JEE Mains · Physics · STD 11 - 7. gravitation
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 \('R'.\)
- A \(\frac{\sqrt{8}}{9} \cdot \frac{ GmM }{ R }\)
- B \(\frac{2 \sqrt{2}}{3} \cdot \frac{ GMm }{ R ^{2}}\)
- C \(\frac{1}{3 \sqrt{8}} \cdot \frac{ GMm }{ R ^{2}}\)
- D \(\frac{\sqrt{8}}{27} \cdot \frac{ GmM }{ R ^{2}}\)
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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