1. First reaction: phenol with concentrated sulphuric acid

Phenol undergoes sulphonation with concentrated sulphuric acid. The $-OH$ group is ortho/para directing, so phenol forms mainly:

$o\text{-hydroxybenzenesulphonic acid} \quad \text{and} \quad p\text{-hydroxybenzenesulphonic acid}$

2. Second reaction: treatment with concentrated nitric acid

Now nitration occurs. In this process, the sulphonic acid group $(-SO_3H)$ is a meta-directing group, while $-OH$ is ortho/para directing.

A well-known result is that sulphonation followed by nitration of phenol is used to control substitution and gives mainly ortho-nitrophenol after replacement/removal of the sulphonic acid group.

Thus, the product asked in standard JEE chemistry is:

$o\text{-nitrophenol}$

3. Checking the options

• A: $2,4,6$-trinitrobenzene — incorrect; this is not formed here.
• B: $o$-nitrophenol — correct.
• C: $p$-nitrophenol — not the expected product in this sequence.
• D: nitrobenzene — incorrect; phenol does not lose its $-OH$ group.

4. Final answer

The correct option is:

$\boxed{\text{B: } o\text{-nitrophenol}}$