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JEE Main202225 Jul 2022Morning ShiftPhysicsGravitationActual

Three identical particle A , B and C of mass 100 kg each are placed in a straight line with A B = B C = 13 m . The gravitational force on a fourth particle P of the same mass is F , when placed at a distance 13 m from the particle B on the perpendicular bisector of the line A C . The value of F will be approximately

Options

  1. A21 G
  2. B100 G
  3. C59 G
  4. D42 G

Correct answer

B. 100 G

Step-by-step solution

Given here, m = 100 kg Gravitational force, F A P = G m 2 13 2 2 , F B P = G m 2 13 2 and F C P = G m 2 13 2 2 Net gravitational force, F n e t = F B P + F A P cos 45 ° + F C P cos 45 ° = G × 10 4 13 2 1 + 1 2 F ≃ 100 G

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