In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf $1.5 \mathrm{~V}$ is found to be $60 \mathrm{~cm}$. If this cell is replaced by another cell of emf E, the length-of null point increases by $40 \mathrm{~cm}$. The value of $E$ is $\frac{x}{10} V$. The value of $x$ is ____________.
Answer (integer)
25
Solution
E<sub>1</sub> = 1.5 V, l<sub>1</sub> = 60 cm, l<sub>2</sub> = 40 cm + 60 cm = 100 cm
<br/><br/>$E \propto l$
<br/><br/>$$
\begin{aligned}
& \frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}} \\\\
& \frac{1.5}{E}=\frac{60}{100} \\\\
& E=\frac{150}{60}=\frac{5}{2}=\frac{25}{10} \\\\
& \text { so } x=25
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Current Electricity · Topic: EMF and Internal Resistance
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