JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 15)
In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes $$x$$ times its initial resonant frequency $$\omega_0$$. The value of $$x$$ is :
1/4
1/16
4
16
Explanation
The resonance frequency of LC oscillations circuit is
$$ \begin{aligned} & \omega_0=\frac{1}{\sqrt{\mathrm{LC}}} \\\\ & \mathrm{L} \rightarrow 2 \mathrm{~L} \\\\ & \mathrm{C} \rightarrow 8 \mathrm{C} \\\\ & \omega=\frac{1}{\sqrt{2 \mathrm{~L} \times 8 \mathrm{C}}}=\frac{1}{4 \sqrt{\mathrm{LC}}} \\\\ & \omega=\frac{\omega_0}{4} \end{aligned} $$
So $\mathrm{x}=\frac{1}{4}$
$$ \begin{aligned} & \omega_0=\frac{1}{\sqrt{\mathrm{LC}}} \\\\ & \mathrm{L} \rightarrow 2 \mathrm{~L} \\\\ & \mathrm{C} \rightarrow 8 \mathrm{C} \\\\ & \omega=\frac{1}{\sqrt{2 \mathrm{~L} \times 8 \mathrm{C}}}=\frac{1}{4 \sqrt{\mathrm{LC}}} \\\\ & \omega=\frac{\omega_0}{4} \end{aligned} $$
So $\mathrm{x}=\frac{1}{4}$
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