Q.A positively charged glass rod is brought close to a metallic rod isolated from ground. The charge on the side of the metallic rod away from the glass rod will be (A) same as that on the glass rod and equal in quantity (B) opposite to that on the glass rod and equal in quantity (C) same as that on the glass rod but lesser in quantity (D) same as that on the glass rod but more in quantity

[!TLDR] Being isolated (not earthed), the induced charge on the far side is opposite to the near side but only a fraction of the inducing charge -- same sign as the rod, lesser in quantity. [!ANSWER] (C) same as that on the glass rod but lesser in quantity.

When a positively charged glass rod is brought near (but not touching) an isolated metallic rod, the free electrons in the metal are attracted toward the near end (nearest the positive rod), leaving that near end with an induced NEGATIVE charge and the far end with an induced POSITIVE charge -- the far side's charge is the SAME sign as the glass rod, since it is left short of electrons in exactly the way the rod itself is short of them. Because the conductor is isolated (never earthed at any stage here, unlike the full induction procedure of Fig. 10.1), no charge is ever added to or removed from the conductor as a whole -- only a REDISTRIBUTION of the conductor's own existing free electrons occurs. Only a portion of the conductor's free electrons are pulled toward the near end by the rod's attraction (the redistribution is limited by the conductor's own internal resistance to further electron migration, and by the finite distance/geometry involved), so the magnitude of induced charge that accumulates on the far side is always LESS than the full charge on the inducing rod itself -- the rod's field is not strong enough, at a distance, to strip away an equal amount of charge from an isolated conductor. [!ANSWER] (C) same as that on the glass rod but lesser in quantity.

Apply the mechanics of electrostatic induction on an isolated (non-earthed) conductor: near side gets the opposite sign, far side gets the same sign as the inducing charge, but only a partial, lesser magnitude since no charge is actually transferred.

Assuming the induced charge equals the inducing charge exactly (confusing this with charging by direct contact/conduction), or getting the sign of the far side backwards.