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

Explanation

To solve the problem step by step, we need to analyze the relationship between the extension produced in the wire (denoted as \( \Delta l \)) and the length of the wire (\( l \)) when subjected to a constant force \( F \).

### Step-by-Step Solution:

1. **Understanding Young's Modulus**:
Young's modulus \( Y \) is defined as:
\[ Y = \frac{F/A}{\Delta l/l} \]
where \( F \) is the force applied, \( A \) is the cross-sectional area, \( \Delta l \) is the extension, and \( l \) is the original length of the wire.

2. **Rearranging Young's Modulus**:
Rearranging the formula gives us:
\[ \Delta l = \frac{F \cdot l}{A \cdot Y} \]
This shows that the extension \( \Delta l \) is directly proportional to the length \( l \) of the wire.

3. **Relating Area and Volume**:
Given that the volume \( V \) of the wire is constant, we can express the area \( A \) in terms of volume and length:
\[ V = A \cdot l \implies A = \frac{V}{l} \]

4. **Substituting Area into the Extension Formula**:
Substitute \( A \) back into the equation for \( \Delta l \):
\[ \Delta l = \frac{F \cdot l}{(V/l) \cdot Y} = \frac{F \cdot l^2}{V \cdot Y} \]
Let \( k = \frac{F}{V \cdot Y} \) (a constant), then:
\[ \Delta l = k \cdot l^2 \]

5. **Identifying the Graph**:
The equation \( \Delta l = k \cdot l^2 \) indicates that \( \Delta l \) is directly proportional to \( l^2 \). Therefore, if we plot \( \Delta l \) on the y-axis and \( l^2 \) on the x-axis, we will get a straight line.

### Conclusion:
The graph that represents a straight line is \( \Delta l \) versus \( l^2 \).

### Final Answer:
The correct option is: \( \Delta l \) versus \( l^2 \).