JEE Advanced 2026 Paper 2, Physics Section 1 Q1: Introduction And Ohm’s Law

JEE Advanced2026Paper 2Physics Section 1

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 $1 Ω$ . 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.]

A
$0.052$
B
$0.104$
C
$0.208$
D
$0.156$

Solution

Step 1 — Resistance of the wire. Using $R = \frac{L}{σ A}$ :

$R = \frac{100}{( 2 \times 1 0 ^{8} ) ( 0.5 \times 1 0 ^{- 6} )} = 1 Ω$ .

Step 2 — Current in the circuit. Total resistance is $R + r = 2 Ω$ , so $I = \frac{2}{2} = 1$ A.

Step 3 — Conduction electron density. With one free electron per atom,

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

Step 4 — Drift velocity from $I = n e A v_{d}$ .

$v_{d} = \frac{I}{n e A} = \frac{1}{( 6 \times 1 0 ^{28} ) ( 1.6 \times 1 0 ^{- 19} ) ( 0.5 \times 1 0 ^{- 6} )} \approx 2.08 \times 1 0^{- 4} m s^{- 1}$ .

$∴ v_{d} \approx 0.208 mm s^{- 1}$ , matching option (C).

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