• # 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) $\frac{2k}{I}\theta$

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