Shells I, II, III are conducting spherical neutral shells of radius R, 2R and 3R respectively. Shell I and shell III are connected. A point charge $q_{0}$ is kept at centre O. Now shell II is also grounded.
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Shell I and III are at same potential.

The charge on surface ‘b’ (outer surface of Shell I) is (2/11) $q_{0}$ .

The potential energy of system decreases when shell II is grounded.

The charge on surface ‘b’ (outer surface of Shell I) is (3/11) $q_{0}$ .

answer is A, B, D.

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Detailed Solution

$- q_{0} + q_{b} - q_{d} + q_{f} = 0 \dots \dots \dots \dots \dots (1) V_{I I} = 0 \frac{K q_{0}}{R_{2}} - \frac{K q_{0}}{R_{2}} + \frac{K q_{b}}{R_{2}} + \frac{K q_{b}}{R_{2}} - \frac{K q_{d}}{R_{3}} + \frac{K q_{f}}{R_{3}} = 0 q_{f} = - q_{d} (\frac{R_{3} - R_{2}}{R_{2}}) \dots \dots \dots \dots \dots \dots . (2) V_{I} - V_{I I I} \frac{K q_{0}}{R} - \frac{K q_{0}}{R} + \frac{K q_{b}}{R} - \frac{K q_{0}}{2 R} - \frac{K q_{d}}{3 R} + \frac{K q_{f}}{3 R} = \frac{K q_{0}}{3 R} - \frac{K q_{0}}{3 R} + \frac{K q_{b}}{3 R} - \frac{K q_{0}}{3 R} - \frac{K q_{b}}{3 R} + \frac{K q_{d}}{3 R} - \frac{K q_{d}}{3 R} + \frac{K q_{f}}{3 r} q_{d} = - 3 q_{b} \dots \dots \dots . (3) S o l v i n g e q n q_{b} = (\frac{2 q}{11})$