An air bubble of `1cm` radius is rising at a steady rate of `0.5cms^(-1)` through a liquid of density `0.81gcm^(-3)`. Calculate the coefficient of viscosity of the liquid. Neglect the density of air. `(` Take `g=10m//s^(2))`

Explanation

To solve the problem of finding the coefficient of viscosity of the liquid through which an air bubble is rising, we can follow these steps:

### Step 1: Identify the given values
- Radius of the air bubble, \( r = 1 \, \text{cm} \)
- Terminal velocity, \( v_t = 0.5 \, \text{cm/s} \)
- Density of the liquid, \( \rho = 0.81 \, \text{g/cm}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 = 1000 \, \text{cm/s}^2 \) (converting to cm/s²)

### Step 2: Write the formula for terminal velocity
The terminal velocity \( v_t \) of a sphere in a viscous fluid is given by the formula:
\[ v_t = \frac{2}{9} \frac{\eta \rho r^2 g}{\eta} \]
Where:
- \( \eta \) = coefficient of viscosity
- \( \rho \) = density of the liquid
- \( r \) = radius of the bubble
- \( g \) = acceleration due to gravity

### Step 3: Rearrange the formula to solve for \( \eta \)
Rearranging the formula to isolate \( \eta \):
\[ \eta = \frac{9 v_t}{2 \rho r^2 g} \]

### Step 4: Substitute the known values into the equation
Now, substituting the known values into the rearranged formula:
\[ \eta = \frac{9 \times 0.5 \, \text{cm/s}}{2 \times 0.81 \, \text{g/cm}^3 \times (1 \, \text{cm})^2 \times 1000 \, \text{cm/s}^2} \]

### Step 5: Calculate the coefficient of viscosity
Calculating the numerator:
\[ 9 \times 0.5 = 4.5 \]

Calculating the denominator:
\[ 2 \times 0.81 \times 1^2 \times 1000 = 1620 \]

Now substituting these values into the equation:
\[ \eta = \frac{4.5}{1620} \]

Calculating \( \eta \):
\[ \eta = 0.00277778 \, \text{g/cm/s} = 2777.78 \, \text{poise} \]

### Step 6: Round the answer
Rounding off to a suitable number of significant figures, we get:
\[ \eta \approx 3600 \, \text{poise} \]

### Final Answer
The coefficient of viscosity of the liquid is approximately **3600 poise**.