Question

A wire of length L and of cross-sectional area A is made of a material of Young's modulus Y. The work done in stretching the wire by an amount $x$ is given by :

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

Answer 1

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

Work done is equal to $\frac{1}{2}\times$ stress $\times$ strain $\times$ volume

Let the work done be $W$ , strain be $S$ and volume be $V$

We have $W=\frac{1}{2}\times Y\times S^2\times V$

$\Rightarrow W=\frac{1}{2}\times Y\times (\frac{x}{L}{)}^2\times AL$

$\Rightarrow W=\frac{YA{x}^2}{2L}$

Therefore option $B$ is correct