Here $E \gt {{ER} \over {R + r}}$, hence the lengths 110 cm and 100 cm are interchanged.

Without being short-circuited through R, only the battery E is balanced.

$E = {V \over L} \times {l_1} = {V \over L} \times 110$   ...(i)

When R is connected across E, $Ri = {V \over L} \times {l_2}$

$\Rightarrow R\left( {{E \over {R + r}}} \right) = {V \over L} \times 100$   ...(ii)

Dividing (i) by (ii), we get

${{R + r} \over R} = {{110} \over {100}}$

$\Rightarrow$ 100 R + 100 r = 110 R

$\Rightarrow$ 10 R = 100 r

$\therefore$ $r = {{10R} \over {100}} = {{10 \times 10} \over {100}}$  ($\because$ R = 10$\Omega$)

$\Rightarrow$ r = 1$\Omega$.