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NTA Abhyas JEE Main2020PhysicsGravitationPractice

A solid sphere of uniform density and radius r applies a gravitational force of attraction equal to F 1 on a particle placed at P , distance 2 R from the centre O of the sphere. A spherical cavity of the radius R 2 is now made in the sphere as shown in the figure. The sphere with cavity now applied a gravitational force F 2 on the same particle placed at P . The ratio F 2 F 1 will be

Options

  1. A1 2
  2. B7 9
  3. C3
  4. D7

Correct answer

B. 7 9

Step-by-step solution

Gravitational force due to solid sphere, F ⁡ 1 = G ⁡ M ⁡ m ⁡ 2 R ⁡ 2 , where M and m are mass of the solid sphere and particle respectively and R is the radius of the sphere. The gravitational force on particle due to sphere with cavity = force due to solid sphere creating cavity, assumed to be present above at that position i.e., F 2 = GMm 4 R 2 = G M / 8 m 3 R / 2 2 = 7 3 6 GMm R 2 So, F 2 F 1 = 7 GMm 3 6 R 2 / GMm 4 R 2 = 7 9

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