JEE Main 2015 · 4 Apr · Shift 1 · Physics Q17

← JEE Main 2015 · 4 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 17 of 90

When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is $2.5\times10^{-4}$ ms $^{-1}$ . If the electron density in the wire is $8\times10^{28}$ m $^{-3}$ , the resistivity of the material is close to :

1.6\times10^{-8}

Ωm

1.6\times10^{-7}

Ωm

1.6\times10^{-6}

Ωm

1.6\times10^{-5}

Ωm

Answer and worked solution

Official answer: D

$J=nev_d$ and $E=\frac VL=50$ V/m:
$\rho=\frac E{nev_d}=\frac{50}{8\times10^{28}\times1.6\times10^{-19}\times2.5\times10^{-4}}\approx1.6\times10^{-5}\ \Omega\text{m}.$

Topics:Drift velocity and current Ohm's law and resistivity

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JEE Main 2015 · 4 Apr · Shift 1, Physics question 17