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(D) The rotational kinetic energy of the disc is given by $KE = \frac{1}{2}I\omega^2 = K	heta^2.$From this,we can express the angular velocity $\omeg

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$\frac{2K}{I}	heta$

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(D) The rotational kinetic energy of the disc is given by $KE = \frac{1}{2}I\omega^2 = K	heta^2.$From this,we can express the angular velocity $\omega$ as:$\omega^2 = \frac{2K	heta^2}{I} \implies \omega = \sqrt{\frac{2K}{I}}	heta \quad \dots(1)$Angular acceleration $\alpha$ is the rate of change of angular velocity with respect to time:$\alpha = \frac{d\omega}{dt} = \frac{d}{dt}\left( \sqrt{\frac{2K}{I}}	heta ight) = \sqrt{\frac{2K}{I}} \frac{d	heta}{dt}$Since $\frac{d	heta}{dt} = \omega,$ we substitute the expression for $\omega$ from equation $(1)$:$\alpha = \sqrt{\frac{2K}{I}} \left( \sqrt{\frac{2K}{I}}	heta ight)$$\alpha = \frac{2K}{I}	heta.$

$A$ 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

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