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JEE Main 2020 · Physics · Gravitation · hard

A uniform solid sphere of mass M (after the removal described) and radius R has a spherical cavity of radius R/2 whose centre lies at a distance R/2…

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A uniform solid sphere of mass MM (after the removal described) and radius RR has a spherical cavity of radius R/2R/2 whose centre lies at a distance R/2R/2 from the centre of the sphere. The magnitude of the gravitational field at the point PP on the surface of the sphere, lying on the line joining the two centres on the side of the cavity, is
  1. A. GMR2\dfrac{GM}{R^2}
  2. B. 8GM7R2\dfrac{8GM}{7R^2}
  3. C. 2GM7R2\dfrac{2GM}{7R^2}
  4. D. 4GM7R2\dfrac{4GM}{7R^2}
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Answer: (D) 4GM7R2\dfrac{4GM}{7R^2}

Use superposition: (full sphere) −- (sphere filling the cavity). The cavity has volume 1/81/8 of the full sphere, so if the remaining mass is MM, the full sphere has M′=8M/7M' = 8M/7 and the cavity mass is M/7M/7. Field of the full sphere at its surface: G(8M/7)/R2G(8M/7)/R^2. The cavity centre is at distance R−R/2=R/2R - R/2 = R/2 from PP, so the removed mass would have given G(M/7)/(R/2)2=4GM/(7R2)G(M/7)/(R/2)^2 = 4GM/(7R^2) towards the cavity, in the same direction. Net =8GM7R2−4GM7R2=4GM7R2= \dfrac{8GM}{7R^2} - \dfrac{4GM}{7R^2} = \dfrac{4GM}{7R^2}.
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