A) \[\frac{YxA}{2L}\]
B) \[\frac{Y{{x}^{2}}A}{L}\]
C) \[\frac{Y{{x}^{2}}A}{2L}\]
D) \[\frac{2Y{{x}^{2}}A}{L}\]
Correct Answer: C
Solution :
Work done \[W=\frac{1}{2}Fx\]where\[x\]is increase in length. Young's modulus \[Y=\frac{FL}{Ax}\] \[F=\frac{TAx}{L}\] Work done, \[W=\frac{1}{2}\frac{YAx}{L}x=\frac{YA{{x}^{2}}}{2L}\]You need to login to perform this action.
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