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\[(a)\;\sqrt {\frac{GM}{2l}} \quad (b)\;\sqrt {\frac{GM}{l}} \quad (c)\;\sqrt {\frac{2GM}{l}} \quad (d)\;\sqrt {\frac{GM}{3l}} \]

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The particle to move in a circular path without chaging distance 'l' will happen if the resultant gravitational force on each particle provides the centripetal force necessary for the motion

$\large\frac{mv^2}{r}$$=\sqrt {F^2+F^2+2 F^2 \cos 60}$

$\large\frac{mv^2}{r}$$=\sqrt {3 }\; F$

Radius of the path is

$r= \large\frac{\sqrt 3}{2}$$l \times \large\frac{2}{3}$

$\quad= \large\frac{l}{3}$

$\therefore v= \sqrt {\large\frac {GM}{l}}$

Hence b is the correct answer.

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