A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force F₁. Now a spherical cavity of radius ( R/2 ) is made in the sphere (as shown in figure) and the force becomes F₂. The value of F₁ : F₂ is

Solution

Physics sir · 4 steps

Fill the hole back in, then subtract it: full sphere minus a small sphere of mass M/8.

1. Superposition. The sphere with a cavity is the full sphere minus a small solid sphere that fills the hole. So F₂ is the pull of the full sphere minus the pull of that small sphere, since both pull along the same line towards the left.
2. Cavity mass and distance. The cavity has radius R/2, so its volume, and hence its mass, is (1/2)³ = 1/8 of the full sphere: M/8. Its centre B is R/2 from O, so it is 3R - R/2 = 5R/2 from the particle at A.

   F₂ = GMm/(3R)² - G(M/8)m/(5R/2)² = GMm/R²(1/9 - 1/50) = 41/450 GMm/R²

3. Compare. The full sphere gives F₁ = GMm/9R² = 50/450 GMm/R². Dividing, F₁ : F₂ = 50 : 41.
4. So. The answer is (c), 50 : 41. F₁ is the larger force, as it must be, because removing mass can only weaken the pull.

The idea to keep. For any cavity, treat it as a negative mass: force of the full body minus force of the removed piece, each measured from its own centre. Mass scales as radius cubed.

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