In potentiometer experiments, two cells of e. m. f. ' E₁ ' and ' E₂ ' are connected in series (E₁ > E₂), the balancing length is 64 cm of the wire. If the polarity of E₂ is reversed, the balancing length becomes 32 cm. The ratio E₁/E₂ is (a) 3: 2 (b) 2: 3 (c) 1: 3 (d) 3: 1

Answer: (d). Balancing length ∝ emf: (E₁+E₂)/(E₁-E₂) = (l₁)/(l₂). (E₁+E₂)/(E₁-E₂) = 64/32 = 2 ⇒ E₁+E₂ = 2E₁-2E₂ E₁ = 3E₂ So the answer is (d) 3:1.

In potentiometer experiments, two cells of e. m. f. ' E₁ ' and ' E₂ '… | MHT CET 2023 (11 May, Shift 1)

Q41MHT CET 2023 · 11 MAY · SHIFT 1Medium

Potentiometer

In potentiometer experiments, two cells of e. m. f. ' $E_1$ ' and ' $E_2$ ' are connected in series $(E_1 > E_2)$ , the balancing length is $64\text{ cm}$ of the wire. If the polarity of $E_2$ is reversed, the balancing length becomes $32\text{ cm}$ . The ratio $\mathrm{E}_1/\mathrm{E}_2$ is

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Answer: (d)

Balancing length $\propto$ emf: $\dfrac{E_1+E_2}{E_1-E_2} = \dfrac{l_1}{l_2}$ .
$\dfrac{E_1+E_2}{E_1-E_2} = \dfrac{64}{32} = 2 \Rightarrow E_1+E_2 = 2E_1-2E_2$
$E_1 = 3E_2$
So the answer is (d) $3:1$ .

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Potentiometer question from MHT CET 2023 (11 May, Shift 1)

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