Atube is completely filled with some incompressible liquid having total mass m. Length of the tube is L. Tube is rotated in horizontal plane, about one of its ends and force exerted by the tube on liquid on the other end is found to be `(1)/(n)m L omega^(2)`. Here `omega` is uniform angular velocity of tube. Calculate the value of n.

Explanation

To solve the problem, we need to analyze the forces acting on the liquid in the tube as it rotates. The tube is rotating about one of its ends, and we are interested in the force exerted by the tube on the liquid at the other end.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a tube of length \( L \) filled with an incompressible liquid of total mass \( m \).
- The tube is rotating in a horizontal plane about one end with a uniform angular velocity \( \omega \).

2. **Identifying Forces**:
- The liquid experiences a centripetal force due to the rotation. The force exerted by the tube on the liquid at the far end (at distance \( L \)) is what we need to calculate.

3. **Centripetal Force Calculation**:
- For a small element of the liquid at a distance \( x \) from the axis of rotation, the mass of this element can be expressed as \( dm \).
- The centripetal force \( dF \) acting on this small mass \( dm \) is given by:
\[ dF = dm \cdot \omega^2 \cdot x \]
- The total mass \( m \) of the liquid can be considered as being uniformly distributed along the length \( L \).

4. **Integrating the Force**:
- To find the total force exerted by the tube on the liquid, we need to integrate \( dF \) from \( x = 0 \) to \( x = L \):
\[ F = \int_0^L \omega^2 \cdot x \cdot dm \]
- Since the mass per unit length \( \lambda \) of the liquid is \( \frac{m}{L} \), we can express \( dm \) as:
\[ dm = \frac{m}{L} \, dx \]
- Substituting this into the integral gives:
\[ F = \int_0^L \omega^2 \cdot x \cdot \frac{m}{L} \, dx \]

5. **Evaluating the Integral**:
- The integral simplifies to:
\[ F = \frac{m \omega^2}{L} \int_0^L x \, dx \]
- The integral \( \int_0^L x \, dx \) evaluates to \( \frac{L^2}{2} \):
\[ F = \frac{m \omega^2}{L} \cdot \frac{L^2}{2} = \frac{m L \omega^2}{2} \]

6. **Relating to Given Expression**:
- According to the problem, the force exerted by the tube on the liquid at the other end is given as:
\[ F = \frac{1}{n} m L \omega^2 \]
- From our calculation, we found:
\[ F = \frac{m L \omega^2}{2} \]
- Equating the two expressions for \( F \):
\[ \frac{1}{n} m L \omega^2 = \frac{m L \omega^2}{2} \]

7. **Solving for \( n \)**:
- Canceling \( m L \omega^2 \) from both sides (assuming they are non-zero), we get:
\[ \frac{1}{n} = \frac{1}{2} \]
- Thus, solving for \( n \) gives:
\[ n = 2 \]

### Final Answer:
The value of \( n \) is \( 2 \).