The potential difference across a metal wire is doubled. The wire becomes warmer, and the relaxation time of its free electrons falls to 0.8 times its earlier value. The ratio of the new drift speed to the earlier drift speed is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.6
The drift speed is \( v_d=eE\tau/m \), so it depends on both the field \( E \) and the relaxation time \( \tau \). The length of the wire does not change. So the field \( E=V/l \) doubles when \( V \) doubles.
\( \begin{aligned} \frac{v_d'}{v_d} &= \frac{E'}{E}\times\frac{\tau'}{\tau} \\ &= 2\times0.8 \\ &= 1.6 \end{aligned} \)
NCERT (2026-27):
Summary – “… If such collisions occur on an average at a time interval τ, vd = aτ = eEτ/m where a is the acceleration of the electron.”
Ohm’s Law – “… Further, if E is the magnitude of uniform electric field in the conductor whose length is l, then the potential difference V across its ends is El. …”