JEE Main 2019 — Current Electricity Question with Solution
From: JEE Main 2019 (Online) 12th January Evening Slot
Question
A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of 4 10–4 A passes through it, its needle ( pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, it should be connected to a resistance of :
Choose an option
Show full solutionCorrect option: A
Correct answer
A200 ohm
Step-by-step explanation
Ig = 4 104 25 = 102 A
2.5 = (50 + R) 102
R = 200
2.5 = (50 + R) 102
R = 200
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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.