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NTA Abhyas JEE Main2020PhysicsGravitationPractice

Three particles P , Q and R are placed on a straight line as shown in the figure. The masses of P , Q and R are 3 m , 3 m and m respectively. The gravitational force on a fourth particle S of mass m is equal to

Options

  1. A3 Gm 2 2 d 2 in ST direction only
  2. B3 Gm 2 2 d 2 in SQ direction and 3 Gm 2 2 d 2 in SU direction
  3. C3 Gm 2 2 d 2 in SQ direction only
  4. D3 Gm 2 2 d 2 in SQ direction and 3 Gm 2 2 d 2 in ST direction

Correct answer

C. 3 Gm 2 2 d 2 in SQ direction only

Step-by-step solution

Along the direction parallel to TU Net force = G 3 mm 1 2 d 2 cos 30 ∘ - Gm 2 4 d 2 cos 60 ∘ = Gm 2 8 d 2 - Gm 2 8 d 2 = 0 Along the direction perpendicular to TU Net force = G 3 m 2 1 2 d 2 cos 60 ∘ + G 3 m 2 3 d 2 + Gm 2 4 d 2 cos 30 ∘ = 3 Gm 2 2 4 d 2 + 3 Gm 2 3 d 2 + 3 Gm 2 8 d 2 = 3 Gm 2 d 2 1 + 8 + 3 2 4 = 3 Gm 2 2 d 2 along SQ

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