Q91 / 100 · BITSAT 2023 (Memory Based Paper 1) · Physics
In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes $\mathrm{x}$ times its initial resonant frequency $\omega_0$. The value of $x$ is:
- Correct answerA$1 / 4$
- B16
- C$1 / 16$
- D4
Solution
The resonance frequency of LC oscillations circuit is
$\omega=\frac{1}{\sqrt{\mathrm{L}^{\prime} \mathrm{C}^{\prime}}} \Rightarrow \mathrm{L}^{\prime} \rightarrow 2 \mathrm{~L}$
$\mathrm{C}^{\prime} \rightarrow 8 \mathrm{C}$
$\omega=\frac{1}{\sqrt{2 \mathrm{~L} \times 8 \mathrm{C}}}=\frac{1}{4 \sqrt{\mathrm{LC}}} 9 \omega_0=\frac{1}{\sqrt{\mathrm{LC}}}$
$\omega=\frac{\omega_0}{4}$ So, $x=\frac{1}{4}$
$\omega=\frac{1}{\sqrt{\mathrm{L}^{\prime} \mathrm{C}^{\prime}}} \Rightarrow \mathrm{L}^{\prime} \rightarrow 2 \mathrm{~L}$
$\mathrm{C}^{\prime} \rightarrow 8 \mathrm{C}$
$\omega=\frac{1}{\sqrt{2 \mathrm{~L} \times 8 \mathrm{C}}}=\frac{1}{4 \sqrt{\mathrm{LC}}} 9 \omega_0=\frac{1}{\sqrt{\mathrm{LC}}}$
$\omega=\frac{\omega_0}{4}$ So, $x=\frac{1}{4}$
Hence, the correct answer is option: 'A'.