Question

Potentiometer wire is $100\text{cm}$ long and a constant potential difference is maintaned across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at $50\text{cm}$ and $10\text{cm}$ 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 $3:2$

let $x$ be the resistance of potentiometer wire per unit length. $E_1+E_2=50x$
and $E_1-E_2=10x$
$\Rightarrow \frac{E_1+E_2}{E_1-E_2}=5$
$\Rightarrow \frac{E_1}{E_2}=\frac{3}{2}$