AP EAMCET · PHYSICS · Current Electricity
Two cells \(A\) and \(B\) are connected in the secondary circuit of a potentiometer one at a time and the balancing lengths are respectively \(360 \mathrm{~cm}\) and \(420 \mathrm{~cm}\). If emf of \(A\) is \(2.4 \mathrm{~V}\), the emf of the second cell \(B\) is
- A \(2.8 \mathrm{~V}\)
- B \(3.2 \mathrm{~V}\)
- C \(3.0 \mathrm{~V}\)
- D \(2.6 \mathrm{~V}\)
Answer & Solution
Correct Answer
(A) \(2.8 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
If \(E_1\) and \(E_2\) are emf's of cells with \(l_1\) and \(l_2\) balance lengths in a potentiometer experiment, then, \(\frac{E_1}{E_2}=\frac{l_1}{l_2}\) or \(\quad E_2=E_1 \cdot \frac{l_2}{l_1}\) Here, \(E_1=24 \mathrm{~V}, l_1=360 \mathrm{~cm}\) and \(l_2=420 \mathrm{~cm}\)…
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