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

  • question_answer
     If a load of 9 kg is suspended on a wire, the increase in length is 4.5 mm. The force constant of the wire is

    A)                 \[0.49\times {{10}^{4}}\,N/m\]       

    B)                 \[1.96\times {{10}^{4}}\,N/m\]

    C)                 \[4.9\times {{10}^{4}}\,N/m\]

    D)                             \[0.196\times {{10}^{4}}\,N/m\]

    Correct Answer: B

    Solution :

        \[F=Kx\Rightarrow K=\frac{F}{x}=\frac{9\times 9.8}{4.5\times {{10}^{-3}}}=1.96\times {{10}^{4}}N/m\]


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