Consider a capacitor circuit charged by a battery to a potential difference $V_O$ as shown in the figure.

When a dielectric slab is inserted into both the capacitors, new potential difference will be

$\frac{8V_0}{5K}$

The correct answer is option (3) : $\frac{8V_0}{5K}$

On inserting dielectric charge on the combination of capacitor remains same $q=cV$

$4C_0V_0=\left(KC_0+\frac{3C_0K}{2}\right)V_1$

$V_1=\frac{8V_0}{5K}$