A wire of length L and cross-sectional area A is made of a material of Young’s modulus Y. It is stretched by an amount x. The work done is

The correct option is Yx²A/2L.

Step-by-step derivation:
When a wire of length $L$ and cross-sectional area $A$ is stretched by an amount $x$, we can find the work done by integrating the stretching force over the extension.

According to Young's modulus ($Y$), the relationship between force ($F$), extension ($y$), original length ($L$), and cross-sectional area ($A$) is given by:
$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{y/L}$

Rearranging the formula to find the restoring force $F$ at a general extension $y$:
$F=\frac{Y·A·y}{L}$

The small work done $dW$ in stretching the wire by an additional small amount $dy$ is:
$dW=F·dy=\frac{Y·A·y}{L}dy$

To find the total work done $W$ to stretch the wire from extension $y=0$ to $y=x$, we integrate the expression:
$W={\int }_0^x\frac{Y·A}{L}y·dy$

Integrating $y$ with respect to $y$ gives:
$W=\frac{Y·A}{L}{\left[\frac{y^2}{2}\right]}_0^x$

Evaluating the limits:
$W=\frac{Y·A}{L}·\frac{x^2}{2}=\frac{Y{x}^{2}A}{2L}$

Therefore, the work done in stretching the wire is $\frac{Y{x}^{2}A}{2L}$.