CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    The Youngs modulus of the material of a wire is\[2\times {{10}^{10}}N{{m}^{-2}}\].If the elongation strain is 1%, then the energy stored in the wire per unit: volume in\[J{{m}^{-3}}\]is

    A)  \[{{10}^{6}}\]                   

    B)         \[{{10}^{8}}\]

    C)  \[2\times {{10}^{6}}\]                  

    D)         \[2\times {{10}^{8}}\]

    E)  \[0.5\times {{10}^{6}}\]

    Correct Answer: A

    Solution :

    Per unit volume energy stored \[=\frac{1}{2}\times Y\times {{(strain)}^{2}}\] \[=\frac{1}{2}\times Y\times {{\left( \frac{l}{L} \right)}^{2}}\] Given   \[l=L\times 1%\] Or           \[l=\frac{L}{100}\] Stored energy \[Y=\frac{1}{2}\times 2\times {{10}^{10}}\times {{\left( \frac{L}{100L} \right)}^{2}}\] \[={{10}^{6}}J{{m}^{-3}}\]


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