A steel ball of radius 2 mm and of relative density 8.2 is falling through a liquid of relative density 1.9. Its terminal velocity is 0.7 m/s. What is the viscosity of the liquid if the acceleration due to gravity is `10m//s^2`?

Options

A. `4xx10^(-2)PI`
B. 8PI
C. 4 PI
D. `8xx10^(-2)` PI

Explanation

To solve the problem, we need to find the viscosity of the liquid using the given parameters. We will use the formula for terminal velocity in a viscous medium, which is given by:

\[ V = \frac{2}{9} \cdot \frac{r^2 (\rho - \sigma) g}{\mu} \]

Where:
- \( V \) is the terminal velocity,
- \( r \) is the radius of the ball,
- \( \rho \) is the density of the ball,
- \( \sigma \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( \mu \) is the viscosity of the liquid.

### Step-by-Step Solution:

1. **Convert the radius of the steel ball to meters:**
\[ r = 2 \text{ mm} = 2 \times 10^{-3} \text{ m} \]

2. **Calculate the density of the steel ball (\( \rho \)):**
The relative density of the steel ball is 8.2, which means:
\[ \rho = 8.2 \times 1000 \text{ kg/m}^3 = 8200 \text{ kg/m}^3 \]

3. **Calculate the density of the liquid (\( \sigma \)):**
The relative density of the liquid is 1.9, thus:
\[ \sigma = 1.9 \times 1000 \text{ kg/m}^3 = 1900 \text{ kg/m}^3 \]

4. **Substitute the known values into the terminal velocity formula:**
Given:
- Terminal velocity \( V = 0.7 \text{ m/s} \)
- Acceleration due to gravity \( g = 10 \text{ m/s}^2 \)

The formula becomes:
\[ 0.7 = \frac{2}{9} \cdot \frac{(2 \times 10^{-3})^2 (8200 - 1900) \cdot 10}{\mu} \]

5. **Calculate \( \rho - \sigma \):**
\[ \rho - \sigma = 8200 - 1900 = 6300 \text{ kg/m}^3 \]

6. **Calculate \( (2 \times 10^{-3})^2 \):**
\[ (2 \times 10^{-3})^2 = 4 \times 10^{-6} \text{ m}^2 \]

7. **Substitute and simplify the equation:**
\[ 0.7 = \frac{2}{9} \cdot \frac{4 \times 10^{-6} \cdot 6300 \cdot 10}{\mu} \]
\[ 0.7 = \frac{2 \cdot 4 \cdot 63000 \times 10^{-6}}{9\mu} \]
\[ 0.7 = \frac{504000 \times 10^{-6}}{9\mu} \]

8. **Rearranging to find \( \mu \):**
\[ \mu = \frac{504000 \times 10^{-6}}{0.7 \cdot 9} \]
\[ \mu = \frac{504000 \times 10^{-6}}{6.3} \]
\[ \mu = 80000 \times 10^{-6} = 0.08 \text{ Pa.s} \]

9. **Convert to Poiseuille:**
\[ 1 \text{ Pa.s} = 10 \text{ Poiseuille} \Rightarrow \mu = 0.08 \text{ Pa.s} = 0.08 \times 10 \text{ Poiseuille} = 0.08 \text{ Poiseuille} \]

### Final Answer:
The viscosity of the liquid is \( 0.08 \text{ Pa.s} \) or \( 8 \times 10^{-2} \text{ Poiseuille} \).