A spherical metal shell $A$ of radius $R_A$ and a solid metal sphere $B$ of radius $R_B(<R_A)$ are kept far apart and each is given $+Q$. Now they are connected by thin metal wire. Then

Inside a conducting shell electric field is always zero. Therefore, Option (1) is correct. When the two are connected, their potentials become the same.

$$\begin{gathered} \Rightarrow V_A=V_B \\ \Rightarrow \frac{Q_A}{R_A}=\frac{Q_B}{R_B} \left\{\because V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}\right\} \end{gathered}$$

Since, $R_A>R_B$

$\Rightarrow Q_A>Q_B$

So, Option (2) is correct.

Potential is equal to, $V=\frac{\sigma R}{{\epsilon }_0}$

Since, $V_A=V_B$

$$\begin{gathered} \Rightarrow {\sigma }_A{R}_A={\sigma }_B{R}_B \\ \Rightarrow \frac{{\sigma }_A}{{\sigma }_B}=\frac{R_B}{R_A} \\ \Rightarrow {\sigma }_A<{\sigma }_B \end{gathered}$$

So, Option (3) is correct.

Electric field on surface, $E=\frac{\sigma }{{\epsilon }_0}$

$\Rightarrow E\propto \sigma$

Since, ${\sigma }_A<{\sigma }_B\Rightarrow E_A<E_B$

So, Option (4) is also correct.