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) $2.66$
Concept. Stacked layers: $C = \dfrac{\epsilon_0A}{\sum t_i/K_i}$; air gap $\dfrac d8$.
Step 1. Condition. $C = 2C_0$: $\dfrac18 + \dfrac{3}{8K_1} + \dfrac{1}{2K_2} = \dfrac12$.
Step 2. Substitute $K_2 = \dfrac{K_1}{1.25}$. $\dfrac{3}{8K_1} + \dfrac{5}{8K_1} = \dfrac{3}{8} \Rightarrow K_1 = \dfrac83 \approx 2.66$.
Answer. $2.66$ (Option A).
Solution by Sreeraj P, M.Sc Physics