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ExamsJEE AdvancedPhysics

A solid sphere of radius R attracts a particle located at distance 3R from its centre with gravitational force F1. A spherical cavity of radius R/2 is then carved out of the sphere (with the cavity touching the surface along the line to the particle, as shown), and the force on the same particle becomes F2. Find the ratio F1: F2.

  1. 25: 36
  2. 36: 25
  3. 50: 41
  4. 41: 50

Correct answer: 50: 41

Solution

By superposition, F2 = (force of full sphere) - (force of the removed small sphere treated as a point mass at its center). The small sphere has 1/8 the mass and its center is at distance 5R/2 from the particle. Computing the ratio gives F1: F2 = 50: 41.

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