Two isolated metallic solid spheres of radii R and 2R are charged such that both of these have same charge density $\sigma$. The spheres are located far away from each other, and connected by a thin conducting wire. Then the new charge density on the bigger sphere is $\frac{p\sigma }{q}$. Find the value of $p\times q$  is

 $\sigma =\frac{q_1}{4\pi R^2}$
Or  $q_1=\sigma \times 4\pi R^2$
For sphere of radius 2 R
$\sigma =\frac{q_2}{4\pi {\left(2R\right)}^2}$
or $q_2=\sigma \times 16\pi R^2$
When the two spheres are connected the potential on the two spheres will be the same. There will be a rearrangement of charge for this to happen.
Let  $q{\text{'}}_1$  and $q{\text{'}}_2$  be the new charge on the two spheres. Since the total charge remains the same,
We get
$q{\text{'}}_1+q{\text{'}}_2=q_1+q_2=\sigma \times 20\pi R^2$
Also, since  $V_1=V_2$
$\begin{array}{l}\frac{1}{4\pi {\epsilon }_0}\frac{q{\text{'}}_1}{R}=\frac{1}{4\pi {\epsilon }_0}\frac{q{\text{'}}_2}{2R}\\ q{\text{'}}_1=\frac{q{\text{'}}_2}{2}\end{array}$
Substituting the value of $q{\text{'}}_1$  from Eq(ii) in Eq(i)
$\frac{q{\text{'}}_2}{2}+q{\text{'}}_2=\sigma \times 20\pi R^2$
Or  $\frac{3q{\text{'}}_2}{2}=\sigma \times 20\pi R^2$
Or   $\frac{q{\text{'}}_2}{4\pi {\left(2R\right)}^2}=\frac{\sigma }{3}\times \frac{5}{2}$
New charge density on the bigger sphere is
$\frac{q{\text{'}}_2}{4\pi {\left(2R\right)}^2}=\frac{5\sigma }{6}$