Three masses, each equal to $M$, are placed at the three corners of a square of side $a$. The force of attraction on unit mass at the fourth corner will be

The correct option is C $\frac{GM}{a^2}[\frac{1}{2}+\sqrt{2}]$
$F_1=F_2=\frac{GM}{a^2}$
Resultant of $F_1\text{and}{F}_2\text{is}\sqrt{2}\frac{GM}{a^2}$
Now, $F_3=\frac{GM}{(\sqrt{2}a{)}^2}=\frac{GM}{2a^2}$
Now, $\frac{\sqrt{2}GM}{a^2}$ and $\frac{GM}{2a^2}$ act in the same direction.
Their resultant is $\frac{\sqrt{2}GM}{a^2}+\frac{GM}{2a^2}$
=$\frac{GM}{a^2}[\sqrt{2}+\frac{1}{2}]$