JEE Mains · Physics · STD 12 - 3. current electricity
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 .............. \(\Omega\)
- A \(20\)
- B \(40\)
- C \(60\)
- D \(30\)
Answer & Solution
Correct Answer
(C) \(60\)
Step-by-step Solution
Detailed explanation
\(\frac{R_{1}}{R_{2}}=\frac{2}{3}\) .....\((i)\) \(\frac{R_{1}+10}{R_{2}}=1\) \(\Rightarrow \quad R_{1}+10=R_{2}\) .....\((ii)\) \(\frac{2 \mathrm{R}_{2}}{3}+10=\mathrm{R}_{2} \quad ; \quad 10=\frac{\mathrm{R}_{2}}{3}\)…
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