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JEE Main & Advanced AIEEE Solved Paper-2003

  • 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     AIEEE  Solved  Paper-2003

    A) \[0.2\] J                               

    B) 10 J                        

    C) 20 J                                        

    D) \[0.1\]J

    Correct Answer: D

    Solution :

    Elastic energy stored in the wire is                     \[U=\frac{1}{2}\times \]Stress \[\times \] Strain \[\times \] Volume                     \[=\frac{1}{2}\frac{F}{A}\times \frac{\Delta l}{L}\times AL=\frac{1}{2}F\Delta l\]                 \[=\frac{1}{2}\times 200\times 1\times {{10}^{-3}}=0.1\,\,J\]


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