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

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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.
Updated On: Jun 26, 2026
  • \(2.8\ \text{V}\)
  • \(3.2\ \text{V}\)
  • \(3.0\ \text{V}\)
  • \(2.6\ \text{V}\)
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The Correct Option is A

Solution and Explanation

Step 1: Use the potentiometer principle.
In a potentiometer, the emf of a cell is directly proportional to the balancing length.
Therefore, \[ \frac{E_A}{E_B}=\frac{l_A}{l_B}. \]

Step 2: Substitute the given values.
Given, \[ E_A=2.4\ \text{V}, \] \[ l_A=360\ \text{cm}, \] and \[ l_B=420\ \text{cm}. \] Thus, \[ \frac{2.4}{E_B}=\frac{360}{420}. \]

Step 3: Solve for \(E_B\).
\[ E_B=2.4\times \frac{420}{360}. \] \[ E_B=2.4\times \frac{7}{6}. \] \[ E_B=2.8\ \text{V}. \]

Step 4: Final conclusion.
Therefore, the emf of the second cell \(B\) is \[ \boxed{2.8\ \text{V}} \] Hence, the correct option is \[ \boxed{(1)} \]
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