JEE Main 2021PhysicsAlternating CurrentHardNumerical

JEE Main 2021 — Alternating Current Question with Solution

JEE Main 2021 (25 Feb Shift 1)

Question

A transmitting station releases waves of wavelength 960 m. A capacitor of 2.56 μF is used in the resonant circuit. The self-inductance of coil necessary for resonance is x×10-8 H. find x

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Show full solutionCorrect answer: 10
Correct answer
10

Step-by-step explanation

λ=960 m

C=2.56 μF=2.56×10-6 F

c=3×108 m s-1

L=?

Now at resonance, ω0=1LC [Resonant frequency]

2πf0=1LC

On substituting f0=cλ, we have 2πcλ=1LC

On substituting f0=cλ, we have 2πcλ=1LC

Squaring both sides: 4π2c2λ2=1LC

=4×10×3×10829602=1 L×2.56×10-6

⇒1 L=4×10×9×1016×2.56×10-6960×960

⇒L=10×10-8 H

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About this question

This is a previous-year question from JEE Main 2021, covering the Alternating Current 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.