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Three solid spheres each of mass $1 \mathrm{~kg}$ and radius $2 \mathrm{~m}$ are arranged at the three corners of an equilateral triangle of side $10 \mathrm{~m}$, such that centres of the spheres coincide with the comers of the triangle. When they are released from that position, the speed of any one sphere at the time of collision would be ($\mathrm{G}$ is universal gravitational constant).
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The correct answer is:
$\sqrt{\frac{3 G}{10}}$
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