A cell in secondary circuit gives null deflection for 2.5 m length of potentiometer having 10 m length of wire. If the length of the potentiometer wire is increased by 1 m without changing the cell in the primary, the position of the null point now is:

2.75 m

In first case, Potential gradient, k = $\frac{\varepsilon_0}{l}=\frac{\varepsilon_0}{10}$

Where ε₀ is the emf of the battery in the potentiometer circuit.

As per questions, ε = k × 2.5 = $\frac{\varepsilon_0}{10} × 2.5$                 (i)

In second case, Length of potentiometer wire = 10 + 1 = 11 m

Potential gradient, k' = $\frac{\varepsilon_0}{11}$

If $l'$ is the new balancing length, then

$\varepsilon=k' l'=\frac{\varepsilon_0}{11} × l'$                      (ii)

Equating (i) and (ii), we get

$\frac{\varepsilon_0}{10} × 2.5=\frac{\varepsilon_0}{11} × l'$           or           $l'=\frac{2.5 × 11}{10}$ = 2.75 m