A student takes Cu, Al, Fe and Zn pieces separately in four test tubes labelled as I, II, III, and IV respectively. He adds 10 mL of freshly prepared ferrous sulphate solution to each test tube and observes the colour of the metal residue in each case.

He would observe a black residue in the test tubes :
(1) (I) and (II)
(2) (I) and (III)
(3) (II) and (III)
(4) (II) and (IV)

Copper is reactive than iron. Therefore, no reaction takes place when ferrous sulphate solution is added to copper containing test tube. So, in case one nothing will happen.
When ferrous sulphate is added to aluminium containing test tube then following reaction takes place:
$2\mathrm{Al}\left(\mathrm{s}\right)+3{\mathrm{FeSO}}_4\left(\mathrm{aq}\right)\to {\mathrm{Al}}_2\left({\mathrm{SO}}_4{)}_3\right(\mathrm{aq})+3\mathrm{Fe}(\mathrm{s})$
So the green colour of FeSO₄ disappears. Iron metal is settling down at the bottom of the test tube.
When ferrous sulphate is added to the iron-containing solution, then no change will be observed since metal is same, i.e. iron.
When ferrous sulphate is added to the zinc-containing test tube then the following reaction takes place:
$\mathrm{Zn}\left(\mathrm{s}\right)+{\mathrm{FeSO}}_4\left(\mathrm{aq}\right)\to {\mathrm{ZnSO}}_4\left(\mathrm{aq}\right)+\mathrm{Fe}\left(\mathrm{s}\right)$
So the green colour of FeSO₄ disappears Fe metal is settling down at the bottom of the test tube.

Hence the correct option is 4.