In potentiometer experiment, cells of e.m.f. $E_1$ and $E_2$ are connected in series $\left(E_1 \gt…

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
  1. $1: 2$
  2. $2: 3$
  3. $3: 4$
  4. $5: 3$

Solution

For potentiometer, $\begin{aligned} & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{l_1+l_2}{l_1-l_2} \\ \therefore \quad & \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{80+20}{80-20}=\frac{100}{60}=\frac{5}{3}=5: 3 \end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 1)

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