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(A) Let the three particles of mass $m$ be placed at vertices $A$,$B$,and $C$ of an equilateral triangle with side length $L$. The centre of the trian

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Zero

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(A) Let the three particles of mass $m$ be placed at vertices $A$,$B$,and $C$ of an equilateral triangle with side length $L$. The centre of the triangle is $O$.Each particle produces a gravitational field intensity at the centre $O$ directed towards itself.The magnitude of the gravitational field intensity due to each particle at the centre is $I = \frac{Gm}{r^2}$,where $r$ is the distance from the vertex to the centre.Since the triangle is equilateral,the distance $r$ is the same for all three particles.Thus,the magnitudes of the gravitational field intensities are equal: $|\vec{I}_A| = |\vec{I}_B| = |\vec{I}_C| = I$.The angle between any two of these vectors is $120^\circ$.According to the principle of superposition,the net gravitational field at the centre is the vector sum: $\vec{I}_{net} = \vec{I}_A + \vec{I}_B + \vec{I}_C$.Since these three vectors are equal in magnitude and are oriented at $120^\circ$ to each other,their vector sum is zero.

$3$ particles each of mass $m$ are kept at the vertices of an equilateral triangle of side $L$. The gravitational field at the centre due to these particles is

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