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(D) Given, angular deceleration $\propto \sqrt{\omega}$.So, $-\frac{d\omega}{dt} = k\sqrt{\omega}$, where $k$ is a constant.Rearranging and integratin

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$\frac{\omega_0}{3}$

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(D) Given, angular deceleration $\propto \sqrt{\omega}$.So, $-\frac{d\omega}{dt} = k\sqrt{\omega}$, where $k$ is a constant.Rearranging and integrating: $-\int_{\omega_0}^{\omega} \omega^{-1/2} d\omega = \int_{0}^{t} k dt$.$-[2\sqrt{\omega}]_{\omega_0}^{\omega} = kt \Rightarrow 2(\sqrt{\omega_0} - \sqrt{\omega}) = kt$.$\sqrt{\omega} = \sqrt{\omega_0} - \frac{kt}{2}$.At $\omega = 0$, the total time $	au = \frac{2\sqrt{\omega_0}}{k}$.Mean angular speed $\langle \omega angle = \frac{1}{	au} \int_{0}^{	au} \omega dt$.Since $\sqrt{\omega} = \sqrt{\omega_0} - \frac{kt}{2}$, we have $\omega = (\sqrt{\omega_0} - \frac{kt}{2})^2 = \omega_0 + \frac{k^2t^2}{4} - kt\sqrt{\omega_0}$.Integrating $\omega$ from $0$ to $	au = \frac{2\sqrt{\omega_0}}{k}$:$\int_{0}^{	au} (\omega_0 + \frac{k^2t^2}{4} - kt\sqrt{\omega_0}) dt = [\omega_0 t + \frac{k^2t^3}{12} - \frac{kt^2}{2}\sqrt{\omega_0}]_{0}^{	au}$.Substituting $	au = \frac{2\sqrt{\omega_0}}{k}$:$= \omega_0(\frac{2\sqrt{\omega_0}}{k}) + \frac{k^2}{12}(\frac{8\omega_0\sqrt{\omega_0}}{k^3}) - \frac{k}{2}\sqrt{\omega_0}(\frac{4\omega_0}{k^2}) = \frac{2\omega_0\sqrt{\omega_0}}{k} + \frac{2\omega_0\sqrt{\omega_0}}{3k} - \frac{2\omega_0\sqrt{\omega_0}}{k} = \frac{2\omega_0\sqrt{\omega_0}}{3k}$.Finally, $\langle \omega angle = \frac{2\omega_0\sqrt{\omega_0}/3k}{2\sqrt{\omega_0}/k} = \frac{\omega_0}{3}$.

$A$ body rotates about a stationary axis. If the angular deceleration is proportional to the square root of the angular speed, then the mean angular speed of the body, given $\omega_0$ as the initial angular speed, is

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