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JEE Main202522 Jan 2025Morning ShiftPhysicsGravitationActual

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 F ₁ . 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 F ₂ . The value of ratio F ₁: F ₂ is

Options

  1. A12: 11
  2. B11: 10
  3. C12: 9
  4. D16: 9

Correct answer

A. 12: 11

Step-by-step solution

aligned & F ₁= GMm (2 R )^2 .....(1) & ~F ₂= GMm (2 R )^2 - ( G ( M 27 ) m ( 4 R 3 )^2 ) & F ₂= 11 48 GMm R ^2 ....(2) & ~F ₁: F ₂=12: 11 aligned

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