A lead sphere of `1.0 mm` diameter and relative density `11.20` attains a terminal velocity of `0.7 cm s^(-1)` in a liquid of relative density `1.26`.
Determine the coefficient of dynamic viscosity of the liquid.

Options

A. `0.45N//m^(2)`
B. `0.85N//m^(2)`
C. `0.56N//m^(2)`
D. `0.77N//m^(2)`

Explanation

To determine the coefficient of dynamic viscosity of the liquid, we can use the formula for terminal velocity of a sphere falling through a fluid:

\[ V_t = \frac{2}{9} \frac{R^2 (ρ_s - ρ_l) g}{μ} \]

Where:
- \( V_t \) = terminal velocity
- \( R \) = radius of the sphere
- \( ρ_s \) = density of the sphere
- \( ρ_l \) = density of the liquid
- \( g \) = acceleration due to gravity
- \( μ \) = coefficient of dynamic viscosity

### Step 1: Convert the given values to SI units
- Diameter of the sphere = \( 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \)
- Radius \( R = \frac{1.0 \times 10^{-3}}{2} = 0.5 \times 10^{-3} \, \text{m} \)
- Terminal velocity \( V_t = 0.7 \, \text{cm/s} = 0.7 \times 10^{-2} \, \text{m/s} \)
- Relative density of lead sphere = \( 11.20 \) (Density of lead \( ρ_s = 11.20 \times 1000 \, \text{kg/m}^3 = 11200 \, \text{kg/m}^3 \))
- Relative density of liquid = \( 1.26 \) (Density of liquid \( ρ_l = 1.26 \times 1000 \, \text{kg/m}^3 = 1260 \, \text{kg/m}^3 \))
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

### Step 2: Substitute the values into the formula
Rearranging the terminal velocity formula to solve for \( μ \):

\[ μ = \frac{2}{9} \frac{R^2 (ρ_s - ρ_l) g}{V_t} \]

Substituting the values:

\[ μ = \frac{2}{9} \cdot \frac{(0.5 \times 10^{-3})^2 \cdot (11200 - 1260) \cdot 10}{0.7 \times 10^{-2}} \]

### Step 3: Calculate the values
1. Calculate \( R^2 \):
\[ R^2 = (0.5 \times 10^{-3})^2 = 0.25 \times 10^{-6} \, \text{m}^2 \]

2. Calculate \( ρ_s - ρ_l \):
\[ ρ_s - ρ_l = 11200 - 1260 = 9940 \, \text{kg/m}^3 \]

3. Now substitute these values into the equation:
\[ μ = \frac{2}{9} \cdot \frac{0.25 \times 10^{-6} \cdot 9940 \cdot 10}{0.7 \times 10^{-2}} \]

4. Calculate the numerator:
\[ 0.25 \times 10^{-6} \cdot 9940 \cdot 10 = 2485 \times 10^{-6} \]

5. Now calculate \( μ \):
\[ μ = \frac{2}{9} \cdot \frac{2485 \times 10^{-6}}{0.7 \times 10^{-2}} = \frac{2}{9} \cdot \frac{2485 \times 10^{-6}}{7 \times 10^{-3}} = \frac{2 \cdot 2485 \times 10^{-6}}{63 \times 10^{-3}} = \frac{4970 \times 10^{-6}}{63 \times 10^{-3}} = \frac{4970}{63} \times 10^{-3} \]

6. Finally, calculate \( μ \):
\[ μ \approx 0.079 \, \text{N s/m}^2 \]

### Final Answer
The coefficient of dynamic viscosity of the liquid is approximately \( 0.079 \, \text{N s/m}^2 \).