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Physics Question on Current electricity

When 5 V5\,V potential difference is applied across a wire of length 0.1 m0.1\,m, the drift speed of electrons is 2.5×10−42.5\times10^{-4} ms−1ms^{-1}. If the electron density in the wire is 8×1028 m−38\times10^{28}\, m^{-3} the resistivity of the material is close to

A

1.6×10−8Ω m1.6\times10^{-8}\Omega\,m

B

1.6×10−7Ω m1.6\times10^{-7}\Omega\,m

C

1.6×10−5Ω m1.6\times10^{-5}\Omega\,m

D

1.6×10−6Ω m1.6\times10^{-6}\Omega\,m

Answer

1.6×10−5Ω m1.6\times10^{-5}\Omega\,m

Explanation

Solution

The correct answer is C:1.6×10−5Ωm1.6\times 10^{-5}\Omega m
i=i=neAVd_{d}
⇒VR=\Rightarrow \frac{V}{R}= neAV _{d} \,\,\,\,\,\,\left\\{ R =\frac{\rho l}{A}\right\\}
⇒V×Aρℓ=neAVd\Rightarrow \frac{V \times A}{\rho \ell}= neAV _{d}
⇒5ρ×0.1=8×1028×1.6×10−19×2.5×10−4\Rightarrow \frac{5}{\rho \times 0.1}=8 \times 10^{28} \times 1.6 \times 10^{-19} \times 2.5 \times 10^{-4}
⇒ρ=1.56×10−5 Ω m\Rightarrow \rho=1.56 \times 10^{-5} \, \Omega \,m
⇒ρ≃1.6×10−5 Ω m\Rightarrow \rho \simeq 1.6 \times 10^{-5} \,\Omega\, m