JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2022
- Hard
- Numerical
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 \(.......\Omega\)
Pick an option to check your answer.
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 \((1)\)
\(\frac{ P + x }{ P }=4 \times \frac{3}{2}=6\)
\(1+\frac{ x }{ P }=6 \Rightarrow \frac{ x }{ P }=5\)
\(\therefore x =5 P =5 \times 4=20 \Omega\)
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