A parallel plate capacitor with plate separation 5 mm is charged up by a battery. It is found that on…

Without dielectric, capacitance will be $C=\frac{A{\epsilon }_0}{d}$.

Therefore, charge on it, $Q=\frac{A{\epsilon }_0}{d}V$

Now, with dielectric capacitance will be $C^{\text{'}}=\frac{A{\epsilon }_0}{d-t+{\displaystyle \frac{t}{K}}}$.

Hence, charge on it will be $Q^{\text{'}}=\frac{A{\epsilon }_{0}V}{d-t+{\displaystyle \frac{t}{K}}}$

Given:

$\frac{A{\epsilon }_{0}V}{d-t+{\displaystyle \frac{t}{K}}}=\left(1.25\right)\frac{{\displaystyle A{\epsilon }_{0}V}}{{\displaystyle d}}$

$\Rightarrow 1.25\left(3+\frac{2}{K}\right)=5$

$\Rightarrow K=2$