A small point of mass m is placed at a distance 2 R from the centre ' O ' of a big uniform solid sphere of mass M and radius R . The gravitational force on ' m ' due to M is F₁. A spherical part of radius R / 3 is removed from the big sphere as shown in the figure and the gravitational force on m due to remaining part of M is found to be F₂. The value of ratio F₁: F₂ is

Solution

Physics sir · 3 steps

Treat the hollowed sphere as the full sphere minus a small sphere, then subtract the two forces.

1. Full sphere. A uniform sphere pulls an outside point as if all its mass sat at the centre. With m at distance 2R, F₁ = GMm/4R².
2. The removed part. Mass scales with volume, so a sphere of radius R/3 has mass M/27. It touches the surface on the side facing m, so its centre is R - R/3 = 2R/3 from O and 2R - 2R/3 = 4R/3 from m. Its pull on m would be G(M/27)m/(4R/3)² = GMm/48R².

   F₂ = GMm/4R² - GMm/48R² = (11 GMm)/48R²

3. So. F₁ : F₂ = 12/48 : 11/48 = 12 : 11, which is option (b).

The idea to keep. For a sphere with a cavity, superpose: full sphere minus the removed sphere. Use volume for the removed mass, (r/R)³, and measure each distance from that sphere's own centre.

Answer from the NTA final answer key. Solution by PrepWiser.