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MGIMS WARDHA MGIMS WARDHA Solved Paper-2009

  • question_answer
    A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm. Then the elastic energy stored in the wire is

    A)  20 J                                       

    B)  0.1 J

    C)  0.2 J                                     

    D)  10 J

    Correct Answer: B

    Solution :

                     Elastic energy stored\[=\frac{1}{2}\times F\times l\] \[=\frac{1}{2}\times 200\times {{10}^{-3}}\] \[=0.1J\]


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