KCET · Physics · Current Electricity
A potentiometer has a uniform wire of length \(5 \mathrm{~m}\). A battery of emf \(10 \mathrm{~V}\) and negligible internal resistance is connected between its ends. A secondary cell connected to the circuit gives balancing length at \(200 \mathrm{~cm}\). The emf of the secondary cell is
- A \(4 \mathrm{~V}\)
- B \(6 \mathrm{~V}\)
- C \(2 \mathrm{~V}\)
- D \(8 \mathrm{~V}\)
Answer & Solution
Correct Answer
(A) \(4 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Given, length of potentiometer wire
\(l=5 \mathrm{~m}\)
emf of battery, \(E=10 \mathrm{~V}\)
Potential gradient across the potentiometer wire,
\(K=\frac{E}{l}=\frac{10}{5}=2 \mathrm{Vm}^{-1}\)
When secondary cell is connected to the circuit of potentiometer, then balancing length,
\(l_{1}=200 \mathrm{~cm}=2 \mathrm{~m}\)
\(\therefore\) emf of the secondary cell, \(E_{s}=K l_{1}=2 \times 2=4 \mathrm{~V}\)
\(l=5 \mathrm{~m}\)
emf of battery, \(E=10 \mathrm{~V}\)
Potential gradient across the potentiometer wire,
\(K=\frac{E}{l}=\frac{10}{5}=2 \mathrm{Vm}^{-1}\)
When secondary cell is connected to the circuit of potentiometer, then balancing length,
\(l_{1}=200 \mathrm{~cm}=2 \mathrm{~m}\)
\(\therefore\) emf of the secondary cell, \(E_{s}=K l_{1}=2 \times 2=4 \mathrm{~V}\)
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