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

Option 1 -

21 G

Option 2 -

100 G

Option 3 -

59 G

Option 4 -

42 G

0 5 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    3 months ago
    Correct Option - 2


    Detailed Solution:

    F on P =  G M M r 2 + 2 G M M ( 2 r ) 2

    F     o n     P = G × 1 0 4 1 3 2 ( 1 + 1 2 ) ≈ 1 0 0 G

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