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 the relationship between the extension (ΔL) produced in a wire of length (L) when a constant force (F) is applied. We will utilize the concept of Young's modulus and the properties of materials.

### 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:
- \( F \) = applied force
- \( A \) = cross-sectional area of the wire
- \( \Delta L \) = extension in length
- \( L \) = original length of the wire

2. **Rearranging Young's Modulus**:
From the definition, we can rearrange the formula to express the extension (ΔL):
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]

3. **Relating Volume and Area**:
The volume (V) of the wire remains constant. The volume can be expressed as:
\[ V = A \cdot L \]
From this, we can express the area (A) in terms of volume (V) and length (L):
\[ A = \frac{V}{L} \]

4. **Substituting Area in the Extension Formula**:
Now, substitute the expression for area (A) back into the equation for ΔL:
\[ \Delta L = \frac{F \cdot L}{(V/L) \cdot Y} \]
Simplifying this, we get:
\[ \Delta L = \frac{F \cdot L^2}{V \cdot Y} \]

5. **Identifying Proportionality**:
From the final expression, we can see that the extension (ΔL) is directly proportional to \( L^2 \):
\[ \Delta L \propto L^2 \]

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

### Final Answer:
The correct option is \( L^2 \).

---