Q156 / 180 · KCET 2014 · Physics
A tuned amplifier circuit is used to generate a carrier frequency of \( 2 \mathrm{MHz} \) for the amplitude
modulation. The value of \( \sqrt{\mathrm{LC}} \) is
modulation. The value of \( \sqrt{\mathrm{LC}} \) is
- A\( \frac{1}{2 \pi \times 10^{6}} \)
- B\( \frac{1}{2 \times 10^{6}} \)
- C\( \frac{1}{3 \Pi \times 10^{6}} \)
- Correct answerD\( \frac{1}{4 \Pi \times 10^{6}} \)
Solution
We know \( f=\frac{1}{2 \pi \sqrt{L} C} \)
\( \Rightarrow \sqrt{L} C=\frac{1}{2 \pi f} \)
Given \( f=2 \mathrm{MHz}=2 \times 10^{6} \mathrm{~Hz} \)
Therefore, \( \sqrt{L C}=\frac{1}{2 \Pi \times 2 \times 10^{6}}=\frac{1}{4 \Pi \times 10^{6}} \)
\( \Rightarrow \sqrt{L} C=\frac{1}{2 \pi f} \)
Given \( f=2 \mathrm{MHz}=2 \times 10^{6} \mathrm{~Hz} \)
Therefore, \( \sqrt{L C}=\frac{1}{2 \Pi \times 2 \times 10^{6}}=\frac{1}{4 \Pi \times 10^{6}} \)
Hence, the correct answer is option: 'D'.