JEE Main 2019 — Current Electricity Question with Solution
From: JEE Main 2019 (Online) 8th April Evening Slot
Question
In the circuit shown, a four-wire potentiometer
is made of a 400 cm long wire, which extends
between A and B. The resistance per unit length
of the potentiometer wire is r = 0.01 /cm. If
an ideal voltmeter is connected as shown with
jockey J at 50 cm from end A, the expected
reading of the voltmeter will be :-


This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: A
Correct answer
A0.25 V
Step-by-step explanation
Resistance of wire AB = 400 × 0.01 = 4
i =
Now voltmeter reading = i (Resistance of 50 cm length)
= (0.5 A) (0.01 × 50) = 0.25 volt
i =
Now voltmeter reading = i (Resistance of 50 cm length)
= (0.5 A) (0.01 × 50) = 0.25 volt
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Current Electricity 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 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.