An air bubble of radius 1 mm is allowed to rise through a long cylindrical column of a viscous liquid of radius 5 cm and travels at a steady rate of 2.1 cm per sec. if the density of the liquid is `1.47(g)/(c c)`. Its viscosity is nearly `(n)/(2)` poise. Then find the value of n. Assume `g=980(cm)/(sec^(2))` and neglect the density of air.

Explanation

To solve the problem, we will use the equation for terminal velocity derived from Stokes' law. The terminal velocity \( V \) of a small sphere moving through a viscous fluid is given by:

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

Where:
- \( r \) = radius of the bubble
- \( \rho \) = density of the liquid
- \( \sigma \) = density of the bubble (air, which we neglect)
- \( g \) = acceleration due to gravity
- \( \eta \) = viscosity of the liquid

### Step 1: Identify the given values
- Radius of the bubble \( r = 1 \, \text{mm} = 0.1 \, \text{cm} \)
- Density of the liquid \( \rho = 1.47 \, \text{g/cm}^3 \)
- Density of air \( \sigma \) is neglected
- Terminal velocity \( V = 2.1 \, \text{cm/s} \)
- Acceleration due to gravity \( g = 980 \, \text{cm/s}^2 \)
- Viscosity \( \eta = \frac{n}{2} \, \text{poise} \)

### Step 2: Substitute the known values into the terminal velocity equation
Since we are neglecting the density of air, we can simplify the equation:

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

Substituting the known values:

\[ 2.1 = \frac{2}{9} \cdot \frac{(0.1)^2 \cdot (1.47) \cdot (980)}{\frac{n}{2}} \]

### Step 3: Simplify the equation
Calculating \( (0.1)^2 \):

\[ (0.1)^2 = 0.01 \]

Now substituting this back into the equation:

\[ 2.1 = \frac{2}{9} \cdot \frac{0.01 \cdot 1.47 \cdot 980}{\frac{n}{2}} \]

This can be rewritten as:

\[ 2.1 = \frac{2 \cdot 0.01 \cdot 1.47 \cdot 980}{9 \cdot \frac{n}{2}} \]

### Step 4: Cross-multiply to solve for \( n \)
Cross-multiplying gives:

\[ 2.1 \cdot 9 \cdot \frac{n}{2} = 2 \cdot 0.01 \cdot 1.47 \cdot 980 \]

This simplifies to:

\[ 18.9 \cdot \frac{n}{2} = 2 \cdot 0.01 \cdot 1.47 \cdot 980 \]

Calculating the right side:

\[ 2 \cdot 0.01 \cdot 1.47 \cdot 980 = 28.836 \]

So we have:

\[ 18.9 \cdot \frac{n}{2} = 28.836 \]

### Step 5: Solve for \( n \)
Multiply both sides by 2:

\[ 18.9n = 57.672 \]

Now divide by 18.9:

\[ n = \frac{57.672}{18.9} \approx 3.05 \]

### Final Result
The value of \( n \) is approximately \( 3 \).