Skip to content
dronacharya.ai
← JEE Advanced 2026 · Paper 2
PhysicsSingle correct+3 / -1Question 19 of 54

JEE Advanced 2026 · Paper 2, Physics question 1

A metal wire of cross-sectional area 0.50.5 mm2^2 and length 100100 m is connected across a battery of e.m.f. 22 V and internal resistance 1 Ω1\ \Omega. The density, atomic mass and electrical conductivity of the metal are 6.35×1036.35\times10^3 kg m−3^{-3}, 63.563.5 gm/mole and 2×1082\times10^8 mho m−1^{-1}, respectively. Assuming one conduction electron per atom of the metal, the drift velocity (in mm s−1^{-1}) of the electrons in the wire is:
[Take Avogadro's number as 6×10236\times10^{23} and charge of the electron as 1.6×10−191.6\times10^{-19} C.]

Choose one option
Answer and worked solution

Official answer: C

Current. R=LσA=1002×108×0.5×10−6=1 ΩR=\dfrac{L}{\sigma A}=\dfrac{100}{2\times10^8\times0.5\times10^{-6}}=1\ \Omega, so I=21+1=1I=\dfrac{2}{1+1}=1 A.

Electron density (one per atom):
n=ρNAM=6.35×103×6×102363.5×10−3=6×1028 m−3.n=\frac{\rho N_A}{M}=\frac{6.35\times10^3\times6\times10^{23}}{63.5\times10^{-3}}=6\times10^{28}\ \text{m}^{-3}.

Drift velocity.
vd=IneA=16×1028×1.6×10−19×0.5×10−6=2.08×10−4 m s−1≈0.208 mm s−1.v_d=\frac{I}{neA}=\frac{1}{6\times10^{28}\times1.6\times10^{-19}\times0.5\times10^{-6}}=2.08\times10^{-4}\ \text{m s}^{-1}\approx0.208\ \text{mm s}^{-1}.

Topics:Drift velocity and currentOhm's law and resistivityEMF and internal resistance
← Previous questionNext question →