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 10cmfrom the positive end of the wire in the two cases. The ratio of emf’s is

Suppose two cells have emfs∈1 and ∈2. also ∈1>∈2Potential difference per unit length of the potentiometer wire=k When ε1 and ε2 are in series and support each other then ε1+ε2=50×κ……..(1) When ε1 and ε2 are in opposite direction ε1-ε2=10×k……..(2) On adding eqn. (1) and eqn. (2) 2ε1=60k⇒ε1=30k and ε2=50k-30k=20k∴  ε1ε2=30k20k=32