A) \[\frac{1}{2}\]\[\times \]stress\[\times \]strain
B) stress\[\times \]strain
C) \[\frac{1}{2}\] \[\times \]load \[\times \]strain
D) load \[\times \]strain
Correct Answer: A
Solution :
When a wire is stretched, work is done against the interatomic forces, this work is stored as elastic potential energy in the wire. Work done = average force\[\times \]increase in length \[W=\frac{F}{2}\times l=U\] Let A be area of cross-section of the wire, then \[U=\frac{1}{2}\left( \frac{F}{A} \right)\times \left( \frac{l}{L} \right)\times LA\] \[U=\frac{1}{2}stress\times strain\times volume\text{ }of\text{ }wire\] Hence, energy per unit volume \[U=\frac{1}{2}\times stress\times strain\]
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