JEE Mains · Physics · STD 12 - 3. current electricity
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 ................. \(V\)

- A \(0.4\)
- B \(0.3\)
- C \(0.6\)
- D \(0.5\)
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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