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Potentiometer

In a potentio...

Question

In a potentiometer experiment two cells of emf $E_{1}$ and $E_{2}$ are used in series and in conjunction and the balancing length is found to be 58 cm of the wire. If the polarity of $E_{2}$ is reversed, then the balancing length becomes 29 cm. The ratio $\frac{E_{1}}{E_{2}}$ of the emf is

A

1:1

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B

2:1

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C

3:1

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D

4:1

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Solution

The correct option is
C

3:1

$E_{1} + E_{2} = 58 + 29 E_{1} - E_{2} = 58 - 29 \frac{E_{1}}{E_{2}} = \frac{58 + 29}{58 - 29} = \frac{3}{1}$

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