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JEE Mains · Physics · STD 12 - 3. current electricity
A potentiometer wire \(PQ\) of \(1\,m\) length is connected to a standard cell \(E _{1}\). Another cell \(E _{2}\) of emf \(1.02\, V\) is connected with a resistance \('r'\) and switch \(S\) (as shown in figure). With switch \(S\) open, the null position is obtained at a distance of \(49\, cm\) from \(Q\). The potential gradient in the potentiometer wire is.......\(V/cm\)
- A \(0.02\)
- B \(0.04\)
- C \(0.01\)
- D \(.0.03\)
Answer & Solution
Correct Answer
(A) \(0.02\)
Step-by-step Solution
Detailed explanation
Sol. Balancing length is measured from \(P.\) So \(100-49=51 cm\) \(E _{2}=\phi \times 51\) Where \(\phi=\) Potential gradient \(1.02=\phi \times 51\) \(\phi=0.02 V / cm\)
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