To solve this problem, we need to understand the concept of charge distribution on connected conductors.$\\$When two charged spherical conductors are connected by a wire, they reach the same potential.$\\$Let's denote:$\\$• $V_1$ as the potential of the first sphere with radius $R_1. \\$• $V_2$ as the potential of the second sphere with radius $R_2. \\$The potential $V$ on the surface of a charged sphere is given by:$\\ V = \frac{Q}{4 \pi \varepsilon_0 R} \\$where $Q$ is the charge on the sphere, $R$ is the radius, and $\varepsilon_0$ is the permittivity of free space.$\\$Since the spheres are connected, $V_1 = V_2. \\$Therefore, we have:$\\ \frac{Q_1}{4 \pi \varepsilon_0 R_1} = \frac{Q_2}{4 \pi \varepsilon_0 R_2} \\$This simplifies to:$\\ \frac{Q_1}{R_1} = \frac{Q_2}{R_2} \\$The surface charge density $\sigma$ is defined as $\sigma = \frac{Q}{4 \pi R^2}. \\$Thus, for the two spheres, we have:$\\ \sigma_1 = \frac{Q_1}{4 \pi R_1^2} \quad$and$\quad \sigma_2 = \frac{Q_2}{4 \pi R_2^2} \\$Taking the ratio of the surface charge densities:$\\ \frac{\sigma_1}{\sigma_2} = \frac{\frac{Q_1}{4 \pi R_1^2}}{\frac{Q_2}{4 \pi R_2^2}} \\$This simplifies to:$\\ \frac{\sigma_1}{\sigma_2} = \frac{Q_1 R_2^2}{Q_2 R_1^2} \\$Using the relation $\frac{Q_1}{R_1} = \frac{Q_2}{R_2},$ we can express $Q_1$ in terms of $Q_2: \\ Q_1 = \frac{Q_2 R_1}{R_2} \\$Substitute $Q_1$ in the ratio of surface charge densities:$\\ \frac{\sigma_1}{\sigma_2} = \frac{\left(\frac{Q_2 R_1}{R_2}\right) R_2^2}{Q_2 R_1^2} \\$Simplify the expression:$\\ \frac{\sigma_1}{\sigma_2} = \frac{R_1 R_2^2}{R_2 R_1^2} \\ \frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1} \\$Therefore, the ratio of surface charge densities is $\frac{R_2}{R_1}. \\$This corresponds to Option 2.