1. Given: A particle moves with constant speed in a circle.

   We need the ratio: $\text{instantaneous velocity} : \text{average velocity} = \pi : x\sqrt{2}$

   after the particle has turned through $90^\circ$.

2. Instantaneous speed / velocity magnitude

   Since the particle moves with constant speed $v$ in the circular path, the magnitude of instantaneous velocity is simply $v.$

3. Time taken to turn through $90^\circ$

   If the radius of the circle is $R$, then the arc length for $90^\circ = \frac{\pi}{2}$ rad is $s = R\theta = R\cdot \frac{\pi}{2} = \frac{\pi R}{2}.$

   Since speed is constant and equal to $v$, $t = \frac{s}{v} = \frac{\pi R}{2v}.$

4. Displacement after turning through $90^\circ$

   The displacement is the chord subtending $90^\circ$ at the center: $d = 2R\sin\frac{90^\circ}{2} = 2R\sin45^\circ = 2R\cdot \frac{1}{\sqrt{2}} = R\sqrt{2}.$

5. Average velocity magnitude

   Average velocity magnitude is $\frac{\text{displacement}}{\text{time}} = \frac{R\sqrt{2}}{\pi R/(2v)}.$

   Simplifying, $v_{\text{avg}} = R\sqrt{2}\cdot \frac{2v}{\pi R} = \frac{2\sqrt{2}}{\pi}v.$

6. Required ratio

   Therefore,

   = v : \frac{2\sqrt{2}}{\pi}v.$$ Cancel $v$: $$= 1 : \frac{2\sqrt{2}}{\pi} = \pi : 2\sqrt{2}.$$

7. Compare with given form

   Given ratio is $\pi : x\sqrt{2}.$

   Hence, $x\sqrt{2} = 2\sqrt{2} \implies x=2.$

8. Option check

   • A: $1$ ❌
   • B: $7$ ❌
   • C: $5$ ❌
   • D: $2$ ✅

Therefore, the correct answer is D.