AFMC AFMC Solved Paper-2009

  • question_answer
    A wire of length L and area of cross-section A is made of material of Young's modulus Y. If the wire is stretched by the amount x, the work done is

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

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

    C) \[\frac{YAx}{2L}\]                                           

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

    Correct Answer: A

    Solution :

    If a force F acts along the length L of the wire of corss-section A and stretches it by \[x,\]then \[Y=\frac{stress}{strain}\] \[=\frac{F/A}{x/L}\] \[\frac{FL}{Ax}\] \[F=\frac{YA}{L}x\] So, the work done for an additional small increase \[dx\]in length Hence the total work done in increasing the length by \[l\] \[W=\int_{0}^{x}{dW=\int_{0}^{x}{Fdx}}\] \[=\int_{0}^{x}{\frac{YA}{L}x.dx}\] \[=\frac{1}{2}\frac{YA}{L}{{x}^{2}}\]


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