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MHT CET202618 April 2026Evening ShiftPhysicsCurrent ElectricityActual

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 along with E₁ so as two cells oppose each other, the balancing length is ' l₂ '. The ratio E₁ : E₂ is

Options

  1. A(l₁) : (l₁ + l₂)
  2. B(l₁) : (l₁ - l₂)
  3. C(l₁ + l₂) : (l₁)
  4. D(l₁ + l₂) : (l₁ - l₂)

Correct answer

B. (l₁) : (l₁ - l₂)

Step-by-step solution

Let k be the potential gradient of the potentiometer wire. When the cell of E.M.F. E₁ is connected, the balancing length is l₁ . E₁ = k l₁ When cell E₂ is connected along with E₁ such that they oppose each other, the net E.M.F. is E₁ - E₂ (since E₁ > E₂ ). The balancing length is l₂ . E₁ - E₂ = k l₂ Subtracting the second equation from the first equation: E₁ - (E₁ - E₂) = k l₁ - k l₂ E₂ = k(l₁ - l₂) Taking the ratio of E₁ and E₂ : E₁ E₂ = k l₁ k(l₁ - l₂) = l₁ l₁ - l₂ The ratio E₁ : E₂ is l₁ : (l₁ - l₂) .

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