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MGIMS WARDHA MGIMS WARDHA Solved Paper-2011

  • question_answer
    A wire of length L and cross-sectional area A is made of a material of Young's modulus Y. It is stretched by an amount x. The work done is

    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, \[W=\frac{1}{2}\frac{YAx}{L}x=\frac{YA{{x}^{2}}}{2L}\]


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