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JEE Mains · Physics · STD 11 - Gravitation
  • JEE Mains 2021
  • Hard
  • MCQ

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 \(\sqrt{8}\, R\) is the distance between the centres of a \(mng\) (of mass ' \(m\) ) and a sphere (mass ' \(M\) ) where both have equal radius \('R'.\)

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Answer & Solution

Correct Answer

(D) \(\frac{\sqrt{8}}{27} \cdot \frac{ GmM }{ R ^{2}}\)

Step-by-step Solution

Detailed explanation

Gravitational field of ring

\(=-\frac{\text { Gmx }}{\left(R^{2}+x^{2}\right)^{3 / 2}}\)

Force between sphere and ring

\(=\frac{\operatorname{GmM}(\sqrt{8} R )}{\left( R ^{2}+8 R ^{2}\right)^{3 / 2}}\)

\(=\frac{ GmM }{ R ^{2}} \times \frac{\sqrt{8}}{27}\)

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