Solution:

The incorrect statement is "The charge on the capacitor is not conserved".

Here’s the explanation:

1. Initially, when the capacitor is connected to the battery with emf
   $V$ 

   , it stores a charge $Q_{\text{initial}} = C \times V$ 

   , where $C$ 

   is the capacitance of the air capacitor.

2. After disconnecting the capacitor from the battery, the charge on the capacitor is conserved because the capacitor is isolated. No charge can flow in or out.
3. When the dielectric slab of dielectric constant
   $K$ 

   is inserted, the capacitance of the capacitor increases to $K \times C$ 

   , but the charge remains the same because the capacitor is disconnected from the battery. The charge is now distributed on the new capacitance, and the voltage across the plates decreases.

4. The statement "charge is not conserved" is incorrect because charge is conserved in this isolated system. The only change is in the voltage across the capacitor due to the increased capacitance.

Thus, charge on the capacitor is conserved and the incorrect statement is the one claiming otherwise.