JEE Main 2020 · PhysicsMultiple choiceDiagram basedEasyCalculation
JEE Main 2 September 2020, Shift 2, Physics Q20
Question 20 of 75 in this shift, Physics question 20 of 25.
A potentiometer wire PQ of 1 m length is connected to a standard cell E. Another cell E 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 :
The figure, in words
Potentiometer circuit: wire PQ drawn horizontally in the middle, P at the left and Q at the right; above it a loop wire connecting P and Q through cell E1. A jockey/contact J (arrow pointing up) touches PQ; from J a wire goes right, down to a galvanometer G (circle with an arrow), then left through a parallel combination of resistor r and switch S, then cell E2, then back to P along the left side.- (1)0.01 V/cm
- (2)0.02 V/cmOfficial answer
- (3)0.03 V/cm
- (4)0.04 V/cm
Official answer
Option 2
NTA final key (Sep 2020).
- Chapter
- Current Electricity
- Topic
- Potentiometer
- Idea tested
- Potential Gradient Calculation