A wheel of radius 1 m rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is

Horizontal distance covered by the wheel in half revolution $\pi R.$

So, the displacement of the point which was initially in contact with ground

$=AA\text{'}=\sqrt{{\left(\pi R\right)}^2+{\left(2R\right)}^2}$

$=R\sqrt{n^2+4}=\sqrt{{\pi }^2+4}[\because R=1m]$