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(A) Given: Inductance $L = 10\,mH = 10^{-2}\,H$,Frequency $f = 1\,MHz = 10^6\,Hz$,$\pi^2 = 10$.The resonance condition for an $LCR$ circuit is given b

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$2.5$

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(A) Given: Inductance $L = 10\,mH = 10^{-2}\,H$,Frequency $f = 1\,MHz = 10^6\,Hz$,$\pi^2 = 10$.The resonance condition for an $LCR$ circuit is given by the formula:$f = \frac{1}{2\pi\sqrt{LC}}$Squaring both sides:$f^2 = \frac{1}{4\pi^2LC}$Rearranging to solve for $C$:$C = \frac{1}{4\pi^2f^2L}$Substituting the given values:$C = \frac{1}{4 	imes 10 	imes (10^6)^2 	imes 10^{-2}}$$C = \frac{1}{40 	imes 10^{12} 	imes 10^{-2}}$$C = \frac{1}{40 	imes 10^{10}} = \frac{1}{4} 	imes 10^{-11} = 0.25 	imes 10^{-11}\,F$Converting to $pF$ $(1\,pF = 10^{-12}\,F)$:$C = 2.5 	imes 10^{-12}\,F = 2.5\,pF$.

The tuning circuit of a radio receiver has a resistance of $50\,\Omega$,an inductor of $10\,mH$ and a variable capacitor. $A$ $1\,MHz$ radio wave produces a potential difference of $0.1\,mV$. The value of the capacitor to produce resonance is.......$pF$ (Take $\pi^2 = 10$).

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