Two isolated metallic spheres of radii $R$ and $2R$ are charged such that both have the same charge density $\sigma$. The spheres are then connected by a thin conducting wire. If the new charge density of the larger sphere is $\sigma_1$. The ratio $\sigma_1$ to $\sigma$ is

Let the initial charge on the smaller sphere be $Q_1$ and on the larger sphere be $Q_2$.

$Q_1 = \sigma(4\pi R^2)$

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

The total charge of the system is $Q = Q_1 + Q_2 = 20\pi R^2\sigma$.

When the spheres are connected by a conducting wire, charge flows until their potentials become equal. Let the new charges on the smaller and larger spheres be $q_1$ and $q_2$ respectively.

Equating the potentials:

$\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1}{R} = \dfrac{1}{4\pi\epsilon_0}\dfrac{q_2}{2R}$

$q_2 = 2q_1$

By conservation of charge, $q_1 + q_2 = Q$.

$q_1 + 2q_1 = 20\pi R^2\sigma$

$3q_1 = 20\pi R^2\sigma$

$q_1 = \dfrac{20\pi R^2\sigma}{3}$

The new charge on the larger sphere is $q_2 = 2q_1 = \dfrac{40\pi R^2\sigma}{3}$.

The new charge density of the larger sphere is $\sigma_1 = \dfrac{q_2}{4\pi(2R)^2}$.

$\sigma_1 = \dfrac{\dfrac{40\pi R^2\sigma}{3}}{16\pi R^2} = \dfrac{40}{48}\sigma = \dfrac{5}{6}\sigma$

Therefore, the ratio $\dfrac{\sigma_1}{\sigma}$ is $\dfrac{5}{6}$.