Question

Two cells \(A\) and \(B\) are connected in the secondary circuit of a potentiometer one at a time and the balancing lengths are respectively \(360\ \text{cm}\) and \(420\ \text{cm}\). If emf of \(A\) is \(2.4\ \text{V}\), the emf of the second cell \(B\) is

• A. \(2.8\ \text{V}\)
• B. \(3.2\ \text{V}\)
• C. \(3.0\ \text{V}\)
• D. \(2.6\ \text{V}\)

Hint:

In a potentiometer comparison of emfs, \[ \frac{E_1}{E_2}=\frac{l_1}{l_2}. \] The cell with the greater balancing length has the greater emf.

Answer 1

The Correct Option is A

Step 1: Use potentiometer proportionality.
In a potentiometer, EMF is proportional to balancing length: \( \frac{E_A}{E_B} = \frac{l_A}{l_B} \).

Step 2: Solve for \( E_B \).
\( E_B = E_A\times\frac{l_B}{l_A} = 2.4\times\frac{420}{360} = 2.4\times\frac{7}{6} = 2.8\text{ V} \)

\[ \boxed{E_B = 2.8\text{ V}} \]