Two isolated metallic solid spheres of radii $$R$$ and $$2R$$ are charged such that both have same charge density $$\sigma$$. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $$\sigma'$$. The ratio $$\frac{\sigma'}{\sigma}$$ is:

Two metallic spheres of radii $$R$$ and $$2R$$ with same surface charge density $$\sigma$$ are connected by a wire.

$$Q_1 = \sigma \cdot 4\pi R^2$$

$$Q_2 = \sigma \cdot 4\pi(2R)^2 = 16\sigma\pi R^2$$

Total charge: $$Q = Q_1 + Q_2 = 4\sigma\pi R^2 + 16\sigma\pi R^2 = 20\sigma\pi R^2$$

$$\frac{Q_1'}{4\pi\epsilon_0 R} = \frac{Q_2'}{4\pi\epsilon_0 \cdot 2R}$$

$$\frac{Q_1'}{R} = \frac{Q_2'}{2R} \implies Q_2' = 2Q_1'$$

$$Q_1' + Q_2' = 20\sigma\pi R^2$$

$$Q_1' + 2Q_1' = 20\sigma\pi R^2$$

$$Q_1' = \frac{20\sigma\pi R^2}{3}, \quad Q_2' = \frac{40\sigma\pi R^2}{3}$$

$$\sigma' = \frac{Q_2'}{4\pi(2R)^2} = \frac{40\sigma\pi R^2}{3 \times 16\pi R^2} = \frac{40\sigma}{48} = \frac{5\sigma}{6}$$

$$\frac{\sigma'}{\sigma} = \frac{5}{6}$$

The correct answer is Option 4: $$\frac{5}{6}$$.