In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf 1.5 V is found to be 60 cm. [2023]

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Solution

Q.
In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf 1.5 V is found to be 60 cm. If this cell is replaced by another cell of emf E, the length of null point increases by 40 cm. The value of E is $\frac{x}{10}$ V. The value of $x$ is ______. [2023]

Ans.
(25)

$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$

$\frac{1.5}{E_{2}} = \frac{60}{60 + 40} = \frac{6}{10} = \frac{3}{5}$

$E_{2} = \frac{5}{2} = \frac{x}{10} \Rightarrow x = 25$

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