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

  • question_answer
    When a 4 kg mass is hung vertically on a light spring that obeys Hooke's law, the spring stretches by 2cms. The work required to be done by an external agent in stretching this spring by 5cms will be \[(g=9.8\text{ }m/se{{c}^{2}})\]

    A) 4.900joule

    B) 2.450joule

    C) 0.495 joule

    D) 0.245 joule

    Correct Answer: B

    Solution :

    [b] \[K=\frac{F}{x}=\frac{4\times 9.8}{2\times {{10}^{-2}}}=19.6\times {{10}^{2}}\] Work done \[=\frac{1}{2}19.6\times {{10}^{2}}\times {{(0.05)}^{2}}=2.45J\]


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