question_answer

9. Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is :   JEE Main  Solved  Paper-2014

A) \[\sqrt{\frac{GM}{R}\left( 1+2\sqrt{2} \right)}\]               

B) \[\frac{1}{2}\sqrt{\frac{GM}{R}\left( 1+2\sqrt{2} \right)}\]

C) \[\sqrt{\frac{GM}{R}}\]                

D) \[\sqrt{2\sqrt{2}\frac{GM}{R}}\]

Correct Answer: B

Solution :

\[\left( \sqrt{2}F+\frac{F}{2} \right)=\frac{m{{v}^{2}}}{R}\] \[\left( \frac{2\sqrt{2}+1}{2} \right)\left( \frac{G{{m}^{2}}}{{{a}^{2}}} \right)=\frac{m{{v}^{2}}}{\frac{a}{\sqrt{2}}}\] \[{{V}^{2}}=\left( \frac{2\sqrt{2}+1}{2\sqrt{2}} \right)\frac{Gm}{a}=\left( \frac{2\sqrt{2}+1}{2\sqrt{2}} \right)\frac{Gm}{\sqrt{2}R}\] \[{{V}^{2}}=\left( \frac{2\sqrt{2}+1}{4} \right)\frac{Gm}{R}\]