Two cells of e.m.f. E₁ and E₂ ( E₁ > E₂) are connected as shown in… | MHT CET 2026 (20 Apr, Shift 1)

Q13MHT CET 2026 · 20 APR · SHIFT 1Medium

Potentiometer

Two cells of e.m.f. $E_1$ and $E_2$ ( $E_1 > E_2$ ) are connected as shown in figure.

When a potentiometer is connected between points A and B the balancing length of potentiometer wire is $412$ cm. When same potentiometer wire is connected between points A and C the balancing length is $103$ cm. The ratio $E_1 : E_2$ is

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Answer: (c)

From the figure, $E_2$ is reversed relative to $E_1$ , so between A and C the cells oppose.
A–B: $E_1 \propto 412$ ; A–C: $E_1 - E_2 \propto 103$ .
$\frac{E_1}{E_1 - E_2} = 4 \Rightarrow 3E_1 = 4E_2$ .
So the answer is (c) $4:3$ .

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