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(D) Since the shell $B$ is earthed,its potential $V_B = 0$. Due to induction,a charge $-Q$ is induced on the inner surface of shell $B$. For $a \leq r

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All of the above.

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(D) Since the shell $B$ is earthed,its potential $V_B = 0$. Due to induction,a charge $-Q$ is induced on the inner surface of shell $B$. For $a \leq r \leq b$,the electric field is due to the charge $Q$ on sphere $A$ only: $E = \frac{1}{4\pi \varepsilon_0} \frac{Q}{r^2}$. Thus,option $A$ is correct. The potential at distance $r$ is the sum of potentials due to charge $Q$ on $A$ and induced charge $-Q$ on $B$: $V(r) = \frac{1}{4\pi \varepsilon_0} \frac{Q}{r} + \frac{1}{4\pi \varepsilon_0} \frac{-Q}{b} = \frac{1}{4\pi \varepsilon_0} Q \left( \frac{1}{r} - \frac{1}{b} \right)$. Thus,option $B$ is correct. The potential of sphere $A$ is $V_A = V(a) = \frac{1}{4\pi \varepsilon_0} Q \left( \frac{1}{a} - \frac{1}{b} \right)$. Since $V_B = 0$,the potential difference $V_A - V_B = \frac{1}{4\pi \varepsilon_0} Q \left( \frac{1}{a} - \frac{1}{b} \right)$. Thus,option $C$ is correct. Therefore,all statements are correct.

$A$ conducting sphere $A$ of radius $a$,with charge $Q$,is placed concentrically inside a conducting shell $B$ of radius $b$. $B$ is earthed. $C$ is the common centre of the $A$ and $B$.

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