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 1: Identify the forces acting on the block
When the block of mass \( M \) is suspended, the force acting on it due to gravity is given by:
\[ F = mg \]
where \( g \) is the acceleration due to gravity.

### Step 2: Determine the stress in the wire
Stress (\( \sigma \)) in the wire can be defined as the force per unit area. The formula for stress is:
\[ \sigma = \frac{F}{A} \]
Substituting the force \( F = mg \):
\[ \sigma = \frac{mg}{A} \]
where \( A \) is the cross-sectional area of the wire.

### Step 3: Calculate the strain in the wire
Strain (\( \epsilon \)) is defined as the change in length divided by the original length. The change in length when the wire is stretched is \( l \), and the original length of the wire is \( L \). Thus, the strain is:
\[ \epsilon = \frac{\Delta L}{L} = \frac{l}{L} \]

### Step 4: Use the formula for elastic potential energy
The elastic potential energy (\( U \)) stored in the wire can be calculated using the formula:
\[ U = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume} \]
The volume of the wire can be expressed as:
\[ \text{Volume} = A \times L \]
Now substituting the expressions for stress, strain, and volume into the energy formula:
\[ U = \frac{1}{2} \times \frac{mg}{A} \times \frac{l}{L} \times (A \times L) \]

### Step 5: Simplify the expression
Notice that the area \( A \) cancels out:
\[ U = \frac{1}{2} \times mg \times \frac{l}{L} \times L \]
This simplifies to:
\[ U = \frac{1}{2} mg l \]

### Final Answer
The elastic potential energy stored in the extended wire is:
\[ U = \frac{1}{2} mg l \]

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