JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2019
- Hard
- Numerical
In the experimental setup of meter bridge shown in the figure, the null point is obtained at a distance of \(40\,cm\) from \(A\). If a \(10\,\Omega \) resistor is connected in series with \(R_1\), the null point shifts by \(10\,cm\). The resistance that should be connected in parallel with \(\left( {{R_1} + 10} \right)\,\Omega \) such that the null point shifts back to its initial position is .............. \(\Omega\)

Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(60\)
Step-by-step Solution
Detailed explanation
\(\frac{R_{1}}{R_{2}}=\frac{2}{3}\) .....\((i)\)
\(\frac{R_{1}+10}{R_{2}}=1\)
\(\Rightarrow \quad R_{1}+10=R_{2}\) .....\((ii)\)
\(\frac{2 \mathrm{R}_{2}}{3}+10=\mathrm{R}_{2} \quad ; \quad 10=\frac{\mathrm{R}_{2}}{3}\)
\(\Rightarrow \quad \mathrm{R}_{2}=30\, \Omega \quad \& \quad \mathrm{R}_{1}=20\, \Omega\)
\(\frac{\frac{30 \times \mathrm{R}}{30+\mathrm{R}}}{30}=\frac{2}{3}\)
\(\mathrm{R}=60 \,\Omega\)
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