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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2021
  • Hard
  • MCQ

In the given potentiometer circuit arrangement, the balancing length \({AC}\) is measured to be \(250\) \({cm}\). When the galvanometer connection is shifted from point \((1)\) to point \((2)\) in the given diagram, the balancing length becomes \(400\, {cm}\). The ratio of the emf of two cells, \(\frac{\varepsilon_{1}}{\varepsilon_{2}}\) is -

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Answer & Solution

Correct Answer

(C) \(\frac{5}{3}\)

Step-by-step Solution

Detailed explanation

\(E_{1}=k l_{1} \ldots .\) (i)

\(E_{1}+E_{2}=k l_{2} \ldots\) (ii)

\(\frac{E_{1}}{E_{1}+E_{2}}=\frac{l_{1}}{l_{2}}=\frac{250}{400}=\frac{5}{8}\)

\(8 E_{1}=5 E_{1}+5 E_{2}\)

\(3 E_{1}=5 E_{2}\)

\(\frac{E_{1}}{E_{2}}=\frac{5}{3}\)

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