Three particles of identical masses $m$ are kept at the vertices of an equilateral triangle of each side length $a$ . The gravitational force of attraction on any one of the particles is:

\sqrt{2} \frac{G m^{2}}{a^{2}}

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\sqrt{3} \frac{G m^{2}}{a^{2}}

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\frac{3 G m^{2}}{a^{2}}

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\frac{2 G m^{2}}{a^{2}}

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Solution

The correct option is C $\sqrt{3} \frac{G m^{2}}{a^{2}}$
$F_{n e t} = \sqrt{(F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} c o s θ)}$
Here, $F_{1} = F_{2} = \frac{G m m}{r^{2}}$ and $θ = 60^{0}$
Therefore, on solving we get, $F_{n e t} = \sqrt{3} \frac{G m^{2}}{a^{2}}$

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Three particles of identical masses m are kept at the vertices of an equilateral triangle of each side length a. The gravitational force of attraction on any one of the particles is:

The correct option is C √3Gm2a2Fnet=√(F21+F22+2F1F2cosθ)Here, F1=F2=Gmmr2 and θ=600Therefore, on solving we get, Fnet=√3Gm2a2

Three particles of identical masses $m$ are kept at the vertices of an equilateral triangle of each side length $a$ . The gravitational force of attraction on any one of the particles is:

The correct option is C $\sqrt{3} \frac{G m^{2}}{a^{2}}$
$F_{n e t} = \sqrt{(F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} c o s θ)}$
Here, $F_{1} = F_{2} = \frac{G m m}{r^{2}}$ and $θ = 60^{0}$
Therefore, on solving we get, $F_{n e t} = \sqrt{3} \frac{G m^{2}}{a^{2}}$