A 2.5 V battery is connected to a potentiometer wire. A cell of e.m.f. 1.08 V is balanced by the voltage…

Potential gradient determination

The known cell with $E = 1.08\text{ V}$ balances at length $l = 2.16\text{ m}$, establishing the potential gradient $k = E/l$.

$k = \frac{1.08}{2.16} = 0.5\text{ V/m}$

Total wire length calculation

The battery voltage $V = 2.5\text{ V}$ equals the potential drop across the entire wire length $L$: $V = kL$.

$2.5 = 0.5L$

$L = \frac{2.5}{0.5} = 5\text{ m}$

The potentiometer wire length is $\boxed{\text{C}}$