JEE Mains · Physics · STD 12 - 3. current electricity
Resistance are connected in a meter bridge circuit as shown in the figure. The balancing length \(l_{1}\) is \(40\,cm\). Now an unknown resistance \(x\) is connected in series with \(P\) and new balancing length is found to be \(80\,cm\) measured from the same end. Then the value of \(x\) will be \(.......\Omega\)

- A \(2.2\)
- B \(22\)
- C \(200\)
- D \(20\)
Answer & Solution
Correct Answer
(D) \(20\)
Step-by-step Solution
Detailed explanation
Initially, \(\frac{ P }{ Q }=\frac{40 cm }{60 cm }=\frac{2}{3}\) Finally, \(\frac{ P + x }{ Q }=\frac{80 cm }{20 cm }=\frac{4}{1} \ldots(2)\) Divide \((2)\) by \((1)\) \(\frac{ P + x }{ P }=4 \times \frac{3}{2}=6\) \(1+\frac{ x }{ P }=6 \Rightarrow \frac{ x }{ P }=5\)…
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