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JEE Main2014PhysicsGravitationActual

From a sphere of mass M and radius R , a smaller sphere of radius 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:

Options

  1. A41 GM ^2 3600 R ^2
  2. B41 GM ^2 450 R ^2
  3. C59 GM ^2 450 R ^2
  4. DGM ^2 225 R ^2

Correct answer

A. 41 GM ^2 3600 R ^2

Step-by-step solution

Volume of removed sphere V_ remo = 4 3 ( R 2 )^3= 4 3 R^3 ( 1 8 ) Volume of the sphere (remaining) aligned &V_ remain = 4 3 R^3- 4 3 R^3 ( 1 8 ) &= 4 3 R^3 ( 7 8 ) aligned Therefore mass of sphere carved and remaining sphere are at respectively 1 8 M and 7 8 M Therefore, gravitational force between these two sphere, F= G M m r^2 aligned &= G 7 M 8 1 8 M (3 R)^2 = 7 64 9 G M^2 R^2 & 41 3600 GM ^2 R ^2 aligned

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