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MGIMS WARDHA MGIMS WARDHA Solved Paper-2006

  • question_answer
    A one metre long steel wire of cross-sectional area 1 mm2 is extended by 1 mm. If \[\text{Y=2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{11}}}\,\text{N/}{{\text{m}}^{\text{2}}}\text{,}\]then the work done is :

    A) 0.1J                                       

    B) 0.2J

    C) 0.3J                                       

    D) 0.4J

    Correct Answer: A

    Solution :

    \[F=\frac{YA.\Delta L}{L}\] Work done \[=\frac{1}{2}F.\Delta L\]                                 \[=\frac{1}{2}\frac{YA.\,\Delta {{L}^{2}}}{L}\]                                 \[=\frac{2\times {{10}^{11}}\times {{10}^{-6}}\times {{10}^{-6}}}{2\times 1}\]                                 \[=0.1\,J\]


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