The composite arc is in equilibrium on a horizontal surface. For equilibrium, the center of mass (COM) of the entire arc must lie directly above the point of contact with the surface. The arc is symmetric, so the COM lies along the vertical line through the center of the semicircle. We find the COM by considering the two segments.

The arc AB is a quarter-circle of radius R and density ρ. Its mass is m_AB = ρ(πR/2). Its COM is at a distance $\frac{2\sqrt{2}R}{\pi }$ from the center O along the line y=x.

The arc BC is a quarter-circle of radius 2R and density σ. Its mass is m_BC = σ(π(2R)/2) = σπR. Its COM is at a distance $\frac{4\sqrt{2}R}{\pi }$ from O, but on the opposite side of the vertical axis.

Balancing the moments about the vertical axis through O for the COM to lie on it gives the equilibrium condition. The horizontal distance of COM of AB is $\frac{2\sqrt{2}R}{\pi }\times \frac{1}{\sqrt{2}}=\frac{2R}{\pi }$. For BC, it is $\frac{4\sqrt{2}R}{\pi }\times \frac{1}{\sqrt{2}}=\frac{4R}{\pi }$ but in the opposite direction.

Setting the net moment to zero: $\left(\rho \frac{\pi R}{2}\right)\frac{2R}{\pi }=\left(\sigma \pi R\right)\frac{4R}{\pi }$. Simplifying, ρR² = σ(4R²), so σ/ρ = 1/4. This value is not among the given options.

Final Answer: none of these