A) \[0.5y{{x}^{2}}\]
B) \[0.5{{y}^{2}}x\]
C) \[2y{{x}^{2}}\]
D) \[y{{x}^{2}}\]
Correct Answer: A
Solution :
When a wire is stretched work is done against the interatomic forces. This work is stored in the wire in the form of elastic potential energy. Elastic potential energy per unit volume is \[U=\frac{1}{2}\times stress\times strain\] From definition of Youngs modulus of wire \[\gamma =\frac{stress}{strain}\] \[\Rightarrow \] \[stress=Y\times strain\] Given, strain = X Therefore, \[U=\frac{1}{2}Y{{X}^{2}}\] \[\Rightarrow \] \[U=0.5Y{{X}^{2}}\]
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