A wire of length $L$, area $A$ and Young's modulus $Y$ is stretched by hanging a mass $m$ from its free end. The elastic potential energy stored in the wire is
A$\dfrac{m^2 g^2 L}{2 A Y}$
B$\dfrac{mgL}{2AY}$
C$\dfrac{m^2 g^2 L}{A Y}$
D$\dfrac{mgL}{AY}$
Answer & Solution
Correct answer: A. $\dfrac{m^2 g^2 L}{2 A Y}$
Stress $\sigma = mg/A$, strain $\varepsilon = \sigma/Y = mg/(AY)$, energy density $u = \tfrac{1}{2}\sigma\varepsilon = \dfrac{m^2g^2}{2A^2Y}$. Total energy $U = u\cdot V = u\cdot AL = \dfrac{m^2 g^2 L}{2 A Y}$.
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