NEET AIPMT SOLVED PAPER SCREENING 2010

  • question_answer
    A circular disk of moment of inertia \[{{\text{I}}_{\text{t}}}\] is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed \[{{\omega }_{\text{t}}}.\]Another disk of moment of inertia \[{{\text{I}}_{\text{b}}}\]is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed \[{{\omega }_{f}}.\]The energy lost by the initially rotating disc due to friction is

    A) \[\frac{1}{2}\frac{I_{b}^{2}}{({{I}_{t}}+{{I}_{b}})}\omega _{i}^{2}\]         

    B) \[\frac{1}{2}\frac{I_{t}^{2}}{({{I}_{t}}+{{I}_{b}})}\omega _{i}^{2}\]

    C)        \[\frac{1}{2}\frac{{{I}_{b}}-{{I}_{t}}}{({{I}_{t}}+{{I}_{b}})}\omega _{i}^{2}\]

    D)        \[\frac{1}{2}\frac{{{I}_{b}}{{I}_{t}}}{({{I}_{t}}+{{I}_{b}})}\omega _{i}^{2}\]

    Correct Answer: D

    Solution :

    Loss of energy,\[\Delta E=\frac{1}{2}{{I}_{t}}\omega _{i}^{2}-\frac{1}{2}\frac{I_{t}^{2}\omega _{i}^{2}}{2({{I}_{t}}+{{I}_{b}})}\] \[=\frac{1}{2}\frac{I_{b}^{{}}{{I}_{t}}\omega _{i}^{2}}{({{I}_{t}}+{{I}_{b}})}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner