Step 1: Understanding the Question.
We need the product of phenol with dilute nitric acid. Two things matter here: where the nitro group goes, and how many nitro groups enter.
Step 2: Number of nitro groups.
Dilute HNO$_3$ is a weak nitrating medium. The OH group already makes the ring very reactive, so even this mild acid can add one NO$_2$ group. Adding three needs concentrated HNO$_3$.
So the option with three nitro groups can be dropped, which removes option (2).
Step 3: Position of the nitro group.
Look at the resonance forms of phenol. The lone pair on oxygen puts negative charge at carbons 2, 4 and 6 of the ring.
The attacking NO$_2^+$ ion goes to these carbons. Carbons 3 and 5 (meta) get no extra negative charge.
So any structure with a meta nitro group, which are options (1) and (4), cannot be right.
Step 4: What is left.
Only option (3) remains. It shows 2-nitrophenol and 4-nitrophenol, the ortho and para isomers.
Final Answer:
The products are ortho and para nitrophenol, option (3).
\[ \boxed{\text{Option (3)}} \]
