A cell can be balanced against $110\,\, cm$ and $100\,\, cm$ of potentiometer wire, respectively with and without being short circuited through a resistance of $10 \,\,\Omega$. Its internal resistance is ............... $\Omega$
AIPMT 2008, Medium
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In the question, the length $110\, \mathrm{cm}$ and
${100\,\,{\mkern 1mu} {\text{cm}}}$ are interchanged as ${\varepsilon > \frac{{\varepsilon R}}{{R + r}}}$
Without being short circuited through $R,$ only the battery $\varepsilon$ is balanced.
$\varepsilon=\frac{V}{L} \times l_{1}=\frac{V}{L} \times 110 \,\mathrm{cm}$ ....$(i)$
When $R$ is connected across $\varepsilon$,
$R i=R \cdot\left(\frac{\varepsilon}{R+r}\right)=\frac{V}{L} \times l_{2} \Rightarrow \frac{R \varepsilon}{R+r}=\frac{V}{L} \times 100$ ......$(ii)$
Dividing eqn. $(i)$ and $(ii)$, $\frac{(R+r)}{R}=\frac{110}{100}$
$\Rightarrow 1+\frac{r}{R}=\frac{110}{100} \Rightarrow \frac{r}{R}=\frac{110}{100}-\frac{100}{100}$
$\Rightarrow r=R \cdot \frac{10}{100}=\frac{R}{10} \cdot$ As $R=10\, \Omega ; r=1\, \Omega$
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