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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 \(.......\Omega\)

  1. A \(2.2\)
  2. B \(22\)
  3. C \(200\)
  4. D \(20\)
Verified Solution

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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