In potentiometer experiment, cells of e.m.f. ' E 1 ' and ' E 2 ' are connected in series ( E 1 E 2 ) the balancing length is 64 cm of the wire. If the polarity of E 2 i

Question

In potentiometer experiment, cells of e.m.f. ' $E_{1}$ ' and ' $E_{2}$ ' are connected in series $(E_{1} > E_{2})$ the balancing length is $64 cm$ of the wire. If the polarity of $E_{2}$ is reversed, the balancing length becomes $32 cm$ . The ratio $\frac{E _{1}}{E _{2}}$ is

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Answer:

C

Solution:

$\frac{E _{1}}{E _{2}} = \frac{ℓ _{1} + ℓ _{2}}{ℓ _{1} - ℓ _{2}} = \frac{64 + 32}{64 - 32} = \frac{96}{32} = 3$

TestHub · Accepted Answer

E 2 ​ E 1 ​ ​ = ℓ 1 ​ − ℓ 2 ​ ℓ 1 ​ + ℓ 2 ​ ​ = 64 − 32 64 + 32 ​ = 32 96 ​ = 3

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