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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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Starting from rest, a particle rotates in a circle of radius $R=\sqrt 2 m$ with an angular acceleration $a=\large\frac{\pi}{4} $$rad/s^2$. The magnitude of average velocity of particle over time when it rotates quarter circle is

\[(a)\;1.5 m/s \quad (b)\;2 m/s \quad (c)\;1.25m/s \quad (d)\;1 m/s\]

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$\qquad=\large\frac{\sqrt 2 R}{t}$
$\qquad=\large\frac{2}{t}$ (since $ R=\sqrt 2 m)$
now $\theta=\large\frac{1}{2} \times a t^2$
$\qquad \large \frac{\pi}{2}=\large \frac{1}{2}$$ \times \large \frac{\pi}{4} $$t^2$
$\qquad t^2=4$
$\qquad t=2$
$v_{av}=\large \frac{2}{t}=\frac{2}{2}$$=1 m/s$
Hence d is the correct answer.


answered Jun 27, 2013 by meena.p
edited Mar 14, 2014 by thagee.vedartham

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