KCET · Physics · Atomic Physics
In a potentiometer experiment of a cell of emf \( 1.25 \mathrm{~V} \) gives balancing length of \( 30 \mathrm{~cm} \). If the
cell is replaced by another cell, balancing length is found to be \( 40 \mathrm{~cm} \). What is the emf of
second cell ?
- A \( \simeq 1.57 \mathrm{~V} \)
- B \( \simeq 1.67 \mathrm{~V} \)
- C \( \simeq 1.47 \mathrm{~V} \)
- D \( \simeq 1.37 \mathrm{~V} \)
Answer & Solution
Correct Answer
(B) \( \simeq 1.67 \mathrm{~V} \)
Step-by-step Solution
Detailed explanation
Given, for potentiometer experiment for cell of emf \(E_{1}=1.25\), balancing length \(L_{1}=30 \mathrm{~cm}\).
For cell of emf \(E_{2}\), balancing length \(L_{2}=40 \mathrm{~cm}\).
We know \(E \propto L\)
\(\Rightarrow \frac{E_{1}}{E_{2}}=\frac{L_{1}}{L_{2}} \Rightarrow \frac{1.25}{E_{2}}=\frac{30}{40}\)
\(\Rightarrow E_{2}=1.25 \times \frac{4}{3}=1.666 \mathrm{~V}\)
Therefore \(E_{2} \sim 1.67 \mathrm{~V}\)
Emf of second cell is \(1.67\)
For cell of emf \(E_{2}\), balancing length \(L_{2}=40 \mathrm{~cm}\).
We know \(E \propto L\)
\(\Rightarrow \frac{E_{1}}{E_{2}}=\frac{L_{1}}{L_{2}} \Rightarrow \frac{1.25}{E_{2}}=\frac{30}{40}\)
\(\Rightarrow E_{2}=1.25 \times \frac{4}{3}=1.666 \mathrm{~V}\)
Therefore \(E_{2} \sim 1.67 \mathrm{~V}\)
Emf of second cell is \(1.67\)
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