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JEE Mains · Physics · STD 11 - 3.2 motion in plane

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

  1. A \(2\)
  2. B \(5\)
  3. C \(1\)
  4. D \(7\)
Verified Solution

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