A mass m is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:

Options

A. inclined at a angle of `60^(@)` from vertical
B. the mass is at the highest point
C. the wire is horizontal
D. the mass is at the lowest point

Explanation

To solve the problem of determining when the wire is most likely to break while a mass \( m \) is whirled in a vertical circle, we need to analyze the tension in the wire at different positions of the mass. The wire is most likely to break when the tension is at its maximum.

### Step-by-Step Solution:

1. **Identify the Positions**: The mass can be at four different positions in the vertical circle:
- A: At an angle of 60 degrees from the vertical.
- B: At the highest point of the circle.
- C: At the horizontal position.
- D: At the lowest point of the circle.

2. **Analyze Tension at Each Position**:
- **Position A (60 degrees from vertical)**:
- The forces acting on the mass are its weight \( mg \) acting downward and the tension \( T \) in the wire acting inward.
- The centripetal force required for circular motion is provided by the net force towards the center:
\[ T + mg \cos(60^\circ) = \frac{mv^2}{R} \]
- Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ T + \frac{mg}{2} = \frac{mv^2}{R} \]
\[ T = \frac{mv^2}{R} - \frac{mg}{2} \]

- **Position B (Highest Point)**:
- At the highest point, both the weight and tension act downward:
\[ T + mg = \frac{mv^2}{R} \]
\[ T = \frac{mv^2}{R} - mg \]

- **Position C (Horizontal Position)**:
- At this position, the weight acts downward and tension acts horizontally:
\[ T = \frac{mv^2}{R} \]

- **Position D (Lowest Point)**:
- At the lowest point, the weight acts downward and tension acts upward:
\[ T - mg = \frac{mv^2}{R} \]
\[ T = mg + \frac{mv^2}{R} \]

3. **Compare Tensions**:
- From the equations derived:
- Position A: \( T = \frac{mv^2}{R} - \frac{mg}{2} \)
- Position B: \( T = \frac{mv^2}{R} - mg \)
- Position C: \( T = \frac{mv^2}{R} \)
- Position D: \( T = mg + \frac{mv^2}{R} \)

4. **Determine Maximum Tension**:
- Among all positions, the tension is maximum at Position D (Lowest Point):
\[ T = mg + \frac{mv^2}{R} \]
- This indicates that the wire is most likely to break when the mass is at the lowest point of the vertical circle.

### Conclusion:
The wire is most likely to break when the mass is at the **lowest point** of the vertical circle. Therefore, the correct answer is **Option D**.