Young's modulus of the material of a wire is Y. ON pulling the wire by a force F, the increase in its length is x. The potential energy of the stretched wire is

Options

A. `(1)/(2) Fx`
B. `(1)/(2) Yx`
C. `(1)/(2) Fx^(2)`
D. None of these

Explanation

To find the potential energy of a stretched wire, we can use the relationship between stress, strain, and the potential energy stored in the wire. Let's go through the solution step by step.

### Step-by-Step Solution:

1. **Understanding Young's Modulus**:
Young's modulus (Y) is defined as the ratio of stress to strain. Stress is the force applied per unit area, and strain is the relative change in length.
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]

2. **Expressing Stress and Strain**:
- Stress = \( \frac{F}{A} \)
- Strain = \( \frac{\Delta L}{L} \) where \( \Delta L \) is the change in length (given as \( x \)) and \( L \) is the original length of the wire.

3. **Potential Energy Formula**:
The potential energy (U) stored in a stretched wire can be expressed as:
\[ U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume} \]

4. **Calculating Volume**:
The volume (V) of the wire can be expressed as:
\[ V = A \times L \]
where \( A \) is the cross-sectional area and \( L \) is the original length of the wire.

5. **Substituting Values**:
Now substituting the expressions for stress, strain, and volume into the potential energy formula:
\[ U = \frac{1}{2} \times \left(\frac{F}{A}\right) \times \left(\frac{x}{L}\right) \times (A \times L) \]

6. **Simplifying the Equation**:
In the equation above, the area \( A \) and the length \( L \) will cancel out:
\[ U = \frac{1}{2} \times F \times x \]

7. **Final Result**:
Therefore, the potential energy of the stretched wire is:
\[ U = \frac{1}{2} F x \]

### Final Answer:
The potential energy of the stretched wire is \( U = \frac{1}{2} F x \).

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