Search any question & find its solution
Question:
Answered & Verified by Expert
Two cells $A$ and. $B$ are connected in the secondary circuit of a potentiometer one at a time and the balancing length are respectively $400 \mathrm{~cm}$ and $440 \mathrm{~cm}$. The emf of the cell $A$ is 1.08 volt. The emf of the second cell $B$ in volts is
Options:
Solution:
1291 Upvotes
Verified Answer
The correct answer is:
$1.188$
$\begin{array}{lc}\frac{E_1}{E_2}=\frac{l_1}{l_2} & \\ \Rightarrow & \frac{1.08}{E_2}=\frac{400}{440} \\ \Rightarrow & E_2=\frac{440 \times 1.08}{400}=1.188 \mathrm{~V}\end{array}$
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.