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MGIMS WARDHA MGIMS WARDHA Solved Paper-2015

  • question_answer
    A wire suspended vertically from one of its ends is stretched by attaching a weight 100 N to the lower end. The wire stretches by 1mm. The elastic energy stored in the wire is

    A)  0.02J                    

    B)  0.05J  

    C)  0.1 J                     

    D)  0.2J

    Correct Answer: B

    Solution :

                    The elastic potential energy is given by \[U=\frac{1}{2}\times stress\times strain\times volume\] \[=\frac{1}{2}\times \frac{F}{A}\times \frac{\Delta l}{l}\times Al\] \[=\frac{1}{2}\times 100\times {{10}^{-3}}=\frac{0.1}{2}=0.05J\]


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