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
In potentiometer experiment, cells of e.m.f. ' $E_1$ ' and ' $E_2$ ' are connected in series $\left(E_1>E_2\right)$ the balancing length is $64 \mathrm{~cm}$ of the wire. If the polarity of $\mathrm{E}_2$ is reversed, the balancing length becomes $32 \mathrm{~cm}$. The ratio $\frac{E_1}{E_2}$ is
PhysicsCurrent ElectricityMHT CETMHT CET 2021 (23 Sep Shift 2)
Options:
  • A $1: 1$
  • B $6: 1$
  • C $3: 1$
  • D $2: 1$
Solution:
2753 Upvotes Verified Answer
The correct answer is: $3: 1$
$\frac{E_1}{E_2}=\frac{\ell_1+\ell_2}{\ell_1-\ell_2}=\frac{64+32}{64-32}=\frac{96}{32}=3$

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.