A wheel of radius $1\text{m}$ 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

The correct option is C $\sqrt{{\pi }^2+4}$
Consider point $A$ as the point of contact of the wheel with ground.

We know for one full revolution, the wheel traces $2\pi R$ distance. So for half the revolution it will trace $\pi R$ distance.

So, the displacement:
$A{A}^{\prime }=\sqrt{(2R{)}^2+(\pi R{)}^2}A{A}^{\prime }=\sqrt{4+{\pi }^2}$