JEE Main 2019PhysicsCurrent ElectricityEasyMCQ

JEE Main 2019 — Current Electricity Question with Solution

JEE Main 2019 (09 Apr Shift 2)

Question

In a conductor, if the number of conduction electrons per unit volume is 8.5×1028  m-3 and mean free time is 25 fs (femto second), it’s approximate resistivity is: me=9.1×10-31 kg

 

Choose an option

Show full solutionCorrect option: D
Correct answer
D10-8 Ω m

Step-by-step explanation

We know i=neAvd
⇒i=VR=VAρl=EAρ
ρ=EAi=EAneAvd
Drift velocity can be written as vd=eEmeτ
ρ=EAmeneAeEτ=mene2τ
=9.1×10-318.5×1028×1.6×10-192×25×10-15
≈1.65×10-8 Ω m

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About this question

This is a previous-year question from JEE Main 2019, 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.