c
The current through the potentiometer wire is

\(I=\frac{E_{0}}{\left(r+r_{1}\right)}\)

and the potential difference across the wire is

\(V=I r=\frac{E_{0} r}{\left(r+r_{1}\right)}\)

The potential gradient along the potentiometer wire is

\(k=\frac{V}{L}=\frac{E_{0} r}{\left(r+r_{1}\right) L}\)

As the unknown e.m.f. \(E\) is balanced against length \(l\) of the potentiometer wire,

\(\therefore \quad E=k l=\frac{E_{0} r}{\left(r+r_{1}\right)} \frac{l}{L}\)