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

  • question_answer
    The diameter of a brass rod is 4 mm and Young's modulus of brass is \[9\times {{10}^{10}}\,N/{{m}^{2}}\]. The force required to stretch by 0.1% of its length is             [MP PET 1991; BVP 2003]

    A)                 \[360\,\pi N\]

    B)                             36 N

    C)                 \[144\pi \times {{10}^{3}}N\]

    D)                             \[36\pi \times {{10}^{5}}N\]

    Correct Answer: A

    Solution :

        \[F=\frac{YAl}{L}=\frac{9\times {{10}^{10}}\times \pi \times 4\times {{10}^{-6}}\times 0.1}{100}=360\ \pi \ N\]


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