QJEEM19Ph6085: JEE Main 2019 Physics - Class 11 JEEM19Ph6085 Solution: Four identical particles of mass M are located at the corners…

Net force on mass M at position B towards centre of circle is

${F_{BO\,net}} = {F_{BD}} + {F_{BA}}\sin 45^\circ + {F_{BC}}\sin 45^\circ$

$= {{G{M^2}} \over {{{(\sqrt 2 a)}^2}}} + {{G{M^2}} \over {{a^2}}}\left( {{1 \over {\sqrt 2 }}} \right) + {{G{M^2}} \over {{a^2}}}\left( {{1 \over {\sqrt 2 }}} \right)$

[where, diagonal length BD is $\sqrt2$ a]

$= {{G{M^2}} \over {2{a^2}}} + {{G{M^2}} \over {{a^2}}}\left( {{2 \over {\sqrt 2 }}} \right) = {{G{M^2}} \over {{a^2}}}\left( {{1 \over 2} + \sqrt 2 } \right)$

This force will act as centripetal force.

Distance of particle from centre of circle is ${a \over {\sqrt 2 }}$ .

Here, ${F_{centripetal}} = {{M{v^2}} \over r} = {{M{v^2}} \over {{a \over {\sqrt 2 }}}} = {{\sqrt 2 M{v^2}} \over a}$ ( $\because$ $r = {a \over {\sqrt 2 }}$ )

So, for rotation about the centre,

${F_{centripetal}} = {F_{BO(net)}}$

$\Rightarrow \sqrt 2 {{M{v^2}} \over a} = {{G{M^2}} \over {{a^2}}}\left( {{1 \over 2} + \sqrt 2 } \right)$

$\Rightarrow {v^2} = {{GM} \over a}\left( {1 + {1 \over {2\sqrt 2 }}} \right) = {{GM} \over a}(1.35)$

$\Rightarrow v = 1.16\sqrt {{{GM} \over a}}$