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JEE Main & Advanced JEE Main Paper (Held On 11 April 2014)

  • question_answer
    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:   [JEE Main Online Paper ( Held On 11 Apirl  2014 )

    A) \[\frac{41G{{M}^{2}}}{3600{{R}^{2}}}\] 

    B) \[\frac{41G{{M}^{2}}}{450{{R}^{2}}}\]

    C) \[\frac{59\,G{{M}^{2}}}{450{{R}^{2}}}\]                

    D) \[\frac{\,G{{M}^{2}}}{225{{R}^{2}}}\]

    Correct Answer: A

    Solution :

                    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, \[F=\frac{GMm}{{{r}^{2}}}\] \[=\frac{G\frac{7M}{8}\times \frac{1}{8}M}{{{(3R)}^{2}}}=\frac{7}{64\times 9}\frac{G{{M}^{2}}}{{{R}^{2}}}\] \[\simeq \frac{41}{3600}\frac{G{{M}^{2}}}{{{R}^{2}}}\]                


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