JEE Main 2021 — Alternating Current Question with Solution
From: JEE Main 2021 (Online) 27th July Evening Shift
Question
A 100 resistance, a 0.1 F capacitor and an inductor are connected in series across a 250 V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 Hz.
Choose an option
Show full solutionCorrect option: D
Correct answer
D70.3 H
Step-by-step explanation
C = 0.1 F = 107 F
Resonant frequency = 60 Hz.
by putting values Hz.
Resonant frequency = 60 Hz.
by putting values Hz.
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Alternating Current chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →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.