The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K1 and K2 with thickness 3d/8 and 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 K₁ = 1.25 K₂, the value of K1 is:

The correct option is 2.66.

Step 1: Understand the initial state of the capacitor
Let a parallel plate capacitor have a plate area $A$ and plate separation $d$. When there is vacuum or air between the plates, its capacitance $C_0$ is given by:
$C_0=\frac{{\epsilon }_{0}A}{d}$
where ${\epsilon }_0$ is the permittivity of free space.

Step 2: Identify the thicknesses of the dielectric slabs and the remaining air gap
Two slabs of dielectric constants $K_1$ and $K_2$ are inserted into the capacitor:
- Thickness of the first slab: $t_1=\frac{3d}{8}$
- Thickness of the second slab: $t_2=\frac{d}{2}$
The remaining space is filled with air (dielectric constant $K_3=1$). The thickness of this remaining air gap is:
$t_3=d-\left(t_1+t_2\right)=d-\left(\frac{3d}{8}+\frac{d}{2}\right)=d-\frac{7d}{8}=\frac{d}{8}$

Step 3: Set up the formula for equivalent capacitance
When multiple dielectric slabs are placed parallel to the capacitor plates, the system can be represented as a series combination of individual capacitors. The equivalent capacitance $C$ is given by:
$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$
Using the formula for capacitance of a filled capacitor $C_i=\frac{K_i{\epsilon }_{0}A}{t_i}$, we write:
$\frac{1}{C}=\frac{t_1}{K_1{\epsilon }_{0}A}+\frac{t_2}{K_2{\epsilon }_{0}A}+\frac{t_3}{1·{\epsilon }_{0}A}$
Substituting the thicknesses $t_1$, $t_2$, and $t_3$:
$\frac{1}{C}=\frac{1}{{\epsilon }_{0}A}\left(\frac{3d}{8K_1}+\frac{d}{2K_2}+\frac{d}{8}\right)$
Factor out $d$ from the numerator:
$\frac{1}{C}=\frac{d}{{\epsilon }_{0}A}\left(\frac{3}{8K_1}+\frac{1}{2K_2}+\frac{1}{8}\right)$

Step 4: Use the given relations to solve for $K_1$
We are given that the new capacitance is two times larger than the original capacitance:
$C=2C_0=\frac{2{\epsilon }_{0}A}{d}\Rightarrow \frac{1}{C}=\frac{d}{2{\epsilon }_{0}A}$
Equating the two expressions for $\frac{1}{C}$:
$\frac{d}{2{\epsilon }_{0}A}=\frac{d}{{\epsilon }_{0}A}\left(\frac{3}{8K_1}+\frac{1}{2K_2}+\frac{1}{8}\right)$
Dividing both sides by $\frac{d}{{\epsilon }_{0}A}$:
$\frac{1}{2}=\frac{3}{8K_1}+\frac{1}{2K_2}+\frac{1}{8}$
Subtract $\frac{1}{8}$ from both sides:
$\frac{1}{2}-\frac{1}{8}=\frac{3}{8K_1}+\frac{1}{2K_2}$
$\frac{3}{8}=\frac{3}{8K_1}+\frac{1}{2K_2}$
We are also given that $K_1=1.25K_2=\frac{5}{4}$, which gives:
$K_2=\frac{4}{5}$
Substitute this relation into the equation:
$\frac{3}{8}=\frac{3}{8K_1}+\frac{1}{2\left(0.8K_1\right)}$
$\frac{3}{8}=\frac{3}{8K_1}+\frac{1}{1.6K_1}$
$\frac{3}{8}=\frac{3}{8K_1}+\frac{5}{8K_1}$
$\frac{3}{8}=\frac{3+5}{8K_1}$
$\frac{3}{8}=\frac{8}{8K_1}$
$\frac{3}{8}=\frac{1}{K_1}$
Solving for $K_1$:
$K_1=\frac{8}{3}\approx 2.66$