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}
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}
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In a potentiometer, the balancing length is directly proportional to the emf.
Updated On: May 3, 2026
- 1:1
- 2:1
- 3:1
- 4:1
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The Correct Option is C
Solution and Explanation
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.
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.
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