Question

With a certain cell the balance point is obtained at 0.60 m from one end of the potentiometer. With another cell whose emf differs from that of the first by 0.1 V, the balance point is obtained at 0.55 m. Then, the two emf’s are

• A. 1.2 V, 1.1 V
• B. 1.2 V, 1.3 V
• C. – 1.1 V, – 1.0 V
• D. None of the above

Answer 1

Answer: (A)

1.2 V, 1.1 V

Answer 2

E₁ = x (0.6) and E₂ = E₁ – 0.1 = x (0.55)

⇒ $\frac{E_{1}}{E_{1} - 0.1} = \frac{0.6}{0.55}$

or 55 E₁ = 60 E₁ – 6 ⇒ $E_{1} = \frac{6}{5} = 1.2V$ thus E₂ = 1.1 V