An air bubble of 1 cm radius is rising at a steady rate of `2.00ms^-1` through a liquid of density `1.5gcm^-3`. Neglect density of air. If `g=1000cms^-2`, then the coeffieciet of viscosity of the liquid is

Explanation

To find the coefficient of viscosity of the liquid through which an air bubble is rising, we can use the formula derived from Stokes' law for the motion of a sphere through a viscous medium. The formula is:

\[ \eta = \frac{2r^2 \rho g}{9v} \]

where:
- \( \eta \) is the coefficient of viscosity,
- \( r \) is the radius of the bubble,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( v \) is the velocity of the bubble.

### Step-by-Step Solution:

1. **Identify and convert the given values:**
- Radius of the bubble, \( r = 1 \, \text{cm} = 0.01 \, \text{m} \) (or \( 1 \, \text{cm} \) directly in calculations).
- Velocity of the bubble, \( v = 2 \, \text{m/s} = 200 \, \text{cm/s} \).
- Density of the liquid, \( \rho = 1.5 \, \text{g/cm}^3 = 1.5 \times 1000 \, \text{kg/m}^3 = 1500 \, \text{kg/m}^3 \).
- Acceleration due to gravity, \( g = 1000 \, \text{cm/s}^2 = 10 \, \text{m/s}^2 \).

2. **Substitute the values into the viscosity formula:**
\[ \eta = \frac{2 \times (1 \, \text{cm})^2 \times (1.5 \, \text{g/cm}^3) \times (1000 \, \text{cm/s}^2)}{9 \times (200 \, \text{cm/s})} \]

3. **Calculate the numerator:**
- Convert the radius to cm for consistency:
\[ 2 \times (1)^2 \times 1.5 \times 1000 = 3000 \, \text{g cm/s}^2 \]

4. **Calculate the denominator:**
\[ 9 \times 200 = 1800 \, \text{cm/s} \]

5. **Calculate the coefficient of viscosity:**
\[ \eta = \frac{3000 \, \text{g cm/s}^2}{1800 \, \text{cm/s}} = \frac{3000}{1800} = \frac{5}{3} \, \text{g/cm/s}^2 \]

6. **Convert to poise:**
- Since \( 1 \, \text{poise} = 1 \, \text{g/cm/s}^2 \):
\[ \eta = \frac{5}{3} \, \text{poise} \]

### Final Answer:
The coefficient of viscosity of the liquid is \( \frac{5}{3} \, \text{poise} \).