A conducting wire of length \( l \) and area of cross-section \( (A) \) is connected to a battery of \( V \) volt. The drift velocity of electrons is \( v_{1} \). If another wire of the same material but length \( 2l \) and area of cross-section \( (A/2) \) is connected to the same battery, then the drift velocity of electrons is \( v_{2} \). The value of \( v_{1}/v_{2} \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
The drift velocity depends on the electric field inside the wire, \( E = V/l \), and not on its area of cross-section.
\( \displaystyle v_{d} = \frac{eE\tau}{m} = \frac{eV\tau}{ml} \)
Both wires are of the same material (same \( \tau \)) and are connected to the same battery (same \( V \)), so \( v_{d} \propto 1/l \):
\( \displaystyle \frac{v_{1}}{v_{2}} = \frac{l_{2}}{l_{1}} = \frac{2l}{l} = 2 \)