JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Gravitation

Each corner is a distance $r=\frac{a}{\sqrt{2}}$ from the centre, and each mass pulls along its own diagonal. Initially one diagonal carries $4M$ and $2M$, whose fields give a resultant proportional to $(4-2)M=2M$, while the other diagonal carries $3M$ and $M$, giving $(3-1)M=2M$; as the diagonals are perpendicular, $F_1=\frac{GM}{r^2}\sqrt{2^2+2^2}=2\sqrt{2}\frac{GM}{r^2}$. Interchanging $3M$ and $4M$ pairs $3M$ with $2M$ (difference $M$) and $4M$ with $M$ (difference $3M$), so $F_2=\frac{GM}{r^2}\sqrt{1^2+3^2}=\sqrt{10}\frac{GM}{r^2}$. Therefore $\frac{F_1}{F_2}=\frac{2\sqrt{2}}{\sqrt{10}}=\frac{2}{\sqrt{5}}$, and $\alpha =2$.