The plates of a parallel plate capacitor are separated by $d$. Two slabs of different dielectric constant $K_1$ and $K_2$ with thickness $\dfrac{3}{8}d$ and $\dfrac{d}{2}$, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates.
If $K_1 = 1.25K_2$, the value of $K_1$ is :
Answer: (A) $2.66$
The slabs fill $\dfrac{3d}{8} + \dfrac{d}{2} = \dfrac{7d}{8}$, leaving an air gap of $\dfrac{d}{8}$. For slabs stacked between the plates:
$$C = \frac{\epsilon_0 A}{\dfrac{d}{8} + \dfrac{3d}{8K_1} + \dfrac{d}{2K_2}}$$
$C = 2C_0 = \dfrac{2\epsilon_0 A}{d}$, so the denominator equals $\dfrac{d}{2}$:
$$\frac{1}{8} + \frac{3}{8K_1} + \frac{1}{2K_2} = \frac{1}{2} \;\Rightarrow\; \frac{3}{8K_1} + \frac{1}{2K_2} = \frac{3}{8}$$
With $K_2 = \dfrac{K_1}{1.25}$: $\dfrac{1}{2K_2} = \dfrac{1.25}{2K_1} = \dfrac{5}{8K_1}$.
$$\frac{3}{8K_1} + \frac{5}{8K_1} = \frac{1}{K_1} = \frac{3}{8} \;\Rightarrow\; K_1 = \frac{8}{3} \approx 2.66$$
Solution by Sreeraj P, M.Sc Physics