MHT CET · Physics · Current Electricity
In potentiometer experiments, two 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 \(E_2\) is reversed, the balancing length becomes \(32 \mathrm{~cm}\). The ratio \(\mathrm{E}_1 / \mathrm{E}_2\) is
- A \(3: 2\)
- B \(2: 3\)
- C \(1: 3\)
- D \(3: 1\)
Answer & Solution
Correct Answer
(D) \(3: 1\)
Step-by-step Solution
Detailed explanation
For potentiometer,
\(
\begin{aligned}
\frac{\mathrm{E}_1}{\mathrm{E}_2} & =\frac{l_1+l_2}{l_1-l_2} \\
\therefore \quad \frac{\mathrm{E}_1}{\mathrm{E}_2} & =\frac{64+32}{64-32} \quad=\frac{96}{32}=\frac{3}{1}=3: 1
\end{aligned}
\)
\(
\begin{aligned}
\frac{\mathrm{E}_1}{\mathrm{E}_2} & =\frac{l_1+l_2}{l_1-l_2} \\
\therefore \quad \frac{\mathrm{E}_1}{\mathrm{E}_2} & =\frac{64+32}{64-32} \quad=\frac{96}{32}=\frac{3}{1}=3: 1
\end{aligned}
\)
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