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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2015
  • Medium
  • MCQ

When \(5\ V\) potential difference is applied across a wire of length \(0.1\ m\), the drift speed of electrons is \(2.5 \times 10^{-4}\) \(ms^{-1}\) . If the electron density in the wire is \(8 \times 10^{28}\) \(m^{-3}\) the resistivity of the material is close to

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Answer & Solution

Correct Answer

(B) \(1.6 \times 10^{-5} \) \(Ω m\)

Step-by-step Solution

Detailed explanation

\(V=I R=\left(n e A v_{d}\right) \rho \frac{\ell}{A}\)

\(\therefore \quad \rho=\frac{\mathrm{V}}{\mathrm{V}_{\mathrm{d}} \mathrm{lne}}\)

Here \(\mathrm{V}=\) potential difference

\(\mathrm{I}=\) length of wire

\(\mathrm{n}=\mathrm{no} .\) of electrons per unit volume of conductor.

\(\mathrm{e}=\mathrm{no.}\) of electrons

Placing the value of above parameters we get resistivity

\(\rho=\frac{5}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 2.5 \times 10^{-4} \times 0.1}\)

\(=1.6 \times 10^{-5} \,\Omega m\)

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