Question

Three particles each of mass 'm₁' are placed at the comers of an equilateral triangle of side `L/3.` A particle of mass 'm₂' is placed at the mid point of any one side of triangle. Due to the system of particles the force acting on 'm₂' is ______.

(G = Universal constant of gravitation).

Options

• `(4"Gm"_1"m"_2)/"L"^2`

• `(12"Gm"_1"m"_2)/"L"^2`

• `(2"Gm"_1"m"_2)/"L"^2`

• `(8"Gm"_1"m"_2)/"L"^2`

Answer 1

Three particles each of mass 'm₁' are placed at the comers of an equilateral triangle of side `L/3.` A particle of mass 'm₂' is placed at the mid point of any one side of triangle. Due to the system of particles the force acting on 'm₂' is `underline((12"Gm"_1"m"_2)/"L"^2)`.

Explanation:

Forces on mass m₂ due to masses at B and C will be equal and opposite and cancel each other.

`h=L/3cos30^circ=L/3sqrt3/2=L/(2sqrt3)`

Force on m2 due to mass m₁ at A is given by

`F = G("m"_1"m"_2)/(L/(2sqrt3))^2=(12"G"  "m"_1"m"_2)/L^2`