A metal wire of cross-sectional area \(0.5 \mathrm{~mm}^{2}\) and length 100 m is connected across a battery of e.m.f. 2 V and internal resistance \(1 \Omega\). The density, atomic mass and electrical conductivity of the metal are \(6.35 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\), \(63.5 \mathrm{gm}/\)mole and \(2 \times 10^{8} \mathrm{mho} \mathrm{m}^{-1}\), respectively. Assuming one conduction electron per atom of the metal, the drift velocity (in \(\mathrm{mm} \mathrm{s}^{-1}\)) of the electrons in the wire is: [Take Avogadro's number as \(6 \times 10^{23}\) and charge of the electron as \(1.6 \times 10^{-19} \mathrm{C}\).]

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Current Electricity

A metal wire of cross-sectional area

0.5 mm^{2}

and length 100 m is connected across a battery of e.m.f. 2 V and internal resistance

1Ω

. The density, atomic mass and electrical conductivity of the metal are

6.35 \times 1 0^{3} kg m^{- 3}

63.5 gm /

mole and

2 \times 1 0^{8} mho m^{- 1}

, respectively. Assuming one conduction electron per atom of the metal, the drift velocity (in

mm s^{- 1}

) of the electrons in the wire is:
[Take Avogadro's number as

6 \times 1 0^{23}

and charge of the electron as

1.6 \times 1 0^{- 19} C

Answer: (C)

First, calculate the resistance of the wire:

R = \frac{1}{σ} \frac{L}{A} = \frac{100}{2 \times 1 0 ^{8} \times 0.5 \times 1 0 ^{- 6}} = 1Ω

.

Total resistance:

R_{total} = R + r = 1 + 1 = 2Ω

.

Current:

I = \frac{E}{R _{total}} = \frac{2}{2} = 1 A

.

Electron number density:

n = \frac{ρ N _{A}}{M} = \frac{6.35 \times 1 0 ^{3} \times 6 \times 1 0 ^{23}}{0.0635} = 6 \times 1 0^{28} m^{- 3}

.

Drift velocity:

v_{d} = \frac{I}{n e A} = \frac{1}{6 \times 1 0 ^{28} \times 1.6 \times 1 0 ^{- 19} \times 0.5 \times 1 0 ^{- 6}} = 2.08 \times 1 0^{- 4} m/s = 0.208 mm/s

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Single correct option