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JEE Main & Advanced Physics Wave Mechanics Question Bank Self Evaluation Test - Mechanical Properties of Solids

  • question_answer
    A 5 metre long wire is fixed to the ceiling. A weight of 10 kg is hung at the lower end and is 1 metre above the floor. The wire was elongated by t mm. The energy stored in the wire due to stretching is

    A) Zero

    B) 0.05 joule

    C) 100 joule

    D) 500 joule

    Correct Answer: B

    Solution :

    [b] \[W=\frac{1}{2}\times F\times l=\frac{1}{2}mgl\] \[=\frac{1}{2}\times 10\times 10\times 1\times {{10}^{-3}}=0.05J\]


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