1. Initial charges on the spheres

Since both metallic solid spheres initially have the same surface charge density $\sigma$:

• For sphere of radius $R$: $q_1 = \sigma \cdot 4\pi R^2$

• For sphere of radius $2R$: $q_2 = \sigma \cdot 4\pi (2R)^2 = 16\pi \sigma R^2$

Thus, $q_1 = 4\pi \sigma R^2, \qquad q_2 = 16\pi \sigma R^2$

So total charge is $Q = q_1 + q_2 = 20\pi \sigma R^2$

2. Condition after connecting by a conducting wire

When two conducting spheres are connected by a wire, their potentials become equal.

For isolated spheres: $V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$

Let final charges be $q_1'$ on radius $R$ sphere and $q_2'$ on radius $2R$ sphere.

Equal potentials give: $\frac{q_1'}{R} = \frac{q_2'}{2R}$ $2q_1' = q_2'$

Hence, $q_2' : q_1' = 2:1$

Let $q_1' = x, \qquad q_2' = 2x$

Using conservation of charge: $x + 2x = 20\pi \sigma R^2$ $3x = 20\pi \sigma R^2$ $x = \frac{20\pi \sigma R^2}{3}$

Therefore, $q_2' = 2x = \frac{40\pi \sigma R^2}{3}$

3. New surface charge density on bigger sphere

Surface area of bigger sphere (radius $2R$) is $4\pi (2R)^2 = 16\pi R^2$

So its new charge density is $\sigma' = \frac{q_2'}{16\pi R^2} = \frac{\frac{40\pi \sigma R^2}{3}}{16\pi R^2}$

$\sigma' = \frac{40}{48}\sigma = \frac{5}{6}\sigma$

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

4. Checking options

• A: $\frac{5}{3}$ ❌
• B: $\frac{5}{6}$ ✅
• C: $\frac{9}{4}$ ❌
• D: $\frac{4}{3}$ ❌

Therefore, the correct option is B.