Charge Distribution Between Connected Spherical Shell and Solid Sphere

To determine the correct statements, let us analyze the electrostatic behavior of the spherical metal shell $\text{A}$ and the solid metal sphere $\text{B}$ after they are connected by a thin conducting wire.

1. Potential Equilibrium

Since the two metallic conductors are placed far apart (so that inductive interactions between them are negligible) and connected by a conducting wire, charge flows between them until they reach the same electrostatic potential:
$V_{\text{A}} = V_{\text{B}}$

The potential on the surface of a conducting sphere/shell carrying charge $Q$ and of radius $R$ is given by:
$V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R}$

Equating the potentials:
$\frac{1}{4\pi\varepsilon_0} \frac{Q_{\text{A}}}{R_{\text{A}}} = \frac{1}{4\pi\varepsilon_0} \frac{Q_{\text{B}}}{R_{\text{B}}} \implies \frac{Q_{\text{A}}}{R_{\text{A}}} = \frac{Q_{\text{B}}}{R_{\text{B}}}$

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2. Analysis of Options

• Option (A): Electric field inside Shell A
  For any hollow spherical conductor in electrostatic equilibrium, charge resides entirely on its outer surface. By Gauss's Law, the electric field inside the hollow cavity of shell A is zero everywhere:
  $E_{\text{A}}^{\text{inside}} = 0$
  Hence, Option (A) is correct.

• Option (B): Charge distribution
  From the equal potential condition, we have:
  $Q_{\text{A}} = Q_{\text{B}} \left( \frac{R_{\text{A}}}{R_{\text{B}}} \right)$
  Given that $R_{\text{A}} > R_{\text{B}}$, it directly follows that:
  $Q_{\text{A}} > Q_{\text{B}}$
  Hence, Option (B) is correct.

• Option (C): Ratio of surface charge densities
  The surface charge density $\sigma$ for a spherical surface of radius $R$ carrying charge $Q$ is:
  $\sigma = \frac{Q}{4\pi R^2}$

  Taking the ratio of surface charge densities:
  $\frac{\sigma_{\text{A}}}{\sigma_{\text{B}}} = \frac{\frac{Q_{\text{A}}}{4\pi R_{\text{A}}^2}}{\frac{Q_{\text{B}}}{4\pi R_{\text{B}}^2}} = \left( \frac{Q_{\text{A}}}{Q_{\text{B}}} \right) \left( \frac{R_{\text{B}}}{R_{\text{A}}} \right)^2$

  Substituting $\frac{Q_{\text{A}}}{Q_{\text{B}}} = \frac{R_{\text{A}}}{R_{\text{B}}}$:
  $\frac{\sigma_{\text{A}}}{\sigma_{\text{B}}} = \left( \frac{R_{\text{A}}}{R_{\text{B}}} \right) \left( \frac{R_{\text{B}}}{R_{\text{A}}} \right)^2 = \frac{R_{\text{B}}}{R_{\text{A}}}$
  Hence, Option (C) is correct.

• Option (D): Comparison of Electric Fields on the Surfaces
  The electric field just outside the surface of a conductor is given by $E = \frac{\sigma}{\varepsilon_0}$, or equivalently in terms of potential $V$:
  $E = \frac{V}{R}$

  Comparing the electric fields on the surfaces of A and B:
  $E_{\text{A}}^{\text{on surface}} = \frac{V_{\text{A}}}{R_{\text{A}}} \quad \text{and} \quad E_{\text{B}}^{\text{on surface}} = \frac{V_{\text{B}}}{R_{\text{B}}}$

  Since $V_{\text{A}} = V_{\text{B}}$ and $R_{\text{A}} > R_{\text{B}}$, we have:
  $E_{\text{A}}^{\text{on surface}} < E_{\text{B}}^{\text{on surface}}$
  Hence, Option (D) is correct.

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Conclusion

All four options (A), (B), (C), and (D) are correct.