A particle is moving with constant speed in a circular path. When particle turns by an angle 90^{cir

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3-2.Motion in Plane

hard

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$7$

(JEE MAIN-2023)

Solution

Let instantaneous velocity be v. time,
$t =\frac{\text { Arc length }}{ v }=\frac{2 \pi \frac{ R }{4}}{ v }=\frac{\pi R }{2 v }$ average velocity,
$\langle v \rangle =\frac{ AB }{ t }=\frac{ R \sqrt{2}(2 v )}{\pi R }=\frac{2 \sqrt{2} V }{\pi}$
$\Rightarrow \frac{ V }{\langle V \rangle} =\frac{\pi}{2 \sqrt{2}} .$

Standard 11

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