Question

Two cells of emf $E_{1}$ and $E_{2}\left(E_{1} > E_{2}\right)$ are connected individually to a potentiometer and their corresponding balancing length are $625 \,cm$ and $500 \,cm$, then the ratio $\frac{E_{1}}{E_{2}}$ is

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

Answer 1

Answer: (A)

$E_{1}=k l_{1}$
$k$ is potential gradient
$E_{2}=k l_{2}$
$\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}=\frac{625}{500}$
$=\frac{25}{20}=\frac{5}{4}$