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

  • question_answer
    To double the length of a iron wire having \[0.5\,c{{m}^{2}}\] area of cross-section, the required force will be \[(Y={{10}^{12}}\,dyne/c{{m}^{2}})\]                             [MP PMT 1987]

    A)                 \[1.0\times {{10}^{-7}}N\]              

    B)                 \[1.0\times {{10}^{7}}N\]

    C)                 \[0.5\times {{10}^{-7}}N\]

    D)                             \[0.5\times {{10}^{12}}\]dyne

    Correct Answer: D

    Solution :

         If length of wire doubled then strain = 1             \[Y=\text{stress}\] Þ \[F=Y\times A\]\[={{10}^{12}}\times 0.5\]\[=0.5\times {{10}^{12}}dyne\]


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