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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 a distance 3 R from the centre 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 the cavity now applies a gravitational force F 2 on the same particle. The ratio F 2 F 1 is

Options

  1. A9 5 0
  2. B4 1 5 0
  3. C3 2 5
  4. D2 2 2 5

Correct answer

B. 4 1 5 0

Step-by-step solution

From the superposition principle, F 1   =   F r   +   F c Here, F r = force due to remaining part = F 2 and F c = force due to cavity Now, F 1 = GMm 3 R 2 = GMm 9 R 2 F c = G M 8 m 5 2 R 2 = GMm 5 0 R 2 ∴     F 2 = F 1 - F c = 4 1 GMm 4 5 0 R 2 ∴     F 2 F 1 = 4 1 5 0

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