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 $x /10$ V.
The value of $x$ is

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TestHub · Accepted Answer

According to the principle of potentiometer, the emf of a cell is directly proportional to its balancing length.

$E \propto l \Rightarrow \frac{E _{1}}{E _{2}} = \frac{l _{1}}{l _{2}}$

Given $E_{1} = 1.5$ V and $l_{1} = 60$ cm.

When the cell is replaced by another cell of emf $E$ , the balancing length increases by $40$ cm.

$l_{2} = 60 + 40 = 100$ cm.

Substituting the values:

$\frac{1.5}{E} = \frac{60}{100}$

$E = \frac{1.5 \times 100}{60} = 2.5$ V.

Given $E = \frac{x}{10}$ V.

$\frac{x}{10} = 2.5 \Rightarrow x = 25$

Answer: $25$

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