Three identical particle A , B and C of mass 100   kg each are placed in a straight line with A B = B C…

Three identical particle A,B and C of mass 100 kg each are placed in a straight line with AB=BC=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 AC. The value of F will be approximately
  1. 21G
  2. 100G
  3. 59G
  4. 42G

Solution

Given here, m=100 kg

Gravitational force, FAP=Gm21322,

FBP=Gm2132 and FCP=Gm21322

Net gravitational force, 

Fnet=FBP+FAPcos45°+FCPcos45°

 =G×1041321+12

F≃100G

Asked in: JEE Main 2022 (25 Jul Shift 1)

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