A gas bubble of 2 cm diameter rises through a liquid `1.75 gm cm^(-3)` with a fixed speed of `0.35 cm s^(-1)`.Naglect the density of the gas. The cofficient of viscosity of the liquid is

Options

A. 870 poise
B. 1120 poise
C. 982 poise
D. 1089 poise

Explanation

To find the coefficient of viscosity of the liquid in which a gas bubble rises, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Diameter of the gas bubble, \( d = 2 \, \text{cm} \)
- Radius of the bubble, \( r = \frac{d}{2} = \frac{2}{2} = 1 \, \text{cm} \)
- Density of the liquid, \( \rho' = 1.75 \, \text{g/cm}^3 \)
- Terminal velocity of the bubble, \( v = 0.35 \, \text{cm/s} \)
- Gravitational acceleration, \( g = 980 \, \text{cm/s}^2 \)

2. **Use the Formula for Coefficient of Viscosity**:
The formula for the coefficient of viscosity \( \eta \) is given by:
\[ \eta = \frac{2}{9} \cdot \frac{r^2 \cdot (\rho' \cdot g)}{v} \]
where:
- \( r \) is the radius of the bubble,
- \( \rho' \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( v \) is the terminal velocity.

3. **Substitute the Values into the Formula**:
Substitute \( r = 1 \, \text{cm} \), \( \rho' = 1.75 \, \text{g/cm}^3 \), \( g = 980 \, \text{cm/s}^2 \), and \( v = 0.35 \, \text{cm/s} \):
\[ \eta = \frac{2}{9} \cdot \frac{(1)^2 \cdot (1.75 \cdot 980)}{0.35} \]

4. **Calculate the Numerator**:
Calculate \( 1.75 \cdot 980 \):
\[ 1.75 \cdot 980 = 1715 \, \text{g cm/s}^2 \]

5. **Complete the Calculation**:
Now substitute back into the formula:
\[ \eta = \frac{2}{9} \cdot \frac{1715}{0.35} \]
Calculate \( \frac{1715}{0.35} \):
\[ \frac{1715}{0.35} = 4900 \, \text{g/cm/s} \]
Now substitute this back:
\[ \eta = \frac{2}{9} \cdot 4900 \]
Calculate \( \frac{2 \cdot 4900}{9} \):
\[ \eta = \frac{9800}{9} \approx 1088.89 \, \text{poise} \]

6. **Final Result**:
The coefficient of viscosity of the liquid is approximately:
\[ \eta \approx 1088 \, \text{poise} \]