We have assumed the centre of the two concentric circles as O and have joined O with P and T , where OT is perpendicular on PQ (Radius is always perpendicular on tangent).

We know $\(\mathrm{OT}=2\)$ and $\(\mathrm{OP}=5\)$ are the radii of the two circles.

Now, applying Pythagoras theorem i.e. Hypotenuse $\({ }^2=\)$ Perpendicular $\(^2+\)$ Base $\(^2\)$ in the triangle POT, we get $\(\mathrm{PT}=\sqrt{\mathrm{PO}^2-\mathrm{OT}^2}=\sqrt{5^2-2^2}=\sqrt{2 \mathrm{I\)$

Note: - Perpendicular drawn from the centre of the circle to the chord always bisects the chord and as PQ is a chord to the bigger circle and OT is perpendicular drawn from the centre of the circle, we get $\(\mathrm{PT}=\mathrm{TQ}\)$

Thus we get $\(P Q=2(P T)=2 \sqrt{21}\)$, so the answer is $\((B)\)$.