MHT CET · Physics · Current Electricity
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
- A 2:3
- B 1:3
- C 3:1
- D 3:2
Answer & Solution
Correct Answer
(D) 3:2
Step-by-step Solution
Detailed explanation
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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