JEE Mains · Physics · STD 11 - 7. gravitation
From a sphere of mass \(M\) and radius \(R\), a smaller sphere of radius \(\frac{R}{2}\) is carved out such that the cavity made in the original sphere is between its centre and the periphery (See figure) . For the configuration in the figure where the distance between the centre of the original sphere and the removed sphere is \(3R\), the gravitational force between the two sphere is
- A \(\frac{{41G{M^2}}}{{3600{R^2}}}\)
- B \(\frac{{41G{M^2}}}{{450{R^2}}}\)
- C \(\frac{{59G{M^2}}}{{450{R^2}}}\)
- D \(\frac{{G{M^2}}}{{225{R^2}}}\)
Answer & Solution
Correct Answer
(A) \(\frac{{41G{M^2}}}{{3600{R^2}}}\)
Step-by-step Solution
Detailed explanation
Volume of removed sphere \({V_{remo}} = \frac{4}{3}\pi {\left( {\frac{R}{2}} \right)^3} = \frac{4}{3}\pi {R^3}\left( {\frac{1}{8}} \right)\) Volume of the sphere (remaining) \({V_{remain}} = \frac{4}{3}\pi {R^3} - \frac{4}{3}\pi {R^3}\left( {\frac{1}{8}} \right)\)…
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