What does the test indicate?
The sequence

$\text{(i) } NaOH \quad \text{(ii) dil. } HNO_3 \quad \text{(iii) } AgNO_3$

is the standard test for halide ions released after hydrolysis.

• First, heating with $NaOH$ can hydrolyse a halo-compound and convert the halogen into $X^-$.

• Then acidification with dilute $HNO_3$ removes excess $OH^-$.

• Finally, $AgNO_3$ gives precipitate of silver halide:

  $AgCl = \text{white}, \quad AgBr = \text{pale yellow/cream}, \quad AgI = \text{yellow}$

So a yellow precipitate means formation of $AgI$, hence the compound must release iodide ion on hydrolysis.

Which type of compounds respond?
Haloalkanes hydrolyse with aqueous $NaOH$ relatively easily, but haloarenes generally do not under these mild conditions because the $C-X$ bond in aryl halides has partial double bond character due to resonance.

Hence, the required compound must be:

• an iodo compound, and
• one that can be hydrolysed by $NaOH$ under the given conditions.

From the given set I–IV
The compound labeled IV is the one that fits this requirement, so it will produce $I^-$ on hydrolysis and then give:

$Ag^+ + I^- \rightarrow AgI \downarrow$

which is a yellow precipitate.

Therefore the correct option is:

$\boxed{\text{IV}}$

which corresponds to Option B.