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JEE Advanced 2023
Paper-1 2023
Question
A series LCR circuit is connected to a $45 \sin (\omega t)$ Volt source. The resonant angular frequency of the circuit is $10^5 \mathrm{rad} \mathrm{s}^{-1}$ and current amplitude at resonance is $I_0$. When the angular frequency of the source is $\omega=8 \times 10^4$ rad $\mathrm{s}^{-1}$, the current amplitude in the circuit is $0.05 I_0$. If $L=50 \mathrm{mH}$, match each entry in List-I with an appropriate value from List-II and choose the correct option.
Select the correct option:
A
𝑃 → 2, 𝑄 → 3, 𝑅 → 5, 𝑆 → 1
B
𝑃 → 3, 𝑄 → 1, 𝑅 → 4, 𝑆 → 2
C
𝑃 → 4, 𝑄 → 5, 𝑅 → 3, 𝑆 → 1
D
𝑃 → 4, 𝑄 → 2, 𝑅 → 1, 𝑆 → 5
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Question Tags
JEE Advance
Physics
Easy
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