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

The correct option is L².

To understand why the extension is proportional to the square of the length, let us analyze the relation between force, deformation, and the material properties of the wire using the concept of elasticity.

Young's modulus ($Y$) of a material is defined as the ratio of tensile stress to tensile strain:
$Y=\frac{\text{Stress}}{\text{Strain}}$

Here, the tensile stress is the restoring force per unit cross-sectional area ($A$):
$\text{Stress}=\frac{F}{A}$

The tensile strain is the fractional change in length, where $x$ is the extension and $L$ is the original length:
$\text{Strain}=\frac{x}{L}$

Substituting these definitions into the formula for Young's modulus gives:
$Y=\frac{F/A}{x/L}=\frac{F\cdot L}{A\cdot x}$

Rearranging the equation to solve for the extension $x$, we get:
$x=\frac{F\cdot L}{A���Y}$

We are given that the iron wire has a fixed volume ($V$). The volume of a cylindrical wire is the product of its cross-sectional area and its length:
$V=A\cdot L$

From this relation, we can express the cross-sectional area $A$ in terms of the constant volume $V$ and length $L$:
$A=\frac{V}{L}$

Substituting this expression for $A$ back into our equation for the extension $x$ yields:
$x=\frac{F\cdot L}{\left(\frac{V}{L}\right)\cdot Y}=\frac{F\cdot L^2}{V\cdot Y}$

Since the applied force ($F$), the volume of the iron ($V$), and the Young's modulus of iron ($Y$) are all constant, the extension $x$ depends only on the length $L$:
$x\propto L^2$

Therefore, the extension produced in the wire is directly proportional to the square of its length ($L^2$).