A spherical metal shell $A$ of radius $R_A$ and a solid metal sphere $B$ of radius $R_B < \left(R_A\right)$…
A spherical metal shell $A$ of radius $R_A$ and a solid metal sphere $B$ of radius $R_B < \left(R_A\right)$ are kept far apart and each is given charge $+Q$. Now, they are connected by a thin metal wire. Then
Inside a conducting shell, electric field is always zero. Therefore, option (a) is correct. When the two are connected, their potentials become the same.
$
\begin{array}{ll}
\therefore & V_A=V_B \\
\text { or } & \frac{Q_A}{R_A}=\frac{Q_B}{R_B} \quad\left(\because V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\right)
\end{array}
$
Since, $R_A>R_B$
$
\therefore \quad Q_A>Q_B
$
$\therefore$ Option (b) is correct.
Potential is also equal to,
$
\begin{array}{rlrl}
V & =\frac{\sigma R}{\varepsilon_0} \\
V_A & =V_B \\
\therefore \quad \sigma_A R_A & =\sigma_B R_B \\
\text { or } \quad \sigma_B & \frac{\sigma_A}{\sigma_B} & =\frac{R_B}{R_A}
\end{array}
$
$\therefore$ Option (c) is correct.
Electric field on surface,
$
E=\frac{\sigma}{E_0} \text { or } E \propto \sigma
$
or $\quad \sigma_A < \sigma_B$
Since, $\quad \sigma_A < \sigma_B$
$\therefore \quad E_A < E_B$
$\therefore$ Option (d) is also correct.
$\therefore$ Correct options are (a), (b), (c) and (d).
Analysis of Question
(i) Question is simple.
(ii) $E=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R^2}=\frac{\sigma}{\varepsilon_0}$
(On surface)
As, $\sigma=$ surface charge density
$
=\frac{Q}{4 \pi R^2}
$
(iii) $V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}=\frac{\sigma R}{\varepsilon_0}$