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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer
    Two coaxial discs, having moments of inertia \[{{I}_{1}}\]and \[\frac{{{I}_{1}}}{2},\]are rotating with respective angular velocities \[{{\omega }_{1}}\]and \[\frac{{{\omega }_{1}}}{2},\]about their common axis. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If \[{{E}_{f}}\]and\[{{E}_{i}}\]are the final and initial total energies, then \[({{E}_{f}}-{{E}_{i}})\]is : [JEE Main 10-4-2019 Morning]

    A) \[\frac{{{I}_{1}}\omega _{1}^{2}}{12}\]                        

    B) \[\frac{3}{8}{{I}_{1}}\omega _{1}^{2}\]

    C) \[\frac{{{I}_{1}}\omega _{1}^{2}}{6}\]                          

    D) \[\frac{{{I}_{1}}\omega _{1}^{2}}{24}\]

    Correct Answer: D

    Solution :

    \[{{E}_{i}}=\frac{1}{2}{{I}_{1}}\times \omega _{1}^{2}+\frac{1}{2}\frac{{{I}_{1}}}{2}\times \frac{\omega _{1}^{2}}{4}\] \[=\frac{{{I}_{1}}\omega _{1}^{2}}{2}\left( \frac{9}{8} \right)=\frac{9}{16}{{I}_{1}}\omega _{1}^{2}\] \[{{I}_{1}}{{\omega }_{1}}+\frac{{{I}_{1}}\omega _{1}^{{}}}{4}=\frac{3{{I}_{1}}}{2}\omega \] \[\frac{5}{4}{{I}_{1}}\omega _{1}^{{}}=\frac{3{{I}_{1}}}{2}\omega \] \[\omega =\frac{5}{6}{{\omega }_{1}}\] \[{{E}_{f}}=\frac{1}{2}\times \frac{3{{I}_{1}}}{2}\times \frac{25}{36}\omega _{1}^{2}\]\[=\frac{25}{48}{{I}_{1}}\omega _{1}^{2}\] \[\Rightarrow \]\[{{E}_{f}}-{{E}_{i}}={{I}_{1}}\omega _{1}^{2}\left( \frac{25}{48}-\frac{9}{16} \right)=\frac{-2}{48}{{I}_{1}}\omega _{1}^{2}\] \[=\frac{-{{I}_{1}}\omega _{1}^{2}}{24}\]


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