Step-by-step Derivations:

1. Understand the Mechanism of Acid-Catalyzed Ether Hydrolysis

   The acid-catalyzed hydrolysis of an ether involves two main steps: a) Reversible Protonation: The ether oxygen atom is protonated by an acid (like $H_3O^+$) in a fast, reversible equilibrium step to form a protonated ether (an oxonium ion). $R-O-R' + H^+ \rightleftharpoons R-\overset{+}{O}H-R'$ b) Rate-Determining Step (RDS): The C-O bond of the protonated ether is cleaved. This can occur via an $S_N1$ or $S_N2$ mechanism, depending on the nature of the R and R' groups. This step is the slowest and therefore determines the overall reaction rate.

   • $S_N1$ pathway: If R or R' can form a stable carbocation (e.g., tertiary or resonance-stabilized), the protonated ether dissociates to form a carbocation and an alcohol.
   • $S_N2$ pathway: If R or R' is a primary or secondary alkyl group, a water molecule attacks the less sterically hindered carbon atom, displacing an alcohol molecule.

2. Factors Affecting the Reaction Rate

   The overall rate of hydrolysis depends on two key factors related to the steps above: a) Basicity of the Ether Oxygen: A more basic oxygen atom will be protonated more readily. This shifts the initial equilibrium to the right, increasing the concentration of the reactive intermediate (the protonated ether). Electron-donating groups (EDGs) attached to the oxygen increase its basicity, while electron-withdrawing groups (EWGs) decrease it. b) Stability of the Transition State in the RDS: The rate of the C-O bond cleavage depends on the activation energy. For an $S_N1$-like mechanism, any factor that stabilizes the developing positive charge (carbocationic character) on the carbon atom will lower the activation energy and increase the rate. EDGs on the group that is cleaved stabilize the transition state.

3. Analysis of the Given Ether (X) and its Modifications

   The starting ether (X) is diphenyl ether, $C_6H_5-O-C_6H_5$.

   • Option D: No structural change is made to X. In diphenyl ether, both groups are phenyl groups. The phenyl group is electron-withdrawing via induction. This makes the oxygen atom not very basic. Furthermore, cleavage of the $C_{aryl}-O$ bond would require forming a very unstable phenyl carbocation. Nucleophilic attack ($S_N2$) on an $sp^2$ carbon of the benzene ring is also extremely difficult. Therefore, diphenyl ether is very unreactive and its hydrolysis is extremely slow.

   • Option A: One phenyl group is replaced by a methyl group. The ether becomes methyl phenyl ether (anisole), $CH_3-O-C_6H_5$. Hydrolysis of anisole proceeds via a different mechanism. After protonation, the C-O bond cleavage occurs by an $S_N2$ attack of water on the less hindered methyl group, not the phenyl group. Cleavage of the $CH_3-O$ bond is much easier than cleavage of the $C_6H_5-O$ bond. This reaction pathway is significantly faster than the hydrolysis of diphenyl ether.

   • Option B: One phenyl group is replaced by a para-methoxyphenyl group. The ether is $p-CH_3O-C_6H_4-O-C_6H_5$. The para-methoxy group ($-OCH_3$) is a strong electron-donating group (+R effect). This has two consequences:

     • It increases the basicity of the ether oxygen compared to diphenyl ether.
     • It strongly stabilizes the transition state for the cleavage of the $(p-CH_3O-C_6H_4)-O$ bond by delocalizing the developing positive charge on the aromatic ring through resonance. Due to these effects, this ether hydrolyzes faster than diphenyl ether (X).

   • Option C: Two phenyl groups are replaced by two para-methoxyphenyl groups. The ether is bis(para-methoxyphenyl) ether, $p-CH_3O-C_6H_4-O-C_6H_4-OCH_3-p$. This molecule has two strong electron-donating groups attached to the ether oxygen.

     • Enhanced Basicity: The presence of two EDGs makes this ether's oxygen the most basic among all the options. This leads to the highest concentration of the protonated intermediate.
     • Maximally Stabilized Transition State: Cleavage of either of the identical $C_{aryl}-O$ bonds is highly favored because the developing positive charge is stabilized by a para-methoxy group. The activation energy for this step is significantly lowered.

4. Conclusion

   Let's compare the relative rates:

   • Rate(D) is the slowest due to the deactivating nature of phenyl groups and the difficulty of breaking the $C_{aryl}-O$ bond.
   • Rate(A) is much faster than Rate(D) because of the accessible $S_N2$ pathway on the methyl group.
   • Rate(B) is faster than Rate(D) due to the activating effect of one $p-methoxy$ group.
   • Rate(C) is faster than Rate(B) because it has two activating groups, making the oxygen more basic and providing stabilization for cleavage on either side.

   The final comparison is between A and C. The rate of reaction can be expressed as Rate $\propto K_{eq} \times k_{RDS}$. For ether C, both the equilibrium constant for protonation ($K_{eq}$) and the rate constant for cleavage ($k_{RDS}$) are enhanced to the maximum extent. The combination of the most basic oxygen (highest $K_{eq}$) and a highly stabilized $S_N1$-like transition state (high $k_{RDS}$) makes the hydrolysis of bis(para-methoxyphenyl) ether the fastest among all the given options. The rate enhancement provided by a para-methoxy group in reactions involving carbocationic intermediates is typically very large (several orders of magnitude), making this pathway extremely efficient.