JEE Main 2019 — Current Electricity Question with Solution
From: JEE Main 2019 (Online) 11th January Morning Slot
Question
The resistance of the meter bridge AB in given figure is 4 . With a cell of emf = 0.5 V and rheostat
resistance Rh = 2 the null point is obtained at some point J. When the cell is replaced by another one of emf = 2 the same null point J is found for Rh = 6 . The emf 2 is, :
This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: A
Correct answer
A0.3 V
Step-by-step explanation
Potential gradient with Rh = 2
is L 100 cm
Let null point be at cm
thus 1 0.5V . . .(1)
Now with Rh 6 new potential gradient is
and at null point
. . .(2)
dividing equation (1) by (2) we get
thus
is L 100 cm
Let null point be at cm
thus 1 0.5V . . .(1)
Now with Rh 6 new potential gradient is
and at null point
. . .(2)
dividing equation (1) by (2) we get
thus
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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.