When a 5 V potential difference is applied across a wire of length 0.1m, the drift speed of electrons is 2.5×10−4ms−1. If the electron density in the wire is 8×1028m−3, the resistivity of the material is close to:
A
1.6×10−8Ωm
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B
1.6×10−7Ωm
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C
1.6×10−6Ωm
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D
1.6×10−5Ωm
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Solution
The correct option is D1.6×10−5Ωm We know, i=neAvd
where symbol have their usual meaning.
From Ohm's law, i=VR=neAvd-------(1)
Where R=ρLA
Putting value of R in equation (1) we get, VρLA=neAvd ρ=VnevdL ρ=58×1028×1.6×10−19×2.5×10−4×0.1 ρ=1.6×10−5Ωm