JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2014
- Hard
- MCQ
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

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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)
\({V_{remain}} = \frac{4}{3}\pi {R^3} - \frac{4}{3}\pi {R^3}\left( {\frac{1}{8}} \right)\)
\( = \frac{4}{3}\pi {R^3}\left( {\frac{7}{8}} \right)\)
Therefore mass of sphere carved and remaining sphere are at respectively \(\frac{1}{8}\,M\,and\,\frac{7}{8}\,M.\)
Therefore, gravitational force between these two sphere
\(F = \frac{{GMm}}{{{r^2}}}\)
\( = \frac{{G\frac{{7M}}{8} \times \frac{1}{8}M}}{{{{\left( {3R} \right)}^2}}} = \frac{7}{{64 \times 9}}\frac{{G{M^2}}}{{{R^2}}}\)
\( = \frac{{41}}{{3600}}\frac{{G{M^2}}}{{{R^2}}}\)
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