Question

In a potentiometer experiment two cells of emf $E_{1}$ and $E_{2}$ are used in series and in conjunction and the balancing length is found to be $58~\mathrm{cm}$ of the wire. If the polarity of $E_{2}$ is reversed, then the balancing length becomes $29~\mathrm{cm}$. The ratio $\frac{E_1}{E_2}$ of the emfs of the two cells is}

• A. 1:1
• B. 2:1
• C. 3:1
• D. 4:1

Hint:

In a potentiometer, the balancing length is directly proportional to the emf.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In a potentiometer, the balancing length is directly proportional to the emf.
Step 2: Detailed Explanation:
$E_1 + E_2 = k \times 58$ and $E_1 - E_2 = k \times 29$. Dividing, $\frac{E_1 + E_2}{E_1 - E_2} = \frac{58}{29} = 2$. Solving gives $E_1 = 3E_2$, so ratio $E_1 : E_2 = 3:1$.
Step 3: Final Answer:
The ratio is 3:1.