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AIIMS AIIMS Solved Paper-2008

  • question_answer
    The adjacent graph shows the extension \[{{\left[ \frac{r}{2G({{m}_{1}}{{m}_{2}})} \right]}^{1/2}}\] of a wire of length 1 m suspended from the top of a root at one end with a load W connected to the other end. If the cross-sectional area of the wire is \[{{10}^{-6}}\ {{m}^{2}},\] calculate the Youngs modulus of the material of the wire.

    A)  \[2\times {{10}^{11}}\,N/{{m}^{2}}\] 

    B) (2)\[2\times {{10}^{-11}}N/{{m}^{2}}\]

    C) (3) \[3\times {{10}^{-12}}\,N/{{m}^{2}}\]

    D) (4) \[2\times {{10}^{-13}}\,N/{{m}^{2}}\]

    Correct Answer: A

    Solution :

    From the graph \[=\frac{3}{2}\times \frac{3}{5}=0.9A\] \[I=\frac{2}{1000}=2mA\] \[I=\frac{2}{2000}=1mA\]    \[\omega \] \[{{I}_{rms}}=\frac{{{V}_{rms}}}{\sqrt{{{R}^{2}}+{{\left( \frac{1}{\omega C} \right)}^{2}}}}\] \[\frac{{{I}_{rms}}}{2}=\frac{{{V}_{rms}}}{\sqrt{{{R}^{2}}+{{\left[ \frac{\frac{1}{\omega c}}{3} \right]}^{2}}}}=\frac{{{V}_{rms}}}{\sqrt{{{R}^{2}}+\frac{9}{{{\omega }^{2}}{{C}^{2}}}}}\]


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