The resistance per centimeter of a meter bridge wire is r , with X Ω resistance in left gap. Balancing…

For first case:

$\frac{X}{r{\mathcal{l}}_1}=\frac{25}{r{\mathcal{l}}_2}$ ... (i)

For second case:

$\frac{X}{\left(2r\right){{\mathcal{l}}^{\text{'}}}_1}=\frac{25}{\left(2r\right)\mathcal{l}{\text{'}}_2}$ ... (ii)

From (i) and (ii), we get

$\frac{\mathcal{l}{\text{'}}_1}{{\mathcal{l}}_1}=\frac{{\displaystyle \mathcal{l}{\text{'}}_2}}{{\displaystyle {\mathcal{l}}_2}}$ also total length is constant.

Therefore, balance length doesn't change and 

$\mathcal{l}{\text{'}}_1={\mathcal{l}}_1=40 \mathrm{cm}$