JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of \[\theta ,\]where \[\theta \] is the angle by which it has rotated, is given as \[k{{\theta }^{2}}.\]If its moment of inertia is I then the angular acceleration of the disc is : [JEE Main 9-4-2019 Morning]

    A) \[\frac{k}{2I}\theta \]                         

    B) \[\frac{k}{I}\theta \]

    C) \[\frac{k}{4I}\theta \]                         

    D) \[\frac{2k}{I}\theta \]

    Correct Answer: D

    Solution :

    Kinetic energy \[KE=\frac{1}{2}I{{\omega }^{2}}=k{{\theta }^{2}}\]             \[\Rightarrow {{\omega }^{2}}=\frac{2k{{\theta }^{2}}}{I}\Rightarrow \omega =\sqrt{\frac{2k}{I}}\theta \]              ?(1) Differentiate (1) wrt time\[\to \] \[\frac{d\omega }{dt}=\alpha =\sqrt{\frac{2k}{I}}\left( \frac{d\theta }{dt} \right)\] \[\Rightarrow \]\[\alpha =\sqrt{\frac{2k}{I}}.\sqrt{\frac{2k}{I}}\theta \{by\,(1)\}\] \[\Rightarrow \alpha =\frac{2k}{I}\theta \] Option

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