A conductor of length 'L' is connected to a battery of emf 'E'. The drift velocity of electron in the conductor is ' $v_{d}$ '. The drift velocity of electron in the conductor on stretching it four times, to length 4 L keeping emf 'E' constant will be:

$0.25 v_d$

The correct answer is Option (2) → $0.25 v_d$

Resistance of a conductor (R),

$R=ρ\frac{L}{A}$

where,

L = length of conductor

A = Cross - sectional area

if the wire is stretched to length 4L then its area of cross - section or A/4 in order to keep the volume constant.

$∴R'=ρ\frac{4L×4}{A}=16R$

and,

$I=\frac{E}{F'}=\frac{E}{16R}=\frac{I}{16}$

and,

$I=neAv_d$

where,

$v_d$ - driff velocity

$∴{v_d}'=\frac{v_d}{16}=0.25v_d$