A metal wire of length `L` is suspended vertically from a rigid support. When a bob of mass `M` is attached to the lower end of wire, the elongation of the wire is `l:`

Options

A. The loss in gravitational potential energy of mass `M` is `Mgl`.
B. The elastic potential energy stored in the wire is `Mgl`
C. The elastic potential energy stored in the wire is `(1)/(2)Mgl`
D. Heat produced is loss of mechanical energy of system is `(1)/(2)Mgl`

Answer

To solve the problem, we need to analyze the situation where a metal wire of length \( L \) is suspended vertically and a bob of mass \( M \) is attached to its lower end, causing an elongation of the wire by \( l \).

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a metal wire of length \( L \) suspended from a rigid support.
- A bob of mass \( M \) is attached to the end of the wire, causing it to stretch by an amount \( l \).

2. **Calculating the Loss in Gravitational Potential Energy**:
- When the bob is attached, it moves down by the elongation \( l \).
- The loss in gravitational potential energy (GPE) can be calculated using the formula:
\[ \text{Loss in GPE} = M \cdot g \cdot h \]
where \( h \) is the distance moved down, which is equal to \( l \).
- Therefore, the loss in GPE is:
\[ \text{Loss in GPE} = M \cdot g \cdot l \]

3. **Calculating the Elastic Potential Energy**:
- The elongation of the wire results in elastic potential energy being stored in the wire.
- The elastic potential energy (EPE) can be expressed as:
\[ \text{EPE} = \frac{1}{2} k l^2 \]
where \( k \) is the spring constant of the wire.
- However, we can also relate it to the force exerted by the bob:
\[ \text{EPE} = \frac{1}{2} M g l \]

4. **Calculating the Heat Produced**:
- The total energy lost by the system (which is the loss in gravitational potential energy) will be converted into elastic potential energy and heat.
- The heat produced can be calculated by the energy balance:
\[ \text{Heat Produced} = \text{Loss in GPE} - \text{EPE} \]
- Substituting the values we found:
\[ \text{Heat Produced} = (M \cdot g \cdot l) - \left(\frac{1}{2} M g l\right) = \frac{1}{2} M g l \]

### Summary of Results:
- Loss in Gravitational Potential Energy: \( M \cdot g \cdot l \)
- Elastic Potential Energy: \( \frac{1}{2} M g l \)
- Heat Produced: \( \frac{1}{2} M g l \)

### Final Answers:
- Loss in GPE: \( M \cdot g \cdot l \)
- Elastic Potential Energy: \( \frac{1}{2} M g l \)
- Heat Produced: \( \frac{1}{2} M g l \)