A uniform wire of length $2\mathrm{L}$ is bent to form an angle of ${60}^{°}$ as shown. The distance of centre of mass from the vertex $O$ is $\frac{\mathrm{L}\sqrt{3}}{\mathrm{n}}$. Where n is

The centre of mass will be at the mid point of line joining the center of masses of wire segments

The centre of mass of each rod will be at a distance of L/2 from 'O'

So the Center of mass of the total system will at a distance of$x=\frac{L}{2}\cos \left({30}^{°}\right)=\frac{L\sqrt{3}}{4}$ from 'O'