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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

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

    E)  \[4(A.L)(YL)\left( \frac{x}{L} \right)\]

    Correct Answer: A

    Solution :

    Work done \[=\frac{1}{2}\times \] stress \[\times \] strain \[\times \] 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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