JEE Main 2017 — Current Electricity Question with Solution
From: JEE Main 2017 (Online) 8th April Morning Slot
Question
A potentiometer PQ is set up to compare two resistances as shown in the figure. The ammeter A in the circuit reads 1.0 A when two way key K3 is open. The balance point is at a length 1 cm from P when two way key K3 is plugged in between 2 and 1, while the balance point is at a length 2 cm from P when key K3 is plugged in between 3 and 1. The ratio of two resistances , is found to be :
This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: D
Correct answer
D
Step-by-step explanation
When key K3 is plugged in between 1 and 2,
V1 = iR1 = x1 . . . . . (1)
When key K3 is plugged in between 3 and 1,
V2 = i(R1 + R2) = x2 . . . . (2)
On dividing (1) and (2),
=
=
V1 = iR1 = x1 . . . . . (1)
When key K3 is plugged in between 3 and 1,
V2 = i(R1 + R2) = x2 . . . . (2)
On dividing (1) and (2),
=
=
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This is a previous-year question from JEE Main 2017, 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.