Copper of fixed volume `V` is drawn into wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is `trianglel`. Which of the following graphs is a straight line?

Options

A. `trianglel` versus`(1)/(l)`
B. `trianglel` versus `l^2`
C. `trianglel` versus `(1)/(l^2)`
D. `trianglel` versus `l`

Explanation

To solve the problem, we need to analyze the relationship between the extension produced in the wire (ΔL) and the length of the wire (L) when a constant force (F) is applied. We will derive the necessary equations 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. Mathematically, it is given by:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
where \( F \) is the force applied, \( A \) is the cross-sectional area, \( \Delta L \) is the extension, and \( L \) is the original length.

2. **Rearranging the Young's Modulus Equation**:
From the definition of Young's modulus, we can express strain in terms of stress:
\[ \frac{\Delta L}{L} = \frac{F}{A \cdot Y} \]

3. **Volume of the Wire**:
Since the volume \( V \) of the copper is fixed, we have:
\[ V = A \cdot L \]
From this, we can express the cross-sectional area \( A \) in terms of volume and length:
\[ A = \frac{V}{L} \]

4. **Substituting Area into the Equation**:
Substitute \( A \) back into the equation for strain:
\[ \frac{\Delta L}{L} = \frac{F}{(V/L) \cdot Y} = \frac{F \cdot L}{V \cdot Y} \]

5. **Solving for ΔL**:
Rearranging the equation gives us:
\[ \Delta L = \frac{F \cdot L^2}{V \cdot Y} \]
This shows that the extension \( \Delta L \) is directly proportional to the square of the length \( L^2 \).

6. **Identifying the Relationship**:
From the equation \( \Delta L = \frac{F}{V \cdot Y} \cdot L^2 \), we can see that if we plot \( \Delta L \) on the y-axis and \( L^2 \) on the x-axis, the relationship is linear. This indicates that the graph of \( \Delta L \) versus \( L^2 \) will be a straight line.

### Conclusion:
The correct answer is that the graph of \( \Delta L \) versus \( L^2 \) is a straight line.