Q. A metallic rod of length $l$ and cross-sectional area $A$ is made of a material of Young modulus $Y$. If the rod is elongated by an amount $y$, then the work done is proportional to
A) $y$
B) $\frac{1}{y}$
C) $y^{2}$
D) $\frac{1}{y^{2}}$
Explanation
(c) Volume $V =$ cross sectional $A \times$ length $l$ or $V = Al$
Strain $= \frac{Elongation}{Original\ length} = \frac{v}{l}$
Young’s modulus $Y = \frac{stress}{strain}$
Work done, $W = \frac{1}{2} \times stress \times strain \times volume$
$W = \frac{1}{2} \times Y \times (strain)^{2} \times Al$
$= \frac{1}{2} \times Y \times \left( \frac{y}{l} \right)^{2} \times Al = \frac{1}{2}\left( \frac{YA}{l} \right)y^{2} \Rightarrow W \propto y^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.