question_answer

3 particles each of mass m are kept at vertices of an equilateral triangle of side L. The gravitational field at centre due to these particles is                             [DCE 2005]              

A) Zero    

B)             \[\frac{3GM}{{{L}^{2}}}\]

C)             \[\frac{9GM}{{{L}^{2}}}\]

D)             \[\frac{12}{\sqrt{3}}\,\frac{GM}{{{L}^{2}}}\]

Correct Answer: A

Solution :

    Due to three particles net intensity at the centre             \[I={{\vec{I}}_{A}}+{{\vec{I}}_{B}}+{{\vec{I}}_{C}}=0\]           Because out of these three intensities one equal in magnitude and the angle between each other is\[120{}^\circ \].