Question

A wire of length 'L' and cross-sectional area 'A' is made of a material of Young's modulus 'Y'. If the wire is stretched by an amount 'x', the work done is

• A. $\frac{YA{x}^2}{L}$
• B. $\frac{Y{A}^{2}x}{2L}$
• C. $\frac{YA{x}^2}{2L}$
• D. $\frac{YAx}{2L}$

Answer 1

The correct option is C $\frac{YA{x}^2}{2L}$

$\begin{array}{c}\text{Given.}\\ \text{length of wire}=L\\ \text{Area of wire}=4\\ \text{Young's modulus}=Y\\ \Delta L=x\end{array}$

$\begin{array}{c}\text{we know}\\ \text{work done}=\frac{1}{2}\times \text{stress}\times \text{strain}\times \text{volume}\\ Y=\frac{\text{stress}}{\text{strain}}\\ \text{stress}=Y\times \text{strain}\end{array}$

$\begin{array}{c}\text{work done}=\frac{1}{2}\times Y\times (\text{strain}{)}^2\times \text{volume}\\ \therefore \text{strain}=\frac{x}{L}\\ \therefore \text{volume}=AL\\ \omega =\frac{1}{2}\times Y\times \frac{x^2}{L^2}\times AL\\ \omega =\frac{Y{x}^{2}A}{2L}\end{array}$