JEE Main 2019 — Current Electricity Question with Solution
From: JEE Main 2019 (Online) 9th April Evening Slot
Question
In a conductor, if the number of conduction
electrons per unit volume is 8.5 × 1028 m–3 and
mean free time is 25ƒs (femto second), it's
approximate resistivity is :-
(me = 9.1 × 10–31 kg)
(me = 9.1 × 10–31 kg)
Choose an option
Show full solutionCorrect option: A
Correct answer
A10–8 m
Step-by-step explanation
= 3.34 × 10–8 m
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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.