Question

Three equal masses each of mass $m$ are placed at the three-corners of an equilateral triangle of side a.
If a fourth particle of equal mass is placed at the centre of triangle, then the net force acting on it, is equal to:

• A. $\frac{G{m}^2}{a^2}$
• B. $\frac{4G{m}^2}{3a^2}$
• C. $\frac{3G{m}^2}{a^2}$
• D. Zero

Answer 1

The correct option is C $\frac{3G{m}^2}{a^2}$

$\therefore \cos {30}^o=\frac{a/2}{r}$

$r=\frac{a}{2\cos {30}^o}=\frac{a}{\sqrt{3}}$

$\therefore |F{|}_{oneparticle}=\frac{G{m}^2}{r^2}=\frac{3G{m}^2}{a^2}$