Question

A potentiometer wire is 100 cm 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:

• A. 5 : 1
• B. 5 : 4
• C. 3 : 4
• D. 3 : 2

Answer 1

The correct option is D 3 : 2

Case I. When cell are connected in series
$E_1+E_2=K\times 50.......\left(1\right)$
Case II. When cell are connected opposite to each other
$E_1-E_2=K\times 10.......\left(2\right)\left(1\right)÷\left(2\right)$
$\frac{E_1+E_2}{E_1-E_2}=\frac{K\times 50}{K\times 10}\Rightarrow \frac{5}{1}\frac{E_1}{E_2}=\frac{3}{2}$