COMEDK 2025 Evening Solved Paper

Concept:Rolling without slipping combines translation of the centre and rotation of the ring; the net displacement of a point is the vector sum of both motions.Explanation:Take the horizontal plane as the X‑axis and the vertical upward direction as the Y‑axis.The initial contact point of the ring is at $(0,0)$.The centre of the ring (radius $R = 2\ \mathrm{cm}$) is initially at $(0,2)$.Half a revolution means the ring rotates by $\pi$ radians.In rolling without slipping, the centre moves horizontally by $s = R\pi = 2\pi\ \mathrm{cm}$.Thus the centre’s final position is $(2\pi,\ 2)$.After half a turn, the marked point is at the top of the ring, i.e., $2\ \mathrm{cm}$ above the centre.Hence its final position is $(2\pi,\ 2+2) = (2\pi,\ 4)$.The displacement vector of the point is $\Delta\mathbf{r} = (2\pi,\ 4)$.Its angle $\theta$ with the X‑axis satisfies $\tan\theta = \dfrac{4}{2\pi} = \dfrac{2}{\pi}$.Therefore $\theta = \tan^{-1}\!\left(\dfrac{2}{\pi}\right)$.Answer:Option B: $\theta = \tan^{-1}\!\left(\dfrac{2}{\pi}\right)$.