$(a)\;q\qquad(b)\;-q\qquad(c)\;zero\qquad(d)\;None$

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Answer : (a) -q

Explanation :

The charge on the inner surface should be $\;-q\;,$ because if we draw a closed gaussion surface through the material of the hollow sphere the total charge encloses by this gaussion surface should be zero . Let $\;q_{1}\;$ be the charge on the outer surface of sphere

Since outer sphere is earthed potential on it is zero . Therefore

$\large\frac{1}{4\;\pi\;\in_{0}}\;(\large\frac{q}{2R}-\large\frac{q}{2R}+\large\frac{q_{1}}{2R})=0$

$q^{|}=0\;.$

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