The drift velocity of electrons in a conductor is 'v' when a current 'i' flows through it. If the area of the cross-section of the conductor is reduced to half of its initial value, then the drift velocity will be:

$2v$

The correct answer is Option (2) → $v$

Drift velocity is given by:

$v_d = \frac{I}{n e A}$

Where $I$ = current, $n$ = electron density, $e$ = electron charge, $A$ = cross-sectional area.

If the area is reduced to half: $A' = \frac{A}{2}$, keeping current $I$ constant, then the new drift velocity $v'$:

$v' = \frac{I}{n e A'} = \frac{I}{n e (A/2)} = \frac{2I}{n e A} = 2v$

Final Answer: $v' = 2v$