Two cells of emf E1 and E2(E1 E2) are connected individually to a potentiometer and their corresponding balancing length are 625 cm and 500 cm, then the ratio (E1/E2) is

Question

Two cells of emf E1 and E2(E1 > E2) are connected individually to a potentiometer and their corresponding balancing length are 625 cm and 500 cm, then the ratio (E1/E2) is (A) 5: 4 (B) 3: 1 (C) 5: 1 (D) 4: 5

Tardigrade Admin · Accepted Answer

E1=k l1 k is potential gradient E2=k l2 (E1/E2)=(l1/l2)=(625/500) =(25/20)=(5/4)

Q. Two cells of emf $E_{1}$ and $E_{2} (E_{1} > E_{2})$ are connected individually to a potentiometer and their corresponding balancing length are $625 c m$ and $500 c m$ , then the ratio $\frac{E _{1}}{E _{2}}$ is

5 : 4

3 : 1

5 : 1

4 : 5

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

Q. Two cells of emf E1​ and E2​(E1​>E2​) are connected individually to a potentiometer and their corresponding balancing length are 625cm and 500cm, then the ratio E2​E1​​ is