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JEE Main2013PhysicsGravitationActual

The gravitational field, due to the 'left over part' of a uniform sphere (from which a part as shown, has been 'removed out'), at a very far off point, P , located as shown, would be (nearly) :

Options

  1. A5 6 G M x^2
  2. B8 9 G M x^2
  3. C7 8 G M x^2
  4. D6 7 G M x^2

Correct answer

C. 7 8 G M x^2

Step-by-step solution

Let mass of smaller sphere (which has to be removed) is m Radius = R 2 ( from figure ) aligned & M 4 3 R ^3 = m 4 3 ( R 2 )^3 & m = M 8 aligned Mass of the left over part of the sphere M ^ = M - M 8 = 7 8 M Therefore gravitational field due to the left over part of the sphere = GM ^ x ^2 = 7 8 GM x ^2

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