A potentiometer wire is $100cm$ long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at $50cm$ and $10cm$ from the positive end of the wire in the two cases. The ratio of emf’s is

Suppose two cells have emfs${\in }_1$ and ${\in }_2$, $also \left({\in }_1>{\in }_2\right)$. Potential difference per unit length of the potentiometer wire$=k$

$\text{ When }{\epsilon }_1\text{ and }{\epsilon }_2\text{ are in series and support each other then }$

${\epsilon }_1+{\epsilon }_2=50\times \kappa$……..(1)

$\text{ When }{\epsilon }_1\text{ and }{\epsilon }_2\text{ are in opposite direction , then}$

${\epsilon }_1-{\epsilon }_2=10\times k$……..(2)

$\text{ On adding eqn. (1) and eqn. (2) }$

$2{\epsilon }_1=60k\Rightarrow {\epsilon }_1=30k\text{ and }{\epsilon }_2=50k-30k=20k$

$\therefore \frac{{\epsilon }_1}{{\epsilon }_2}=\frac{30k}{20k}=\frac{3}{2}$