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A particle starting from rest moves along the circumference of a circle of radius ' $\mathrm{r}^{\prime}$ with angular acceleration ' $\alpha^{\prime}$ '. The magnitude of the average velocity, in the time it completes the small angular displacement ' $\theta$ ' is
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The correct answer is:
$r\left(\frac{\alpha \theta}{2}\right)^{\frac{1}{2}}$
(A)
$S=r \theta$
$\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2}$
$\therefore t=\sqrt{\frac{2 \theta}{\alpha}}$
Average velocity $=\frac{\text { displacement }}{\text { time }}=\frac{r \theta}{\sqrt{\frac{2 \theta}{\alpha}}}$
$=r \sqrt{\frac{\alpha}{2 \theta}} \cdot \theta=r \sqrt{\frac{\alpha \theta}{2}}$
$S=r \theta$
$\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2}$
$\therefore t=\sqrt{\frac{2 \theta}{\alpha}}$
Average velocity $=\frac{\text { displacement }}{\text { time }}=\frac{r \theta}{\sqrt{\frac{2 \theta}{\alpha}}}$
$=r \sqrt{\frac{\alpha}{2 \theta}} \cdot \theta=r \sqrt{\frac{\alpha \theta}{2}}$
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