A point marked on a ring of radius 2 cm is in contact with a horizontal plane. Now the ring is rolled forward half a revolution along the positive X - direction. Then the angle made by the displacement vector of the point with the X - axis is:

Let the radius of the ring be $R = 2$ cm. Initially, the point is at the origin $(0, 0)$ in contact with the horizontal plane (X-axis).

The ring rolls forward by half a revolution. The distance moved by the center of the ring is $d = R \times \theta_{rot} = R \times \pi = 2\pi$.

After half a revolution, the center of the ring is at $(2\pi, 2)$. The point, which was initially at the bottom, is now at the top of the ring. The coordinates of the point after half a revolution are $(2\pi, 2R) = (2\pi, 4)$.

The displacement vector $\vec{S}$ of the point from its initial position $(0, 0)$ to its final position $(2\pi, 4)$ is $\vec{S} = (2\pi - 0)\hat{i} + (4 - 0)\hat{j} = 2\pi\hat{i} + 4\hat{j}$.

The angle $\theta$ made by the displacement vector with the X-axis is given by $\tan \theta = \dfrac{\Delta y}{\Delta x} = \dfrac{4}{2\pi} = \dfrac{2}{\pi}$.

Therefore, $\theta = \tan ^{-1}\left(\dfrac{2}{\pi}\right)$.

Answer: $\theta=\tan ^{-1}\left(\dfrac{2}{\pi}\right)$