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

  • question_answer
    If the interatomic spacing in a steel wire is 3.0Å and \[{{Y}_{steel}}\]= \[20\times {{10}^{10}}N/{{m}^{2}}\] then force constant is [RPET 2001]

    A)             \[6\times {{10}^{-2}}\,N/{\AA}\]

    B)                      \[6\times {{10}^{-9}}N/{\AA}\]

    C)             \[4\times {{10}^{-5}}\,N/{\AA}\]

    D)                      \[6\times {{10}^{-5}}N/{\AA}\]

    Correct Answer: B

    Solution :

                    \[K=Y{{r}_{0}}=20\times {{10}^{10}}\times 3\times {{10}^{-10}}=60\ N/m\]                          \[=6\times {{10}^{-9}}N/{\AA}\]


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