Two nichrome wires are connected in series with a battery. Their lengths are in ratio of 3 : 2 and their resistances are in ratio of 3 : 4, respectively. The ratio of drift velocities of free electrons in them is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1 : 2
In a series circuit the same current flows through both wires, and the wires are of the same material, so the number density n of free electrons is also the same.
\( \displaystyle I = neAv_d \;\Rightarrow\; v_d \propto \frac{1}{A} \)
The area ratio follows from the resistances.
\( \displaystyle R = \frac{\rho l}{A} \;\Rightarrow\; A = \frac{\rho l}{R} \)
\( \displaystyle \frac{A_1}{A_2} = \frac{l_1}{l_2}\times\frac{R_2}{R_1} = \frac{3}{2}\times\frac{4}{3} \)
\( \displaystyle \frac{A_1}{A_2} = 2 \)
\( \displaystyle \frac{v_1}{v_2} = \frac{A_2}{A_1} = \frac{1}{2} \)
Shortcut: The drift velocity is proportional to the electric field in the wire, \( E = V/l \), and in series \( V \propto R \). So \( v_d \propto R/l \), giving \( \frac{3/4}{3/2} = \frac{1}{2} \).