Q 11-06-120JEE MainJEE Main 2023 (13 Apr, Shift 1)Medium
A disc is rolling without slipping on a surface. The radius of the disc is $R$. At $t=0$, the top most point on the disc is $A$ as shown in figure. When the disc completes half of its rotation, the displacement of point $A$ from its initial position is
Answer: (B) $R\sqrt{\pi^2+4}$
In half a rotation the centre moves $\pi R$ forward and $A$ goes from the top to the bottom (down by $2R$):
$$d=\sqrt{(\pi R)^2+(2R)^2}=R\sqrt{\pi^2+4}$$
Solution by Sreeraj P, M.Sc Physics