Q138 / 150 · MHT CET 2023 (11 May Shift 1) · Physics

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
    $1: 3$
  • Correct answer
    D
    $3: 1$

Solution

For potentiometer, $$ \begin{aligned} \frac{\mathrm{E}_1}{\mathrm{E}_2} & =\frac{l_1+l_2}{l_1-l_2} \\ \therefore \quad \frac{\mathrm{E}_1}{\mathrm{E}_2} & =\frac{64+32}{64-32} \quad=\frac{96}{32}=\frac{3}{1}=3: 1 \end{aligned} $$

Hence, the correct answer is option: 'D'.

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