Verified Solution • 2025–26 Session moderate

QJEEM21Phb164: JEE Main 2021 Physics - Class 11 JEEM21Phb164 Solution: Four particles each of mass M, move along a circle of radius R…

Four particles each of mass M, move along a circle of radius R under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is :

JEE Main 2021 (Online) 1st September Evening Shift Physics - Gravitation Question 99 English

Multiple Choice Options

  • A 12GMR(22+1){1 \over 2}\sqrt {{{GM} \over {R(2\sqrt 2 + 1)}}}
  • B 12GMR(22+1){1 \over 2}\sqrt {{{GM} \over R}(2\sqrt 2 + 1)}
  • C 12GMR(22−1){1 \over 2}\sqrt {{{GM} \over R}(2\sqrt 2 - 1)}
  • D GMR\sqrt {{{GM} \over R}}

Step-by-Step Verified Solution

Let us consider the gravitational force acting on each mass M by adjacent particles be F.

and the gravitational force acting on each mass M diagonally be F1.

JEE Main 2021 (Online) 1st September Evening Shift Physics - Gravitation Question 99 English Explanation
Fnet=MV2R{F_{net}} = {{M{V^2}} \over R}

Along the centre of circle,


2F+F1=MV2R\sqrt 2 F + {F_1} = {{M{V^2}} \over R}

2GMM(2R)2+GMM(2R)2=MV2R\sqrt 2 {{GMM} \over {{{\left( {\sqrt 2 R} \right)}^2}}} + {{GMM} \over {{{(2R)}^2}}} = {{M{V^2}} \over R}

GMR(12+14)=V2{{GM} \over R}\left( {{1 \over {\sqrt 2 }} + {1 \over 4}} \right) = {V^2}

GMR(4+242)=V2{{GM} \over R}\left( {{{4 + \sqrt 2 } \over {4\sqrt 2 }}} \right) = {V^2}

V=GM(4+2)R42V = \sqrt {{{GM\left( {4 + \sqrt 2 } \right)} \over {R4\sqrt 2 }}}

V=12GM(22+1)RV = {1 \over 2}\sqrt {{{GM\left( {2\sqrt 2 + 1} \right)} \over R}}

Option (b)

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