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Question

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

Step 1: Current flowing through the wire
We know that
I=nAevd
We also know by ohm's law , I=VR
VR=Anevd .....(1)

Step 2: Resistance of the wire
R=ρ lA .....(2)

Step 3: Solving above equations
From (1) and (2)
V(ρlA)=AnevdVAρl=Anevdρ=Vlnevd=5V0.1m×(8×1028m3)×(1.6×1019C)×(2.5×104ms1)=1.5625×105 Ωm
ρ1.6×105 Ωm
Hence, option (D) is correct.

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