Q 11-07-095JEE MainJEE Main 2023 (15 Apr, Shift 1)Easy
Two identical particles each of mass $m$ go round a circle of radius $a$ under the action of their mutual gravitational attraction. The angular speed of each particle will be
Answer: (C) $\sqrt{\dfrac{Gm}{4a^3}}$
The particles are $2a$ apart: $\dfrac{Gm^2}{(2a)^2}=m\omega^2a\Rightarrow\omega=\sqrt{\dfrac{Gm}{4a^3}}$.
Solution by Sreeraj P, M.Sc Physics