i-A, ii-B, iii-C, iv-D

To match the columns, let’s analyze each scenario in Column-I step by step and understand the corresponding effects in Column-II.

1. **(i)** When a dielectric slab is gradually inserted between the plates of an isolated parallel plate charged capacitor, it increases the capacitance which leads to a decrease in the electric potential energy of the system because the charge remains constant; hence, energy decreases.
**Match: (i) - [A]**

2. **(ii)** When a dielectric slab is gradually inserted between the plates of a parallel plate capacitor while its potential is kept constant (connected to a battery), work is done by the external agent to insert the slab against the attractive force between the charged plates. The energy supplied to the system causes the work done by the external source to be positive.
**Match: (ii) - [B]**

3. **(iii)** When the plates of a parallel capacitor are pulled apart while keeping its potential constant (by connecting to a battery), work is done by the battery as it supplies additional charge to maintain constant potential, thus work done by the battery is positive.
**Match: (iii) - [C]**

4. **(iv)** When the plates of a parallel plate capacitor are pulled apart keeping charge constant (by disconnecting the battery), the external agent must do work against the attractive force due to charges on the plates, which causes the work done by the external agent to be negative.
**Match: (iv) - [D]**

Therefore, the correct pairs are: (i - A), (ii - B), (iii - C), (iv - D).