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(D) The radius of the circular path is $R = 10 \, cm = 0.1 \, m$. The speed of the particle is $v = \sqrt{2} \, m/s$.When the particle covers $\frac{3

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$\frac{4}{3\pi} \, m/s$

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(D) The radius of the circular path is $R = 10 \, cm = 0.1 \, m$. The speed of the particle is $v = \sqrt{2} \, m/s$.When the particle covers $\frac{3}{4}$ of the circular path,its displacement $\Delta r$ is the distance between the initial point $A$ and the final point $B$. This forms the hypotenuse of a right-angled triangle with two sides of length $R$.Displacement $\Delta r = \sqrt{R^2 + R^2} = R\sqrt{2}$.The distance covered is $d = \frac{3}{4} 	imes (2\pi R) = \frac{3\pi R}{2}$.The time taken is $t = \frac{d}{v} = \frac{3\pi R}{2v}$.The magnitude of average velocity is $|v_{avg}| = \frac{	ext{Displacement}}{	ext{Time}} = \frac{R\sqrt{2}}{\frac{3\pi R}{2v}} = \frac{2\sqrt{2}v}{3\pi}$.Substituting $v = \sqrt{2} \, m/s$,we get $|v_{avg}| = \frac{2\sqrt{2} 	imes \sqrt{2}}{3\pi} = \frac{4}{3\pi} \, m/s$.

$A$ particle is moving with a constant speed of $\sqrt{2} \, m/s$ on a circular path of radius $10 \, cm$. Find the magnitude of the average velocity when it has covered $\frac{3}{4}$ of the circular path.

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