Physics · 2025
JEE Main · 7 April 2025, Shift 1 · Q49
For ac circuit shown in figure, R =100 k Ω and C =100 pF and the phase difference between V_in and ( V_B - V_A ) is 90^°. The input signal frequency…
For ac circuit shown in figure, $\displaystyle \mathrm{R}=100 \mathrm{k} \Omega$ and $\displaystyle \mathrm{C}=100 \mathrm{pF}$ and the phase difference between $\displaystyle V_{\text {in }}$ and $\displaystyle \left(\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}\right)$ is $\displaystyle 90^{\circ}$. The input signal frequency is $\displaystyle 10^x \mathrm{rad} / \mathrm{sec}$, where ' $\displaystyle x$ ' is $\displaystyle \_\_\_\_$.

Official answer
From NTA’s final answer key for this paper.
5
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JEE Main 2025 Physics question, with the answer from NTA’s final answer key. Where our answers come from.