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BVP Medical BVP Medical Solved Paper-2003

  • question_answer
    A wire of length L and cross-sectional area A is made of a material of Youngs modulus Y. If the wire is stretched by the amount x. The work done is :

    A) \[\frac{YA{{x}^{2}}}{2L}\]                                            

    B) \[\frac{YA{{x}^{2}}}{L}\]

    C) \[Ya{{x}^{2}}L\]                

    D) \[\frac{Yax}{2L}\]

    Correct Answer: A

    Solution :

                    Energy stored is \[U=\frac{1}{2}\times force\times extension=\frac{1}{2}Fx\] ??.(i)                 and   \[Y=\frac{F}{A}\frac{L}{x}\] or  \[F=\frac{YA{{x}^{2}}}{L}\]                ??.(ii) From equations (i) and (ii), work done                 \[=\frac{1}{2}\left( \frac{YAx}{L} \right)\times x=\frac{1}{2}\frac{YA{{x}^{2}}}{L}\]


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