1. Identify the first reaction: diazotization of aniline

Aniline in dilute HCl reacts with sodium nitrite at $0^\circ\text{C}$ to form benzene diazonium chloride.

So the solution being added dropwise contains benzenediazonium chloride.

2. Now identify the coupling components present in the second solution

The diazonium salt is added to an equimolar mixture of aniline and phenol in dilute HCl.

We must decide which aromatic ring will undergo azo coupling more readily.

3. Effect of acidic medium on aniline and phenol

In dilute HCl:

• Aniline gets protonated to form anilinium ion:

  $$\ce{C6H5NH2 + HCl -> C6H5NH3+Cl-}$$

  The group $\ce{-NH3+}$ is strongly deactivating and meta-directing, so the ring becomes much less reactive toward azo coupling.

• Phenol in acidic medium remains largely as phenol, $\ce{-OH}$ being an activating group. Although less activated than phenoxide in basic medium, it is still much more reactive than protonated aniline.

Therefore, coupling occurs predominantly with phenol, not with aniline.

4. Position of coupling in phenol

The $\ce{-OH}$ group is ortho/para-directing.

During azo coupling of benzenediazonium chloride with phenol, substitution occurs mainly at the para position because:

• para position is activated,
• steric hindrance is lower than at ortho position.

Thus the major product is p-hydroxyazobenzene:

with $\ce{-OH}$ at the para position relative to the azo linkage.

5. Conclusion

So the major product is $p$-hydroxyazobenzene (4-hydroxyazobenzene).

Hence the correct option is C.

6. Comparison with stored correct answer

Stored correct answer = C

My derived answer = C

They agree.