A) \[\frac{1}{2}\times load\times extension\]
B) \[\frac{1}{2}\times stress\times strain\]
C) \[\frac{1}{2}\times load\times strain\]
D) \[\frac{1}{2}\times load\times stress\]
Correct Answer: B
Solution :
When a wire is stretched, work is done against the interatomic forces. This work is stored in the form of elastic potential energy. Let\[Y\]is Youngs 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 all, then work done is \[dW=F\times dl=\frac{YA}{L}ldl\] \[W=\int_{0}^{l}{\frac{YA}{L}l\,\,dl}=\frac{YA}{L}\left[ \frac{{{l}^{2}}}{2} \right]_{0}^{l}=\frac{1}{2}YA\frac{{{l}^{2}}}{L}\] \[=\frac{1}{2}\left( y\frac{l}{L} \right)\left( \frac{l}{L} \right)(A\,\,l)\] \[=\frac{1}{2}\times stress\times strain\times volume\] So, elastic potential energy per unit volume is \[U=\frac{1}{2}\times stress\times strain\]
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