Two concentric conducting spherical shells of radii $R$ and $2 R$ having charges $q_{A}$ and $q_{B}$ and potential $2 v$ and $3 v / 2$ respectively. Now shell $B$ is earthed by closing switch $S .$ Let charges on spherical shells $A & B$ become ${q^{'}}_{A}$ and ${q^{'}}_{B}$ respectively, then

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$q_{A} / q_{B} = \frac{1}{2}$ (magnitude)

${q^{'}}_{A} / {q^{'}}_{B} = 1$ (magnitude)

Potential difference between $A$ and $B (v_{A} - v_{B})$ after earthing becomes $\frac{v}{2}$

Potential of $A$ after earthing becomes $\frac{3 v}{2}$

answer is A, B, D.

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Detailed Solution

$V_{A} = \frac{K q_{A}}{R} + \frac{K q_{B}}{2 R} 2 V, V_{B} = \frac{K (q_{A} + q_{B})}{2 R} = \frac{3}{2} V$
$\Rightarrow \frac{q_{A}}{q_{B}} = \frac{1}{2}$
After earthing of $B, q_{A} = q_{B} & V' B = \frac{K (q'_{A} + q'_{B})}{2 R} = 0$
$\Rightarrow | \frac{q' A}{q B} | = 1 & V'_{A} - V'_{B} = \frac{V}{2}$