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MHT CET202620 April 2026Morning ShiftPhysicsCurrent ElectricityActual

Two cells of e.m.f. E₁ and E₂ ( E₁ > E₂ ) 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 E₁ : E₂ is

Options

  1. A6 : 1
  2. B4 : 1
  3. C4 : 3
  4. D3 : 4

Correct answer

C. 4 : 3

Step-by-step solution

From the circuit diagram, the potential difference between points A and B is V_ AB = E₁ . Given the balancing length for points A and B is 412 cm , we have: E₁ = 412k where k is the potential gradient of the potentiometer wire. The cells E₁ and E₂ are connected in opposition because their negative terminals are joined at point B. The potential difference between points A and C is: V_ AC = E₁ - E₂ Given the balancing length for points A and C is 103 cm , we have: E₁ - E₂ = 103k Dividing the first equation by the sec

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