MHT CET · Physics · Current Electricity
In potentiometer experiment, cells of e.m.f. \(E_1\) and \(E_2\) are connected in series \(\left(E_1 \gt E_2\right)\) the balancing length is 80 cm of the wire. If the polarity of \(\mathrm{E}_2\) is reversed, the balancing length becomes 20 cm. The ratio, \(\mathrm{E}_1 / \mathrm{E}_2\) is
- A \(1: 2\)
- B \(2: 3\)
- C \(3: 4\)
- D \(5: 3\)
Answer & Solution
Correct Answer
(D) \(5: 3\)
Step-by-step Solution
Detailed explanation
For potentiometer,
\(\begin{aligned}
& \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{l_1+l_2}{l_1-l_2} \\
\therefore \quad & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{80+20}{80-20}=\frac{100}{60}=\frac{5}{3}=5: 3
\end{aligned}\)
\(\begin{aligned}
& \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{l_1+l_2}{l_1-l_2} \\
\therefore \quad & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{80+20}{80-20}=\frac{100}{60}=\frac{5}{3}=5: 3
\end{aligned}\)
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