To solve this problem, let's analyze the situation step-by-step.$\\$Initially, the capacitor is charged by a cell of emf $V. \\$The initial charge on the capacitor is given by:$\\ Q = CV \\$The initial energy stored in the capacitor is:$\\ U_{initial} = \frac{1}{2} CV^2 \\$After disconnecting the capacitor from the cell, a dielectric slab with dielectric constant $K \\$is inserted. The capacitance of the capacitor becomes:$\\ C' = KC \\$Since the capacitor is disconnected, the charge remains conserved:$\\ Q' = Q = CV \\$The new potential difference across the capacitor is:$\\ V' = \frac{Q'}{C'} = \frac{CV}{KC} = \frac{V}{K} \\$The new energy stored in the capacitor is:$\\ U_{final} = \frac{1}{2} C' (V')^2 = \frac{1}{2} KC \left(\frac{V}{K}\right)^2 = \frac{1}{2} \frac{CV^2}{K} \\$Now, let's analyze each option:$\\$Option 1: The energy stored in the capacitor decreases $K$ times.$\\$This is correct because $U_{final} = \frac{1}{2} \frac{CV^2}{K}$ is $\frac{1}{K}$ times $U_{initial}. \\$Option 2: The change in energy stored is $\frac{1}{2} CV^2 \left(\frac{1}{K} - 1 \right). \\$The change in energy $\Delta U = U_{final} - U_{initial} = \frac{1}{2} \frac{CV^2}{K} - \frac{1}{2} CV^2 \\ = \frac{1}{2} CV^2 \left(\frac{1}{K} - 1 \right). \\$This is correct.$\\$Option 3: The charge on the capacitor is not conserved.$\\$This is incorrect because the capacitor is disconnected from the cell, so the charge is conserved.$\\$Option 4: The potential difference between the plates decreases $K$ times.$\\$This is correct because $V' = \frac{V}{K}. \\$Therefore, the incorrect option is Option 3.