A wire of length L and cross-sectional area A is made of material of Young's modulus Y. The work done in stretching the wire by an amount x is

Options

A. `(YAx^(2))/(L)`
B. `(YAx^(2))/(2L)`
C. `(2YAx^(2))/(L)`
D. `(4YAx^(2))/(L)`

Answer

To find the work done in stretching a wire of length \( L \) and cross-sectional area \( A \) made of material with Young's modulus \( Y \) by an amount \( x \), we can follow these steps:

### Step-by-Step Solution:

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

2. **Expressing Force in Terms of Young's Modulus**:
From the definition of Young's modulus, we can express the force \( F \) acting on the wire:
\[ Y = \frac{F/A}{x/L} \implies F = Y \cdot \frac{A \cdot x}{L} \]

3. **Calculating the Work Done**:
The work done \( W \) in stretching the wire can be calculated using the formula for work done, which is the integral of force over the distance stretched:
\[ W = \int F \, dx \]
Substituting the expression for \( F \):
\[ W = \int_0^x \left( Y \cdot \frac{A \cdot x'}{L} \right) dx' \]
Here, \( x' \) is the variable of integration.

4. **Integrating the Expression**:
Performing the integration:
\[ W = Y \cdot \frac{A}{L} \int_0^x x' \, dx' = Y \cdot \frac{A}{L} \left[ \frac{x'^2}{2} \right]_0^x = Y \cdot \frac{A}{L} \cdot \frac{x^2}{2} \]

5. **Final Expression for Work Done**:
Thus, the work done in stretching the wire by an amount \( x \) is:
\[ W = \frac{1}{2} \cdot Y \cdot \frac{A}{L} \cdot x^2 \]

### Final Answer:
\[ W = \frac{1}{2} \cdot \frac{Y \cdot A}{L} \cdot x^2 \]