A heavy particle of weight w, attached to a fixed point by a light inextensible string describes a circle in a vertical plane. The tension in the string has the values nw and mw respectively when the particle is at the highest and lowest points in the path. Then :

Options

A. `m + n = 6`
B. `m/n = 2`
C. `m - n = -6`
D. `n - m = -6`

Explanation

To solve the problem, we need to analyze the forces acting on the particle at the highest and lowest points of its circular motion and apply the principles of circular motion and energy conservation.

### Step-by-Step Solution:

1. **Identify the Forces at the Highest Point**:
At the highest point of the circular path, the forces acting on the particle are:
- Weight (downward) = \( w \)
- Tension (upward) = \( T_1 = nw \)

The net centripetal force required to keep the particle moving in a circle is provided by the difference between the tension and the weight:
\[ T_1 - w = \frac{mv_1^2}{L} \]
Substituting \( T_1 \):
\[ nw - w = \frac{mv_1^2}{L} \]
Simplifying gives:
\[ (n-1)w = \frac{mv_1^2}{L} \quad \text{(1)} \]

2. **Identify the Forces at the Lowest Point**:
At the lowest point of the circular path, the forces acting on the particle are:
- Weight (downward) = \( w \)
- Tension (upward) = \( T_2 = mw \)

The net centripetal force required is given by:
\[ T_2 - w = \frac{mv_2^2}{L} \]
Substituting \( T_2 \):
\[ mw - w = \frac{mv_2^2}{L} \]
Simplifying gives:
\[ (m-1)w = \frac{mv_2^2}{L} \quad \text{(2)} \]

3. **Apply the Work-Energy Theorem**:
The particle travels a vertical distance of \( 2L \) (from the highest point to the lowest point). The work done by gravity is:
\[ W = -w \cdot 2L = -2wL \]
The change in kinetic energy is:
\[ \Delta KE = \frac{1}{2}m(v_2^2 - v_1^2) \]
According to the work-energy theorem:
\[ -2wL = \frac{1}{2}m(v_2^2 - v_1^2) \]
Rearranging gives:
\[ v_2^2 - v_1^2 = -\frac{4wL}{m} \quad \text{(3)} \]

4. **Relate the Velocities**:
From equations (1) and (2), we can express \( v_1^2 \) and \( v_2^2 \):
- From (1): \( v_1^2 = \frac{(n-1)wL}{m} \)
- From (2): \( v_2^2 = \frac{(m-1)wL}{m} \)

5. **Substituting into the Work-Energy Equation**:
Substitute \( v_1^2 \) and \( v_2^2 \) into equation (3):
\[ \frac{(m-1)wL}{m} - \frac{(n-1)wL}{m} = -\frac{4wL}{m} \]
Simplifying gives:
\[ \frac{(m - n)wL}{m} = -\frac{4wL}{m} \]
Canceling \( wL \) from both sides (assuming \( wL \neq 0 \)):
\[ m - n = -4 \quad \text{(4)} \]

6. **Final Rearrangement**:
Rearranging equation (4) gives:
\[ m - n = 6 \]

### Conclusion:
Thus, the correct relationship between \( m \) and \( n \) is:
\[ m - n = 6 \]