When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L+l). The elastic potential energy stored in the extended wire is

Options

A. `(1)/(2)MgL`
B. `Mgl`
C. `MgL`
D. `(1)/(2)Mgl`

Explanation

To find the elastic potential energy stored in the extended wire when a block of mass \( M \) is suspended, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Mass of the block: \( M \)
- Original length of the wire: \( L \)
- Extended length of the wire: \( L + l \)
- Elongation of the wire: \( l \)

2. **Calculate the Force Applied on the Wire**:
- The force applied due to the weight of the block is given by:
\[ F = Mg \]
where \( g \) is the acceleration due to gravity.

3. **Calculate the Stress in the Wire**:
- Stress is defined as force per unit area. If \( A \) is the cross-sectional area of the wire, then:
\[ \text{Stress} = \frac{F}{A} = \frac{Mg}{A} \]

4. **Calculate the Strain in the Wire**:
- Strain is defined as the change in length (elongation) divided by the original length. Thus:
\[ \text{Strain} = \frac{l}{L} \]

5. **Calculate the Elastic Potential Energy Stored**:
- The elastic potential energy stored in the wire can be calculated using the formula:
\[ U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times V \]
where \( V \) is the volume of the wire. The volume \( V \) can be expressed as:
\[ V = A \times L \]
Substituting the values of stress and strain into the energy formula:
\[ U = \frac{1}{2} \times \left(\frac{Mg}{A}\right) \times \left(\frac{l}{L}\right) \times (A \times L) \]

6. **Simplify the Expression**:
- The area \( A \) cancels out:
\[ U = \frac{1}{2} \times Mg \times l \]

7. **Final Result**:
- Therefore, the elastic potential energy stored in the extended wire is:
\[ U = \frac{1}{2} M g l \]