JEE Main 2024 — 1 February, Shift 1 — Question 52

The identical spheres each of mass 2 M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 4 m each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is $\frac{4 \sqrt{2}}{x}$, where the In air $\tan \frac{\theta}{2}=\frac{F}{m g}=\frac{q^{2}}{4 \pi \varepsilon_{0} r^{2} m g}$ value of $x$ is $\_\_\_\_\_\_\_$