An air bubble of radius 1 cm rises with terminal velocity 0.21 cm/s in liquid column. If the density of liquid is `1.47 xx 10^(3) kg//m^(3).` Then the value of coefficient of viscosity of liquid ignoring the density of air, will be

Options

A. `1. 71 xx 10 ^(4)` poise
B. `1. 82 xx 10 ^(4) ` poise
C. `1. 78 xx 10 ^(4)` poise
D. `1. 52 xx 10 ^(4)` poise

Explanation

To find the coefficient of viscosity (η) of the liquid in which an air bubble rises, we can use Stokes' law. The formula for terminal velocity (Vt) of a sphere in a viscous fluid is given by:

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

Where:
- \( V_t \) = terminal velocity
- \( r \) = radius of the bubble
- \( \rho \) = density of the liquid
- \( \sigma \) = density of the bubble (air, which we will ignore)
- \( g \) = acceleration due to gravity
- \( \eta \) = coefficient of viscosity

### Step-by-step Solution:

1. **Identify the given values**:
- Radius of the bubble, \( r = 1 \text{ cm} = 1 \times 10^{-2} \text{ m} \)
- Terminal velocity, \( V_t = 0.21 \text{ cm/s} = 0.21 \times 10^{-2} \text{ m/s} \)
- Density of the liquid, \( \rho = 1.47 \times 10^{3} \text{ kg/m}^3 \)
- Acceleration due to gravity, \( g = 9.8 \text{ m/s}^2 \)

2. **Substituting values into the formula**:
Since we are ignoring the density of air (σ), we can simplify the equation to:
\[ V_t = \frac{2r^2 \rho g}{9\eta} \]

3. **Rearranging the formula to solve for η**:
\[ \eta = \frac{2r^2 \rho g}{9V_t} \]

4. **Substituting the known values**:
\[ \eta = \frac{2 \times (1 \times 10^{-2})^2 \times (1.47 \times 10^{3}) \times (9.8)}{9 \times (0.21 \times 10^{-2})} \]

5. **Calculating the numerator**:
- Calculate \( 2 \times (1 \times 10^{-2})^2 = 2 \times 1 \times 10^{-4} = 2 \times 10^{-4} \)
- Calculate \( 2 \times 10^{-4} \times 1.47 \times 10^{3} = 2.94 \times 10^{-1} \)
- Calculate \( 2.94 \times 10^{-1} \times 9.8 = 2.8832 \)

6. **Calculating the denominator**:
- Calculate \( 9 \times (0.21 \times 10^{-2}) = 1.89 \times 10^{-2} \)

7. **Final calculation of η**:
\[ \eta = \frac{2.8832}{1.89 \times 10^{-2}} \approx 152.52 \text{ Pa.s} = 1.5252 \times 10^{4} \text{ poise} \]

### Final Answer:
The coefficient of viscosity of the liquid is approximately \( 1.52 \times 10^{4} \text{ poise} \).