radius to meters
The radius of the copper ball is given as \( 2 \, \text{mm} \). We need to convert this to meters:

\[ r = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \]

### Step 2: Write down the known values
- \( r = 2 \times 10^{-3} \, \text{m} \)
- \( g = 9.8 \, \text{m/s}^2 \)
- Density of copper \( \rho = 8.9 \times 10^{3} \, \text{kg/m}^3 \)
- Density of oil \( \sigma = 1.5 \times 10^{3} \, \text{kg/m}^3 \)
- Terminal velocity \( v_t = 6.5 \, \text{cm/s} = 6.5 \times 10^{-2} \, \text{m/s} \)

### Step 3: Calculate \( \rho - \sigma \)
Now, we calculate the difference between the densities:

\[ \rho - \sigma = 8.9 \times 10^{3} - 1.5 \times 10^{3} = 7.4 \times 10^{3} \, \text{kg/m}^3 \]

### Step 4: Substitute the values into the viscosity formula
Now we can substitute all the values into the viscosity formula:

\[ \eta = \frac{2}{9} \times (2 \times 10^{-3})^2 \times 9.8 \times \frac{(7.4 \times 10^{3})}{(6.5 \times 10^{-2})} \]

### Step 5: Calculate \( (2 \times 10^{-3})^2 \)
Calculating \( (2 \times 10^{-3})^2 \):

\[ (2 \times 10^{-3})^2 = 4 \times 10^{-6} \]

### Step 6: Substitute and simplify
Now substitute this back into the equation:

\[ \eta = \frac{2}{9} \times 4 \times 10^{-6} \times 9.8 \times \frac{(7.4 \times 10^{3})}{(6.5 \times 10^{-2})} \]

### Step 7: Calculate the entire expression
Calculating the expression step by step:

1. Calculate \( \frac{(7.4 \times 10^{3})}{(6.5 \times 10^{-2})} \):

\[ \frac{7.4 \times 10^{3}}{6.5 \times 10^{-2}} = \frac{7.4}{6.5} \times 10^{3 + 2} = 1.1385 \times 10^{5} \]

2. Now substitute this value back into the viscosity equation:

\[ \eta = \frac{2}{9} \times 4 \times 10^{-6} \times 9.8 \times 1.1385 \times 10^{5} \]

3. Calculate \( \frac{2}{9} \times 4 \times 9.8 \):

\[ \frac{2 \times 4 \times 9.8}{9} = \frac{78.4}{9} \approx 8.7333 \]

4. Finally, multiply:

\[ \eta \approx 8.7333 \times 10^{-6} \times 1.1385 \times 10^{5} \approx 9.9 \times 10^{-1} \, \text{kg/(m·s)} \]

### Final Result
Thus, the viscosity of the oil at \( 20^{\circ}C \) is approximately:

\[ \eta \approx 0.99 \, \text{kg/(m·s)} \]