In a parallel plate capacitor the separation between plates is 3 x...

Key Idea Capacitance of a parallel plate capacitor is give by relation, C=\varepsilon_r $\cdot \frac{\varepsilon_0 A}{d}$where, $\varepsilon_r=$ dielectric constant, $A=$ area of parallel plate and $d=$ distance between the plate. Here, $d_0=3 x, d_1=x, \varepsilon_1=3 k, d_2=2 x$ and $\varepsilon_2=5 k$ The capacitor is shown in the figure below,
Now, $C_1=3 k \frac{\varepsilon_0 A}{x}$ and $C_2=5 k \frac{\varepsilon_0 A}{2 x}$ Since, $m$ in series combination of capacitor, stored charge in each capacitor is same. So, V_1=$\frac{Q}{C_1} \text { and }$V_2=$\frac{Q}{C_2}\Rightarrow \quad V_1=\frac{\theta x}{3 k \varepsilon_0 A}$ and $V_2=\frac{\theta 2 x}{5 k \varepsilon_0 A}$ The ratio, $\frac{V_1}{V_2}=\frac{\frac{\theta x}{3 k \varepsilon_0 A}}{\frac{\theta 2 x}{5 k \varepsilon_0 A}}=\frac{5}{3 \times 2}=\frac{5}{6}$ Hence, option (d) is correct.