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Two cells of e.m.f.'s $E_1$ and $E_2\left(E_1>E_2\right)$ are connected as shown in figure:
When a potentiometer is connected between A and B the balancing length of the potentiometer wire is $300 \mathrm{~cm}$. By connecting the same potentiometer between $\mathrm{A}$ and $\mathrm{C}$, the balancing length is $100 \mathrm{~cm}$. The ratio of $\frac{E_1}{E_2}$ is
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When a potentiometer is connected between A and B the balancing length of the potentiometer wire is $300 \mathrm{~cm}$. By connecting the same potentiometer between $\mathrm{A}$ and $\mathrm{C}$, the balancing length is $100 \mathrm{~cm}$. The ratio of $\frac{E_1}{E_2}$ is
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The correct answer is:
3:2
For balance potentiometer $\frac{E_{A B}}{E_{A C}}=\frac{l_{A B}}{l_{A C}}$
$\begin{aligned}
& \Rightarrow \frac{E_1}{E_1-E_2}=\frac{300}{100} \\
& \Rightarrow 3 E_1-3 E_2=E_1 \\
& \Rightarrow E_1: E_2=3: 2
\end{aligned}$
$\begin{aligned}
& \Rightarrow \frac{E_1}{E_1-E_2}=\frac{300}{100} \\
& \Rightarrow 3 E_1-3 E_2=E_1 \\
& \Rightarrow E_1: E_2=3: 2
\end{aligned}$
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