A fixed volume of iron is drawn into a wire of length `l`. The extension produced in this wire by a constant force F is proportional to

Options

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

Answer

To solve the problem, we need to analyze how the extension produced in a wire of fixed volume changes with respect to its length when a constant force is applied. Let's break this down step by step.

### Step-by-Step Solution:

1. **Understanding Fixed Volume**:
- We start with a fixed volume of iron, which can be expressed as:
\[ V = A \cdot l \]
where \( V \) is the volume, \( A \) is the cross-sectional area, and \( l \) is the length of the wire.

2. **Expressing Area in Terms of Length**:
- Since the volume is fixed, if the length \( l \) changes, the cross-sectional area \( A \) must also change to keep the volume constant. We can express the area as:
\[ A = \frac{V}{l} \]

3. **Using Young's Modulus**:
- Young's modulus \( Y \) relates stress and strain in the material. The formula for Young's modulus is given by:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/l} \]
where \( F \) is the applied force, \( \Delta l \) is the extension, and \( l \) is the original length.

4. **Rearranging Young's Modulus**:
- From the definition of Young's modulus, we can rearrange the equation to find the extension \( \Delta l \):
\[ \Delta l = \frac{F \cdot l}{Y \cdot A} \]

5. **Substituting Area**:
- Now substitute the expression for area \( A \) into the equation:
\[ \Delta l = \frac{F \cdot l}{Y \cdot \left(\frac{V}{l}\right)} = \frac{F \cdot l^2}{Y \cdot V} \]

6. **Identifying Proportionality**:
- From the equation \( \Delta l = \frac{F \cdot l^2}{Y \cdot V} \), we see that the extension \( \Delta l \) is directly proportional to \( l^2 \):
\[ \Delta l \propto l^2 \]

### Conclusion:
Thus, the extension produced in the wire by a constant force \( F \) is proportional to \( l^2 \).