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

  • question_answer
    On applying a stress of \[20\times {{10}^{8}}\] N/\[{{m}^{2}}\] the length of a perfectly elastic wire is doubled. Its Young?s modulus will be [MP PET 2000]

    A)             \[40\times {{10}^{8}}N/{{m}^{2}}\]

    B)                      \[20\times {{10}^{8}}N/{{m}^{2}}\]

    C)             \[10\times {{10}^{8}}N/{{m}^{2}}\]

    D)                      \[5\times {{10}^{8}}N/{{m}^{2}}\]

    Correct Answer: B

    Solution :

                    Young?s modules =\[\frac{\text{stress}}{\text{strain}}\] As the length of wire get doubled therefore strain = 1 Y = strain = 20 ×108 N/ m2


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