Q 11-03-085JEE MainJEE Main 2023 (31 Jan, Shift 2)Easy
A body is moving with constant speed, in a circle of radius $10\ \text{m}$. The body completes one revolution in $4\ \text{s}$. At the end of $3^\text{rd}$ second, the displacement of body (in m) from its starting point is
Answer: (D) $10\sqrt2$
In $3\ \text{s}$ the body turns through $\dfrac34\times360^\circ=270^\circ$. The chord between the start and end points subtends $90^\circ$ at the centre:
$$d=\sqrt{r^2+r^2}=10\sqrt2\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics