Question 289 of 308

A particle is moving with constant speed in a circular path. When the particle turns by an angle $90^{\circ},$ the ratio of instantaneous velocity to its average velocity is $\pi :x\sqrt{2}.$ The value of $x$ will be

A$2$B$5$C$1$D$7$

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Answer: (A) $2$

In circular motion with constant speed, magnitude of instantaneous velocity, ${v}_{ins}=\omega r$.
Time taken to turn by $90^{\circ}$ will be $=\dfrac{\dfrac{\pi }{2}}{\omega }=\dfrac{\pi }{2\omega }$.
Displacement will be equal to the length of $AB=\sqrt{{r}^{2}+{r}^{2}}=r\sqrt{2}$
Average velocity
${v}_{avg}=\dfrac{\text{displacement}}{\text{time}} =\dfrac{r\sqrt{2}\times 2\omega }{\pi }=\dfrac{2\sqrt{2}}{\pi }\omega r$
Hence, required ratio $\dfrac{{v}_{\text{ins}}}{{v}_{\text{avg}}}=\dfrac{\omega r\pi }{2\sqrt{2}\omega r}=\dfrac{\pi }{2\sqrt{2}}$
Therefore, $x=2$.