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WB JEE Medical WB JEE Medical Solved Paper-2007

  • question_answer
    A wire of length L and area of cross-section A is stretched through a distance \[x\]metre by applying a force F along length, then the work done in this process is (Y is Youngs modulus of the material)

    A)  \[\frac{1}{2}(A.L)\left( \frac{Yx}{L} \right)\left( \frac{x}{L} \right)\]

    B)  \[(A.L)(YL)\left( \frac{x}{L} \right)\]

    C)  \[2(A.L)(YL)\left( \frac{x}{L} \right)\]

    D)  \[3(A.L)(YL)\left( \frac{x}{L} \right)\]

    Correct Answer: A

    Solution :

                     Work done \[\text{=}\frac{\text{1}}{\text{2}}\,\,\text{ }\!\!\times\!\!\text{ }\,\text{stress}\,\text{ }\!\!\times\!\!\text{ }\,\text{strain}\,\text{ }\!\!\times\!\!\text{ }\,\text{volume}\] or            \[W=\frac{1}{2}Y\frac{x}{L}\times \frac{x}{L}\times AL\] \[=\frac{1}{2}(AL)\left( Y+\frac{x}{L} \right)\left( \frac{x}{L} \right)\]


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