JEE Main 2021 — Current Electricity Question with Solution
From: JEE Main 2021 (Online) 16th March Evening Shift
Question
The energy dissipated by a resistor is 10 mJ in 1 s when an electric current of 2 mA flows through it. The resistance is ___________. (Round off to the Nearest Integer)
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Show full solutionCorrect answer: 2500
Correct answer
2500
Step-by-step explanation
Given, energy dissipated by a resistor, H = 10 mJ = 10 103 J
Time, t = 1 s
Electric current, I = 2 mA = 2 103 A
Resistance, R = ?
According to Joule's law of heating,
H = I2Rt
....... (i)
Substituting the given values in Eq. (i), we get
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This is a previous-year question from JEE Main 2021, 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.