A) \[\frac{1}{2}\] x load x extension
B) \[\frac{1}{2}\]- x stress x strain
C) \[\frac{1}{2}\]- x load x strain
D) \[\frac{1}{2}\]- x load x stress
Correct Answer: B
Solution :
When a wire is stretched, work is done against the interatomic force. This work is stored in the form of elastic potential energy. Let Y is Young's modulus, F is force, A is area, L is length of wire and \[l\]is increase in length then \[Y=\frac{stress}{strain}=\frac{F/A}{l/L}=\frac{FL}{Al}\] \[\Rightarrow \] \[F=\frac{YA}{L}l.\] When wire is further increased by infinitesimal- length, then work done is \[dW=F\times dl=\frac{YA}{L}ldl\] \[W=\int_{0}^{l}{\frac{YA}{L}l\,dl=\frac{YA}{L}\left[ \frac{{{l}^{2}}}{2} \right]_{0}^{l}=\frac{1}{2}}YA\frac{{{l}^{2}}}{L}\] \[=\frac{1}{2}\left( U\frac{l}{L} \right)\left( \frac{l}{L} \right)(A\,L)\] \[\text{=}\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ stress }\!\!\times\!\!\text{ strain }\!\!\times\!\!\text{ volume}\] So, elastic potential energy per unit volume is \[u=\frac{1}{2}\text{stres }\!\!\times\!\!\text{ strain}\]
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