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(D) For a tuned circuit of a transistor oscillator unit,the given values are:$L = 5 mH = 5 	imes 10^{-3} H$$C = 5 pF = 5 	imes 10^{-12} F$The natura

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$1 MHz$

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(D) For a tuned circuit of a transistor oscillator unit,the given values are:$L = 5 mH = 5 	imes 10^{-3} H$$C = 5 pF = 5 	imes 10^{-12} F$The natural frequency $f$ of the oscillator is given by the formula:$f = \frac{1}{2 \pi \sqrt{LC}}$Substituting the values:$f = \frac{1}{2 \pi \sqrt{5 	imes 10^{-3} 	imes 5 	imes 10^{-12}}}$$f = \frac{1}{2 \pi \sqrt{25 	imes 10^{-15}}}$$f = \frac{1}{2 \pi 	imes 5 	imes 10^{-7.5}}$Wait,correcting the calculation:$f = \frac{1}{2 \pi \sqrt{25 	imes 10^{-15}}} = \frac{1}{2 \pi 	imes 5 	imes 10^{-7} 	imes \sqrt{10^{-1}}} = \frac{1}{10 \pi 	imes 10^{-7} 	imes 0.316}$Actually,using $LC = 25 	imes 10^{-15} = 2.5 	imes 10^{-14}$:$f = \frac{1}{2 \pi \sqrt{2.5 	imes 10^{-14}}} = \frac{1}{2 \pi 	imes 1.581 	imes 10^{-7}}$$f = \frac{10^7}{9.93} \approx 1.006 	imes 10^6 Hz = 1 MHz$.

$A$ tuned circuit of a transistor oscillator unit has an inductance of $5 mH$ and a capacitance of $5 pF$. The natural frequency of the oscillator is

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