JEE Main 2020 — Current Electricity Question with Solution
From: JEE Main 2020 (Online) 2nd September Evening Slot
Question
A potentiometer wire PQ of 1 m length is
connected to a standard cell E1. Another cell E2
of emf 1.02 V is connected with a resistance ‘r’
and switch S (as shown in figure). With switch
S open, the null position is obtained at a
distance of 49 cm from Q. The potential
gradient in the potentiometer wire is :
This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: C
Correct answer
C0.02 V/cm
Step-by-step explanation
Balancing length is measured from P
PQ = 1m
QJ = 49 cm
PJ = 51 cm
Potential drop = Potential gradient() length
1.02 = 51
= 0.02 V/cm
PQ = 1m
QJ = 49 cm
PJ = 51 cm
Potential drop = Potential gradient() length
1.02 = 51
= 0.02 V/cm
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