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JEE Main & Advanced Physics Elasticity Question Bank Work Done in Stretching a Wire

  • question_answer
    When a 4 kg mass is hung vertically on a light spring that obeys Hooke's law, the spring stretches by 2 cms. The work required to be done by an external agent in stretching this spring by 5 cms will be \[(g=9.8\,metres/sex{{c}^{2}})\] [MP PMT 1995]

    A) 4.900 joule           

    B)                 2.450 joule

    C)                 0.495 joule

    D)                             0.245 joule

    Correct Answer: B

    Solution :

    \[K=\frac{F}{x}\]\[=\frac{40}{2\times {{10}^{-2}}}=0.2\ N/m\] Work done\[=\frac{1}{2}K{{x}^{2}}=\frac{1}{2}\times (0.2)\times {{(0.05)}^{2}}=2.5\ J\]


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