Principle of the test with $AgNO_3$

Silver nitrate solution is commonly used to test whether a halide ion can be released from an organic halide.

• If the compound easily ionizes to give $Cl^-$, $Br^-$, etc., then a precipitate of silver halide forms.
• For chloride: $Ag^+ + Cl^- \rightarrow AgCl\downarrow$

Thus, compounds that undergo easy nucleophilic substitution / ionization with release of halide ion give a precipitate.

Examine each option

Option A: $CHCl_3$ (chloroform)

• The chlorine atoms are attached to a carbon bearing hydrogen and three chlorines.
• It does not ionize easily to release $Cl^-$ in $AgNO_3$ solution.
• Hence, it is not likely to give a precipitate.

Option B: $(CH_3)_3CCl$ (tert-butyl chloride)

• This is a tertiary alkyl halide.
• Tertiary halides undergo substitution very easily by the $S_N1$ mechanism because they form a stable tertiary carbocation: $(CH_3)_3CCl \rightarrow (CH_3)_3C^+ + Cl^-$
• The released $Cl^-$ reacts with $Ag^+$: $Ag^+ + Cl^- \rightarrow AgCl\downarrow$
• Therefore, this compound will give a precipitate.

Option C: $CCl_4$ (carbon tetrachloride)

• This compound does not undergo hydrolysis/ionization easily.
• No easy release of $Cl^-$ occurs.
• So it is not likely to give a precipitate.

Option D: $CH_2=CH{-}Cl$ (vinyl chloride)

• Here chlorine is attached to an $sp^2$ hybridized carbon.
• Vinyl halides are very resistant to nucleophilic substitution because the $C{-}Cl$ bond has partial double bond character due to resonance.
• Hence, it does not readily release $Cl^-$.
• So it is not likely to give a precipitate.

Conclusion

Only tert-butyl chloride readily gives $Cl^-$ with $AgNO_3$ solution.

Therefore, the correct option is: $\boxed{B}$