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MHT CET202012 Oct 2020Evening ShiftPhysicsGravitationActual

Three particles each of mass ' m ₁ ' are placed at the corners of an equilateral triangle of side L 3 '. A particle of mass 'm ₂ ' is placed at the mid point of any one side of triangle. Due to the system of particles the force acting on ' m ₂ ' is ( G = Universal constant of gravitation )

Options

  1. A12 G m ₁ ~m ₂ ~L ²
  2. B2 G m ₁ ~m ₂ ~L ²
  3. C4 G m ₁ ~m ₂ ~L ²
  4. D8 G m ₁ ~m ₂ ~L ²

Correct answer

A. 12 G m ₁ ~m ₂ ~L ²

Step-by-step solution

Forces on mass m ₂ due to masses at B and C will be equal and opposite and cancel each other. h = L 3 30^ = L 3 3 2 = L 2 3 Force on m₂ due to mass m₁ at A is given by F=G m₁ m₂ ( L 2 3 )² = 12 G m₁ m₂ L²

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