Q. If in a wire of Young’s modulus Y, longitudinal strain X is produced then the potential energy stored in its unit volume will be
A) $0.5\ {YX}^{2}$
B) $0.5\ Y^{2}X$
C) $2\ {YX}^{2}$
D) $\ {YX}^{2}$
Explanation
(a) Elastic potential energy per unit volume is given as
$U = \frac{1}{2} \times stress \times strain$
From definition of Young’s modulus of wire
$Y = \frac{stress}{strain}$
⟹ $stress = Y\ \times strain$
Given, $strain = X$
Therefore, $U = \frac{1}{2} \times YX^{2}$
⟹ $U = 0.5\ YX^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.