question_answer

Three 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

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 :

The net force acting on a unit mass placed at \[O\]due to three equal masses\[M\]at vertices\[A,\,\,B\]and\[C\]is the gravitational field intensity at point\[O\]. The gravitational force on the particle placed at the point of intersection of three medians. \[\overset{\to }{\mathop{{{F}_{1}}}}\,+\overset{\to }{\mathop{{{F}_{2}}}}\,+\overset{\to }{\mathop{{{F}_{3}}}}\,=0\] Since, the resultant of\[\overset{\to }{\mathop{{{F}_{1}}}}\,\]and\[\overset{\to }{\mathop{{{F}_{2}}}}\,\]is equal and opposite to\[\overset{\to }{\mathop{{{F}_{3}}}}\,\].