JEE Main 2019 — Current Electricity Question with Solution
From: JEE Main 2019 (Online) 12th January Morning Slot
Question
The galvanometer deflection, when key K1 is closed but K2 is open, equals 0 (see figure). On closing K2 also and adjusting R2 to 5, the deflection in galvanometer becomes . The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] :
This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: D
Correct answer
D22
Step-by-step explanation
case I :
ig = = C0 . . .(i)
Case II :
ig = = . . .(ii)
= . . .(ii)
. . .(i)
5500 + 25Rg = 225Rg + 1100
200Rg = 4400
Rg = 22
ig = = C0 . . .(i)
Case II :
ig = = . . .(ii)
= . . .(ii)
. . .(i)
5500 + 25Rg = 225Rg + 1100
200Rg = 4400
Rg = 22
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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.