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Question

When a 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5×104 ms1. If the electron density in the wire is 8×1028 m3, the resistivity of the material is close to:

A
1.6×108 Ωm
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B
1.6×107 Ωm
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C
1.6×106 Ωm
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D
1.6×105 Ωm
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Solution

The correct option is D 1.6×105 Ωm
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=58×1028×1.6×1019×2.5×104×0.1
ρ=1.6×105 Ωm

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