A metal ring of mass m and radius R is placed on a smooth horizontal table and is set rotating about axis, such that ring moves with constant speed. The tension in the ring is `(mv^(2))/(npiR)` Then the value of n is

Answer

To solve the problem, we will analyze the forces acting on a small segment of the metal ring and derive the expression for tension in the ring.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a metal ring of mass \( m \) and radius \( R \) rotating on a smooth horizontal table.
- The ring is moving with a constant speed \( v \).

2. **Consider a Small Segment of the Ring**:
- Let’s take a small segment of the ring with mass \( dm \) and angle \( d\theta \).
- The total mass of the ring is distributed uniformly along its circumference.

3. **Components of Tension**:
- The tension \( T \) in the ring has two components acting on the small segment:
- The radial components \( T \sin d\theta \) (towards the center).
- The tangential components \( T \cos d\theta \) (which cancel out due to symmetry).

4. **Centripetal Force Requirement**:
- The radial components of tension provide the necessary centripetal force for the segment:
\[ 2T \sin d\theta = dm \frac{v^2}{R} \]

5. **Using Small Angle Approximation**:
- For small angles, \( \sin d\theta \approx d\theta \). Thus, we can rewrite the equation as:
\[ 2T d\theta = dm \frac{v^2}{R} \]

6. **Finding the Mass of the Segment**:
- The mass \( dm \) of the small segment can be expressed as:
\[ dm = \frac{m}{2\pi} \cdot d\theta \]
- This is because the total mass \( m \) is distributed uniformly over the circumference \( 2\pi R \).

7. **Substituting \( dm \) into the Equation**:
- Substituting \( dm \) into the centripetal force equation gives:
\[ 2T d\theta = \left(\frac{m}{2\pi} d\theta\right) \frac{v^2}{R} \]

8. **Canceling \( d\theta \)**:
- We can cancel \( d\theta \) from both sides (assuming \( d\theta \neq 0 \)):
\[ 2T = \frac{m v^2}{2\pi R} \]

9. **Solving for Tension \( T \)**:
- Rearranging the equation gives:
\[ T = \frac{m v^2}{4\pi R} \]

10. **Comparing with Given Tension**:
- The problem states that the tension is given by:
\[ T = \frac{m v^2}{n \pi R} \]
- Equating the two expressions for tension:
\[ \frac{m v^2}{4\pi R} = \frac{m v^2}{n \pi R} \]

11. **Finding \( n \)**:
- By comparing the coefficients, we find:
\[ n = 4 \]

### Final Answer:
The value of \( n \) is \( 4 \).