TS EAMCET 18-July-2022 Shift 1 Solved Paper

The situation given in the question is depicted belowLet the speed of particle is $v$ for circular motion. Clearly the gravitational force acting all the 3 masses will be same due to symmetry and let it be $F$ .Now; considering the mass $M$ placed at point $B$ . Since, it is performing circular motion in radius say $R$ , hence the net gravitational pull on it due to the other two masses are providing the required centripetal force.Now, net force on $M$ at $B$, $F' = \sqrt{F^{2}+F^{2}+2F^{2}\cos 60^{\circ}}$ $\Rightarrow \;\; F' = \sqrt{3} F \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(i)$and $\;\; F = \frac{G M^{2}}{l^{2}}$ (gravitational force of Newton)Also, by symmetry of forces the $F'$ will be acting along the bisector of angle at point $B$ i.e. at $30^{\circ}$ . from each force $F$.Now, in $\triangle O B P$ , $\cos 30^{\circ} = \frac{\frac{l}{2}}{R} \;\; \left( \because B P = \frac{B C}{2}\right)$ $\;\frac{\sqrt{3}}{2} = \frac{l}{2 R}$ $\Rightarrow \;\; R = \frac{l}{\sqrt{3}} \;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(ii)$Now, since $F' = F_{\;\text{centripetal }\;}$i.e. $\;\; F' = F_{c}$ $\Rightarrow \;\; \sqrt{3} F = \frac{M v^{2}}{R} \;\;$ [from Eq. (i) $] \left[ \because F_{c} = \frac{M v^{2}}{R}\right]$ $\Rightarrow \sqrt{3} \;\frac{G M^{2}}{l^{2}} = \frac{M v^{2}}{R}$Now, from Eq. (ii), $\sqrt{3} \;\frac{G M^{2}}{l^{2}} = \frac{M v^{2}}{l} \sqrt{3}$ $\Rightarrow \;\; v^{2} = \frac{G M}{l}$ $\Rightarrow \;\; v = \sqrt{\frac{G M}{l}}$