To solve this problem, we need to understand the behavior of aniline in a strong acidic medium during nitration.$\newline$Aniline is an aromatic amine with the structure $C_6H_5NH_2. \newline$The amino group $(NH_2)$ is a strong activating group and ortho/para directing in electrophilic aromatic substitution reactions.$\newline$However, in a strong acidic medium, the amino group gets protonated to form the anilinium ion $(C_6H_5NH_3^+). \newline$Step-by-step explanation:$\newline$1. In a strong acidic medium, aniline is protonated:$\newline C_6H_5NH_2 + H^+ \rightarrow C_6H_5NH_3^+ \newline$2. The anilinium ion $(C_6H_5NH_3^+)$ is a deactivated form of aniline.$\newline$• The positive charge on the nitrogen reduces the electron density on the benzene ring.$\newline$• This makes the ring less reactive towards electrophilic substitution.$\newline$3. The anilinium ion is meta-directing:$\newline$• The positive charge on the nitrogen withdraws electrons from the ortho and para positions.$\newline$• This makes the meta position relatively more electron-rich compared to ortho and para positions.$\newline$4. As a result, nitration in a strong acidic medium leads to the formation of m-nitroaniline.$\newline$Analyzing the options:$\newline$• Option 1: Incorrect, as the nitro group does not always go to the meta position in absence of substituents.$\newline$• Option 2: Incorrect, as the amino group is not meta-directing in its neutral form.$\newline$• Option 3: Incorrect, as the nitro group does not always go to the meta position despite substituents.$\newline$• Option 4: Correct, as in acidic medium, aniline is present as anilinium ion, which is meta-directing.$\newline$Therefore, the correct option is Option 4.