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JEE Main 2023 (1 Feb, Shift 1) · Potentiometer

Potentiometer question from JEE Main 2023 (1 Feb, Shift 1)

Q38JEE Main 2023 · 1 FEB · SHIFT 1Easy

Potentiometer

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

Numerical answer type. Work it out, then open the answer.

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Answer: 25

$E \propto l$ , new length $= 60 + 40 = 100\ \text{cm}$ .
$E = 1.5 \times \frac{100}{60} = 2.5\ \text{V} = \frac{25}{10}\ \text{V}$ .
So $x = 25$ .

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