Question

A wire is subjected to a tensile stress. If A represents area of cross-section, L represents original length, I represents extension and Y is Young's modulus of elasticity, then elastic potential energy of the stretched wire is

• A. $U=\frac{2L}{AY}{I}^2$
• B. $U=\frac{AL}{2Y}{I}^2$
• C. $U=\frac{AY}{2L}{I}^2$
• D. $U=\frac{1}{4}\frac{AY}{L}{I}^2$

Answer 1

The correct option is C $U=\frac{AY}{2L}{I}^2$

Work done $=\frac{1}{2}\times stress\times strain\times volume$

$=\frac{1}{2}\times Y\times {\left(strain\right)}^2\times volume$

$=\frac{1}{2}\times Y\times {\left(\frac{I}{L}\right)}^2\times AL$

$=\frac{1}{2}\times Y\times \frac{I^2}{L^2}\times AL$

$=\frac{Y{I}^{2}A}{2L}$

$=\frac{Y{I}^{2}A}{2L}$

Hence, the answer is $\frac{Y{I}^{2}A}{2L}.$