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JEE Mains · Physics · STD 11 - Motion in plane
  • JEE Mains 2023
  • Hard
  • Numerical

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 \(.........\)

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Answer & Solution

Correct Answer

(A) \(2\)

Step-by-step Solution

Detailed explanation

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,

\(\langle v \rangle =\frac{ AB }{ t }=\frac{ R \sqrt{2}(2 v )}{\pi R }=\frac{2 \sqrt{2} V }{\pi}\)

\(\Rightarrow \frac{ V }{\langle V \rangle} =\frac{\pi}{2 \sqrt{2}} .\)

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