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MGIMS WARDHA MGIMS WARDHA Solved Paper-2008

  • question_answer
    A brass rod of cross-sectional area 1 cm2 and length 0.2 m is compressed lengthwise by a weight of 5 kg of Young's modulus of elasticity - of brass is \[1\times {{10}^{11}}N/{{m}^{2}}\]and \[g=10\text{ }m/{{s}^{2}},\]then increase in the energy of the rod will be'

    A) \[{{10}^{-5}}J\]                                 

    B) \[2.5\times {{10}^{-5}}J\]

    C) \[5\times {{10}^{-5}}J\]                                

    D) \[2.5\times {{10}^{-4}}J\]

    Correct Answer: B

    Solution :

    \[U=\frac{1}{2}\frac{{{(stress)}^{2}}}{Y}\times volume\] \[=\frac{1}{2}\frac{{{F}^{2}}\times A\times L}{{{A}^{2}}\times Y}\] \[=\frac{1}{2}\times \frac{{{F}^{2}}L}{AY}\] \[=\frac{1}{2}\frac{{{(50)}^{2}}\times 0.2}{1\times {{10}^{-4}}\times 1\times {{10}^{11}}}\] \[=2.5\times {{10}^{-5}}J\]


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