A cell can be balanced against $110 \mathrm{~cm}$ and $100 \mathrm{~cm}$ of potentiometer wire, respectively with and without being short circuited through a resistance of $10 \Omega$ Its internal resistance is

Question

Aicharya · Accepted Answer

Key Idea : This problem is based on the application of potentiometer in which we find the internal resistance of a cell.
In potentiometer experiment in which we find internal resistance of a cell, let $E$ be the emf of the cell and $V$ the terminal potential difference, then
$\frac{E}{V}=\frac{l_1}{l_2}$
where $l_1$ and $l_2$ are lengths of potentiometer wire with and without short circuited through a resistance.
Since, $\frac{E}{V}=\frac{R+r}{R}[\because E=I(R+r)$ and $V=I R]$
$\therefore \frac{R+r}{R}=\frac{l_1}{l_2}$
$\begin{aligned}
1+\frac{r}{R} & =\frac{110}{100} \\
\frac{r}{R} & =\frac{10}{100} \\
r & =\frac{1}{10} \times 10=1 \Omega
\end{aligned}$
$r=\frac{1}{10} \times 10=1 \Omega$

A cell can be balanced against $110 \mathrm{~cm}$ and $100 \mathrm{~cm}$ of potentiometer wire, respectively…

Solution