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 $$\pi : x\sqrt{2}$$. The value of $$x$$ will be

We have a particle moving with constant speed $$v$$ in a circular path of radius $$r$$. When it turns by $$90°$$, we need to find the ratio of instantaneous velocity to average velocity.

The instantaneous speed at any point is $$v$$. For a $$90°$$ turn, the displacement is the chord subtending $$90°$$ at the centre. If the particle moves from one point to another $$90°$$ apart, the displacement is

$$|\vec{d}| = \sqrt{r^2 + r^2} = r\sqrt{2}$$

Now, the time taken for a quarter circle is

$$t = \frac{\text{arc length}}{v} = \frac{\frac{2\pi r}{4}}{v} = \frac{\pi r}{2v}$$

So the magnitude of average velocity becomes

$$v_{avg} = \frac{|\vec{d}|}{t} = \frac{r\sqrt{2}}{\frac{\pi r}{2v}} = \frac{2v\sqrt{2}}{\pi}$$

The ratio of instantaneous velocity to average velocity is then

$$\frac{v}{v_{avg}} = \frac{v}{\frac{2v\sqrt{2}}{\pi}} = \frac{\pi}{2\sqrt{2}}$$

This gives the ratio as $$\pi : 2\sqrt{2}$$. Comparing with $$\pi : x\sqrt{2}$$, we get $$x = 2$$. So, the answer is $$2$$.