JEE Main & Advanced Physics Elasticity JEE PYQ-Elasticity

  • question_answer
    A uniformly tapering conical wire is made from a material of Young's modulus Y and has a normal, un extended length L. The radii, at the upper and lower ends of this conical wire , have values R and 3R, respectively, The upper end of the wire is fixed to a rigid support and a mass M is suspended from its lower and The equilibrium extended length, of this wire, would equal                                                                                                          [JEE ONLINE 09-04-2016]

    A) \[L\left( 1+\frac{1}{3}\frac{Mg}{\pi Y{{R}^{2}}} \right)\]

    B)      \[L\left( 1+\frac{2}{3}\frac{Mg}{\pi Y{{R}^{2}}} \right)\]

    C) \[L\left( 1+\frac{1}{9}\frac{Mg}{\pi Y{{R}^{2}}} \right)\]

    D)      \[L\left( 1+\frac{2}{9}\frac{Mg}{\pi Y{{R}^{2}}} \right)\]

    Correct Answer: A

    Solution :

    [a]
    \[\frac{r-R}{x}=\frac{3R-R}{L}\Rightarrow r=R\left( 1+\frac{2x}{L} \right);Y=\frac{Mg}{\pi {{R}^{2}}\frac{dL}{dx}}\]
                \[dL\frac{Mg}{\pi {{R}^{2}}};\frac{dx}{{{\left( 1+\frac{2x}{L} \right)}^{2}}}\]
                \[\Delta L=\frac{Mg}{Y\pi {{R}^{2}}}\int\limits_{0}^{L}{\frac{dx}{{{\left( 1+\frac{2x}{L} \right)}^{2}}}=\frac{MgL}{3\pi {{R}^{2}}Y};}\]
    \[L'=L+\Delta K=L\left( 1+\frac{1}{3}\frac{Mg}{\pi {{R}^{2}}Y} \right)\]


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