Two cells having unknown emfs E₁ and E₂ (E₁ > E₂) are connected in potentiometer circuit, so as to assist each other. The null point obtained is at 490 cm from the higher potential end. When cell E₂ is connected, so as to oppose cell E₁, the null point is obtained at 90 cm from the same end. The ratio of the emfs of two cells ((E₁)/(E₂)) is (a) 0.689 (b) 5.33 (c) 1.45 (d) 0.182

Answer: (c). (E₁+E₂)/(E₁-E₂) = 490/90 = 49/9 9E₁+9E₂ = 49E₁-49E₂ ⇒ 40E₁ = 58E₂ (E₁)/(E₂) = 58/40 = 1.45 So the answer is (c) 1.45.

Two cells having unknown emfs E₁ and E₂ (E₁ > E₂) are connected in… | MHT CET 2020 (19 Oct, Shift 2)

Q66MHT CET 2020 · 19 OCT · SHIFT 2Medium

Potentiometer

Two cells having unknown emfs $E_1$ and $E_2\ (E_1 > E_2)$ are connected in potentiometer circuit, so as to assist each other. The null point obtained is at 490 cm from the higher potential end. When cell $E_2$ is connected, so as to oppose cell $E_1$ , the null point is obtained at 90 cm from the same end. The ratio of the emfs of two cells $\left(\frac{E_1}{E_2}\right)$ is

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Answer: (c)

$\dfrac{E_1+E_2}{E_1-E_2} = \dfrac{490}{90} = \dfrac{49}{9}$
$9E_1+9E_2 = 49E_1-49E_2 \Rightarrow 40E_1 = 58E_2$
$\dfrac{E_1}{E_2} = \dfrac{58}{40} = 1.45$
So the answer is (c) 1.45.

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Potentiometer question from MHT CET 2020 (19 Oct, Shift 2)

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