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 length are respectively \(400 \mathrm{~cm}\) and \(440 \mathrm{~cm}\). The emf of the cell \(A\) is 1.08 volt. The emf of the second cell \(B\) in volts is
- A \(1.08\)
- B \(1.188\)
- C \(11.88\)
- D \(12.8\)
Answer & Solution
Correct Answer
(B) \(1.188\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{lc}\frac{E_1}{E_2}=\frac{l_1}{l_2} & \\ \Rightarrow & \frac{1.08}{E_2}=\frac{400}{440} \\ \Rightarrow & E_2=\frac{440 \times 1.08}{400}=1.188 \mathrm{~V}\end{array}\)
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