Two cells of e.m.f. E1 and E2 are joined in series and the balancing length of the potentiometer wire is 625 cm. If the terminals of E1 are reversed, the balancing length obtained is 125 cm. Given E2 E1, the ratio E1: E2 will be

Question

Two cells of e.m.f. E1 and E2 are joined in series and the balancing length of the potentiometer wire is 625 cm. If the terminals of E1 are reversed, the balancing length obtained is 125 cm. Given E2 > E1, the ratio E1: E2 will be (A) 2: 3 (B) 5: 1 (C) 3: 2 (D) 1: 5

Tardigrade Admin · Accepted Answer

(E1+E2/E1-E2)=(625/125)=5 (E1/E2)=(3/2)

Q. Two cells of e.m.f. $E_{1}$ and $E_{2}$ are joined in series and the balancing length of the potentiometer wire is $625 c m$ . If the terminals of $E_{1}$ are reversed, the balancing length obtained is $125 c m$ . Given $E_{2} > E_{1}$ , the ratio $E_{1} : E_{2}$ will be

2: 3

5: 1

3: 2

1: 5

$\frac{E _{1} + E _{2}}{E _{1} - E _{2}} = \frac{625}{125} = 5$
$\frac{E _{1}}{E _{2}} = \frac{3}{2}$

Q. Two cells of e.m.f. E1​ and E2​ are joined in series and the balancing length of the potentiometer wire is 625cm. If the terminals of E1​ are reversed, the balancing length obtained is 125cm. Given E2​>E1​, the ratio E1​:E2​ will be