When cell of E.M.F. ' E₁ ' is connected to potentiometer wire, the balancing length is l₁. Another cell of E.M.F. ' E₂ ' (E₁ > E₂) is connected so that two cells oppose each other, then the balancing length is l₂. The ratio E₁: E₂ is (a) (l₁)/(l₁ + l₂) (b) (l₁)/(l₁ - l₂) (c) (l₁ + l₂)/(l₁) (d) (l₁ + l₂)/(l₁ - l₂)

Answer: (b). E₁ = k l₁ and E₁ - E₂ = k l₂. So E₂ = k(l₁ - l₂). (E₁)/(E₂) = (l₁)/(l₁ - l₂). So the answer is (b).

When cell of E.M.F. ' E₁ ' is connected to potentiometer wire, the… | MHT CET 2024 (2 May, Shift 2)

Q34MHT CET 2024 · 2 MAY · SHIFT 2Medium

Potentiometer

When cell of E.M.F. ' $E_1$ ' is connected to potentiometer wire, the balancing length is $l_1$ . Another cell of E.M.F. ' $E_2$ ' $(E_1 > E_2)$ is connected so that two cells oppose each other, then the balancing length is $l_2$ . The ratio $E_1 : E_2$ is

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Answer: (b)

$E_1 = k l_1$ and $E_1 - E_2 = k l_2$ .
So $E_2 = k(l_1 - l_2)$ .
$\frac{E_1}{E_2} = \frac{l_1}{l_1 - l_2}$ .
So the answer is (b).

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Potentiometer question from MHT CET 2024 (2 May, Shift 2)

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