Q. A wire of natural length $l$, Young’s modulus Y and area of cross-section A is extended by $x$. Then the energy stored in the wire is given by
A) $\frac{1}{2}\ \frac{YA}{l}x^{2}$
B) $\frac{1}{3}\ \frac{YA}{l}x^{2}$
C) $\frac{1}{2}\ \frac{Yl}{A}x^{2}$
D) $\frac{1}{2}\ \frac{YA}{l^{2}}x^{2}$
Explanation
(a) Energy stored in the wire
$U = \frac{1}{2}Y \times (strain)^{2} \times volume$
or $U = \frac{1}{2}Y \times \left( \frac{x}{l} \right)^{2} \times Al$
or $U = \frac{1}{2}\frac{{Yx}^{2}}{l} \times$A
or $U = \frac{1}{2}\frac{YA}{l}x^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.