Here length of potentiometer wire, $l=10 \mathrm{~m}$ Resistance of potentiometer wire
$R=\rho \times \frac{l}{A}$ or $R=\left(\rho \times \frac{10}{A}\right)$
The value of $2.5 \mathrm{~m}$ length wire
$$
R^{\prime}=\frac{\rho \times 10}{A \times 10} \times 2.5 \Rightarrow R^{\prime}=\left(\frac{2.5 \rho}{A \times 10}\right)
$$
Potential, $V^{\prime}=I \times R^{\prime}=I\left(\frac{2.5 \rho}{A \times 10}\right)$
Again the length of potentiometer wire is increased by $1 \mathrm{~m}$.
Resistance of null position
$$
R^{n}=\left(\frac{\rho \times l}{11 \times A}\right)
$$
$\therefore V^{\prime \prime}=I R^{\prime \prime}$ and $V=V^{\prime}$
$$
\begin{array}{l}
\Rightarrow \frac{I \times 2.5 \rho}{A \times 10}=\frac{\rho \times I}{11 \times A} \times I \\
\text { or } \frac{2.5 \times 11}{10}=I=2.75 \mathrm{~m}
\end{array}
$$