$\text{The plates of a parallel plate capacitor are separated by d. Two slabs of dielectric constants K₁ and K₂ with thickness 3d/8 and d/2 respectively are inserted.}$

$\text{Due to this, capacitance becomes twice the original. If K₁ = 1.25 K₂, find K₁.}$ $$

The correct option is 2.66.

Let us solve the problem step-by-step.

Step 1: Understand the initial capacitance
For a parallel plate capacitor of plate area $A$ and separation $d$ without any dielectric, the capacitance $C$ is given by:
$C=\frac{{\epsilon }_{0}A}{d}$
where ${\epsilon }_0$ is the permittivity of free space.

Step 2: Express capacitance with dielectric slabs
When dielectric slabs of thicknesses $t_1$ and $t_2$ with dielectric constants $K_1$ and $K_2$ are inserted into the capacitor, the equivalent capacitance $C^{\prime }$ is given by the formula:
$C^{\prime }=\frac{{\epsilon }_{0}A}{d-t_1-t_2+\frac{t_1}{K_1}+\frac{t_2}{K_2}}$

Given values for thickness:
$t_1=\frac{3d}{8}$
and
$t_2=\frac{d}{2}$
Substituting these thicknesses into the denominator:
$d-t_1-t_2=d-\frac{3d}{8}-\frac{d}{2}=d-\frac{7d}{8}=\frac{d}{8}$
Thus, the expression for $C^{\prime }$ becomes:
$C^{\prime }=\frac{{\epsilon }_{0}A}{\frac{d}{8}+\frac{3d}{8K_1}+\frac{d}{2K_2}}$

Step 3: Setup the equation based on the given conditions
We are given that the new capacitance is twice the original capacitance:
$C^{\prime }=2C$
Substituting the expressions for $C^{\prime }$ and $C$:
$\frac{{\epsilon }_{0}A}{\frac{d}{8}+\frac{3d}{8K_1}+\frac{d}{2K_2}}=2\left(\frac{{\epsilon }_{0}A}{d}\right)$
Dividing both sides by ${\epsilon }_{0}A$ and taking the reciprocal:
$\frac{1}{8}+\frac{3}{8K_1}+\frac{1}{2K_2}=\frac{1}{2}$
Subtracting $\frac{1}{8}$ from both sides:
$\frac{3}{8K_1}+\frac{1}{2K_2}=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}$

Step 4: Solve for $K_1$
We are given the relationship:
$K_1=1.25K_2=\frac{5}{4}{K}_2$
Therefore, we can express $K_2$ in terms of $K_1$:
$K_2=\frac{4}{5}{K}_1$
Substitute this into our simplified equation:
$\frac{3}{8K_1}+\frac{1}{2\left(\frac{4}{5}{K}_1\right)}=\frac{3}{8}$
Simplify the second term:
$\frac{3}{8K_1}+\frac{5}{8K_1}=\frac{3}{8}$
Combine the fractions:
$\frac{8}{8K_1}=\frac{3}{8}$
Simplifying gives:
$\frac{1}{K_1}=\frac{3}{8}$
Solving for $K_1$:
$K_1=\frac{8}{3}\approx 2.66$