Two concentric shells A and B have radii R and 2R, charges $q_{A}$ and $q_{B}$ and potentials 2V and 3V /2 respectively. Now, the shell B is earthed and the new charges on them become $q_{A}^{1}$ and $q_{B}^{1}$ . Then

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$\frac{q_{A}^{1}}{q_{B}^{1}} = 1$

Potential difference between A and B after earthing becomes $\frac{3 V}{2}$

Potential difference between A and B after earthing becomes $\frac{V}{2}$

$\frac{q_{A}^{}}{q_{B}^{}} = \frac{1}{2}$

answer is A, D.

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Detailed Solution

$\frac{1}{4 π \in_{0}} (\frac{q_{A}}{R} + \frac{q_{B}}{2 R}) = 2 V ............ (1)$
$\frac{1}{4 π \in_{0}} (\frac{q_{A}}{2 R} + \frac{q_{B}}{2 R}) = \frac{3}{2} V ............. (2)$
Solving equations (1) and (2), we get
$\frac{q_{A}}{q_{B}} = \frac{1}{2}$
When B is earthed, potential of B becomes zero. So,
$q_{B}^{1} = - q_{A}^{1} = - q_{A}$ {Charge on A remains same}
$\Rightarrow \frac{q_{A}^{1}}{q_{B}^{1}} = - 1$
Also after earthing,
$V_{A} - V_{B} = \frac{q_{A}}{4 π \in_{0}} (\frac{1}{R} - \frac{1}{2 R}) = \frac{q_{A}}{8 π \in_{0} R}$
Substituting $q_{B} = 2 q_{A}$ in equation (1), we get
$\frac{1}{4 π \in_{0}} \frac{q_{A}}{2 R} = \frac{V}{2}$
$\Rightarrow V_{A} - V_{B} = \frac{V}{2}$
Since B is earthed, so $V_{B} = 0$ and hence $V_{A} = \frac{V}{2}$