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