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

  • question_answer
    A copper wire of length 4.0m and area of cross-section \[1.2\,c{{m}^{2}}\] is stretched with a force of \[4.8\times {{10}^{3}}\] N. If Young?s modulus for copper is \[1.2\times {{10}^{11}}\,N/{{m}^{2}},\] the increase in the length of the wire will be                             [MP PET 2001]

    A) 1.33 mm

    B)                      1.33 cm

    C)             2.66 mm

    D)                      2.66 cm

    Correct Answer: A

    Solution :

                    \[l=\frac{FL}{AY}=\frac{4.8\times {{10}^{3}}\times 4}{1.2\times {{10}^{-4}}\times 1.2\times {{10}^{11}}}=1.33\ mm\]


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