Q154 / 160 · AP EAMCET 2016 · Physics
The natural frequency of an $L-C$ circuit is $125 \mathrm{kHz}$. When the capacitor is totally filled with a dielectric material, the natural frequency decreases by $25 \mathrm{kHz}$. Dielectric constant of the material is nearly
- A$3.33$
- B$2.12$
- Correct answerC$1.56$
- D$1.91$
Solution
The natural frequency, $f_{\mathrm{N}}=\frac{1}{2 \pi \sqrt{K L C}}=\mathrm{Fe}$ When capacitor is totally filled with dielectric material of dielectric constant $\mathrm{K}$ then capacitance $\mathrm{C}^{\mathrm{C}}=\mathrm{KC}$
$\begin{aligned}
& f_{\mathrm{C}^{\prime}}=\frac{1}{2 \pi \sqrt{K L C}} \\
& \frac{f_C}{f_{C^{\prime}}}=\frac{1 / 2 \pi \sqrt{L C}}{1 / 2 \pi \sqrt{K L C}}=\sqrt{K} \\
& \Rightarrow \quad \frac{125 \times 10^3}{100 \times 10^3}=\sqrt{K} \Rightarrow \frac{S}{4}=\sqrt{K} \\
& {\left[\mathrm{f}_{\mathrm{c}}=125 \mathrm{kHz}-25 \mathrm{kHz}=100 \mathrm{kHz}=100 \times 10^3 \mathrm{~Hz}\right]} \\
& \therefore \quad K=\left(\frac{5}{4}\right)^2=\frac{25}{16}=1.562
\end{aligned}$
Hence, the correct answer is option: 'C'.