Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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 the two-way key K3 is open. The balance point is at a length l1 cm from P when the two-way key K3 is plugged in between 2 and 1, while the balance point is at a length l2 cm from P when the key K3 is plugged in between 3 and 1 . The ratio of two resistances R1R2, is found to be 
PhysicsCurrent ElectricityJEE MainJEE Main 2017 (08 Apr Online)
Options:
  • A l1l1-l2
  • B l2l2-l1
  • C l1l1+l2
  • D l1l2-l1
Solution:
1896 Upvotes Verified Answer
The correct answer is: l1l2-l1
When the key is between 2 and 1

Then, emf V1= IR1=xl1

When the key is between 3 and 1

Emf, V2=IR1+R2=xl2

The ratio of both emf is R1R1+R2=l1l2

On simplifying the above relation, we get the ratio of resitances, R1R2=l1l2-l1

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.