A particle is moving with constant speed in a circular path. When the particle turns by an angle 90°, the ratio of instantaneous velocity to its average velocity is π: x√(2). The value of x will be - (a) 1 (b) 7 (c) 5 (d) 2

Answer: (d). For a 90° turn: displacement = r√2, arc = (π r)/(2), time t = (π r)/(2v). vavg = (r√2)/(π r/2v) = (2√2 v)/(π). vvavg = (π)/(2√2) = π: x√2 ⇒ x = 2. So the answer is (d) 2.

A particle is moving with constant speed in a circular path. When the… | JEE Main 2023 (6 Apr, Shift 1)

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Angular Variables

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: (d)

For a $90^\circ$ turn: displacement $= r\sqrt2$ , arc $= \frac{\pi r}{2}$ , time $t = \frac{\pi r}{2v}$ .
$v_{avg} = \frac{r\sqrt2}{\pi r/2v} = \frac{2\sqrt2\,v}{\pi}$ .
$\frac{v}{v_{avg}} = \frac{\pi}{2\sqrt2} = \pi : x\sqrt2 \Rightarrow x = 2$ .
So the answer is (d) 2.

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