Three particles each of mass ' $\mathrm{m}_{1}$ ' are placed at the corners of an equilateral triangle of…

Three particles each of mass ' $\mathrm{m}_{1}$ ' are placed at the corners of an equilateral triangle of side $\frac{\mathrm{L}}{3}$ '. 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 ' $\mathrm{m}_{2}$ ' is ( G = Universal constant of gravitation )
  1. $\frac{12 \mathrm{G} \mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~L}^{2}}$
  2. $\frac{2 \mathrm{G} \mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~L}^{2}}$
  3. $\frac{4 \mathrm{G} \mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~L}^{2}}$
  4. $\frac{8 \mathrm{G} \mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{~L}^{2}}$

Solution

Forces on mass $\mathrm{m}_{2}$ due to masses at $\mathrm{B}$ and $\mathrm{C}$ will be equal and opposite and cancel each other. $\mathrm{h}=\frac{\mathrm{L}}{3} \cos 30^{\circ}=\frac{\mathrm{L}}{3} \frac{\sqrt{3}}{2}=\frac{\mathrm{L}}{2 \sqrt{3}}$ Force on $m_{2}$ due to mass $m_{1}$ at $A$ is given by $F=G \frac{m_{1} m_{2}}{\left(\frac{L}{2 \sqrt{3}}\right)^{2}}=\frac{12 G m_{1} m_{2}}{L^{2}}$

Asked in: MHT CET 2020 (12 Oct Shift 2)

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