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JEE Main & Advanced Physics Elasticity Question Bank Youngs Modulus and Breaking Stress

  • question_answer
     Longitudinal stress of \[1\,kg/m{{m}^{2}}\] is applied on a wire. The percentage increase in length is \[(Y={{10}^{11}}\,N/{{m}^{2}})\]

    A) 0.002     

    B)                 0.001

    C)                 0.003     

    D)                 0.01

    Correct Answer: B

    Solution :

        Longitudinal strain \[\frac{l}{L}=\frac{\text{stress}}{Y}=\frac{{{10}^{6}}}{{{10}^{11}}}={{10}^{-5}}\] Percentage increase in length \[={{10}^{-5}}\times 100=0.001%\]


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