JEE Main 2015PhysicsCurrent ElectricityEasyMCQ

JEE Main 2015 — Current Electricity Question with Solution

JEE Main 2015 (04 Apr)

Question

When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5×10-4 m s-1 . If the electron density in the wire is 8×1028 m-3 , the resistivity of the material is close to:

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Show full solutionCorrect option: A
Correct answer
A1.56×10-5 Ω m

Step-by-step explanation

Potential difference= 5 V, length =0.1 m=l and the electron speed = drift velocity vd=2.5×10-4 m s-1, electron density n = 8 × 1 0 2 8 m - 3  and the charge on each electron(e) =1.6×10-19 C. We know i=nAevd  ...(i). And V=iR        ...(ii)



Resistance R is also equal to, ρlA⇒R=ρlA

ρ=ARl        ρ = Resistivity

=Al×Vi     [from (ii)]=Avl×nA evd & [from (i)]=Vl×n×e×vd

= 5 0.1 × 8 × 1 0 2 8 × 1.6 × 1 0 - 1 9 × 2.5 × 1 0 - 4

=0.156×10-4 Ω m=1.56×10-5 Ω m

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About this question

This is a previous-year question from JEE Main 2015, covering the Current Electricity chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.