A potentiometer circuit is employed to find the electromotive force (e.m.f.) $E$ of a cell. The setup also contains a second cell with e.m.f. $1.5\ \text{V}$ and internal resistance $0.50\ \Omega$, which is joined to a uniform resistance wire $XY$, as shown. The resistance wire $XY$ is $0.96\ \text{m}$ long and has a resistance of $0.50\ \Omega$. The sliding contact Z is shifted along wire $XY$. The galvanometer shows zero when the length $XZ$ is $0.64\ \text{m}$. What is the value of e.m.f. $E$?
- A$0.50\ \text{V}$
- B$0.75\ \text{V}$
- C$1.0\ \text{V}$
- D$1.1\ \text{V}$