Dielectrics question from JEE Main 2019 (9 Jan, Shift 1)

Answer: (b)

1. At distance $x$, dielectric thickness is $y = \frac{d x}{a}$; the strip of width $dx$ is air + dielectric in series: $dC = \frac{\epsilon_0 a\, dx}{d - y + y/K}$.
2. $d - y\left(1 - \frac{1}{K}\right) = d\left[1 - \frac{(K-1)x}{Ka}\right]$
3. $C = \int_0^a \frac{\epsilon_0 a\, dx}{d[1 - (K-1)x/(Ka)]} = \frac{\epsilon_0 a^2 K}{d(K-1)}\ln K$
4. So the answer is (b).