\({100\,\,{\mkern 1mu} {\text{cm}}}\)  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 \(R=10\, \Omega ; r=1\, \Omega\)