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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2019
  • Hard
  • MCQ

The resistance of the meter bridge \(AB\) in given figure is \(4\,\Omega \). With a cell of emf  \(\varepsilon \, = 0.5\,\,V\) and rheostat resistance \(R_h = 2\,\Omega \) the null point is obtained at some point \(J.\) When the cell is replaced by another one of emf \(\varepsilon \, = {\varepsilon _2}\) the same null point \(J\) is found for \(R_h = 6\,\Omega .\) The \(emf\)  \({\varepsilon _2}\) is ................. \(V\)

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Answer & Solution

Correct Answer

(B) \(0.3\)

Step-by-step Solution

Detailed explanation

\(0.5=\frac{6}{(2+\lambda L)} \lambda x\)        ......\((1)\)

\(E_{2}=\frac{6}{(6+\lambda L)} \lambda x\)        .......\((2)\)

So dividing equation \(( 1)\) and \(( 2)\)

\(\frac{E_{2}}{0.5}=\frac{2+4}{6+4}=\frac{3}{5}\)

\(\Rightarrow \quad E_{2}=0.3\) \(volt\)

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