Verified Solution • 2025–26 Session moderate

QJEEM21Ph0670: JEE Main 2021 Physics - Class 11 JEEM21Ph0670 Solution: Find the gravitational force of attraction between the ring and …

Find the gravitational force of attraction between the ring and sphere as shown in the diagram, where the plane of the ring is perpendicular to the line joining the centres. If 8\sqrt 8 R is the distance between the centres of a ring (of mass 'm') and a sphere (mass 'M') where both have equal radius 'R'.

JEE Main 2021 (Online) 26th February Morning Shift Physics - Gravitation Question 121 English

Multiple Choice Options

  • A 223.GMmR2{{2\sqrt 2 } \over 3}.{{GMm} \over {{R^2}}}
  • B 89.GmMR{{\sqrt 8 } \over 9}.{{GmM} \over R}
  • C 827.GmMR2{{\sqrt 8 } \over {27}}.{{GmM} \over {{R^2}}}
  • D 138.GMmR2{1 \over {3\sqrt 8 }}.{{GMm} \over {{R^2}}}

Step-by-Step Verified Solution

JEE Main 2021 (Online) 26th February Morning Shift Physics - Gravitation Question 121 English Explanation
Force on sphere by the ring,

F=ME=M(Gm8R(R2+(8R)2)3/2)F = ME = M\left( {{{Gm\sqrt {8R} } \over {{{\left( {{R^2} + {{(\sqrt 8 R)}^2}} \right)}^{3/2}}}}} \right)

F=GMm8R(9R2)3/2=22GmM(9R2)3/2F = {{GMm\sqrt 8 R} \over {{{\left( {9{R^2}} \right)}^{3/2}}}} = {{2\sqrt 2 GmM} \over {{{\left( {9{R^2}} \right)}^{3/2}}}}

=22GmM27R2= {{2\sqrt 2 GmM} \over {27{R^2}}}

F=8GMm27R2F = {{\sqrt 8 GMm} \over {27{R^2}}}

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