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JEE Main2015PhysicsGravitationActual

From a solid sphere of mass M and radius R , a spherical portion of radius R 2 is removed as shown in the figure. Taking gravitational potential V = 0 at r = ∞ , the potential at the centre of the cavity thus formed is ( G = gravitational constant)

Options

  1. A- 2 G M R
  2. B- G M 2 R
  3. C- G M R
  4. D- 2 G M 3 R

Correct answer

C. - G M R

Step-by-step solution

Central idea is, consider the cavity as negative a mass and apply the superposition of gravitational potential. Consider the cavity formed in a solid sphere as shown in the figure. The gravitational potential at infinite distance, V = - G M r ⇒ V ∞ = 0 According to the question, we can write potential at an internal point P due to complete solid sphere, V s = - G M 2 R 3 3 R 2 - R 2 2 = - G M 2 R 3 3 R 2 - R 2 4 = - G M 2 R 3 11 R 2 4 = - 11 G M 8 R Mass of removed part = M 4 3 × π R 3 ×

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