Question

The natural frequency of an $L-C$ circuit is $125 \,kHz$. When the capacitor is totally filled with a dielectric material, the natural frequency decreases by $25\, kHz$. Dielectric constant of the material is nearly

• A. 3.33
• B. 2.12
• C. 1.56
• D. 1.91

Answer 1

Answer: (C)

Given, natural frequency of an $L-C$ circuit
$=125 kHz =125 \times 10^{3} \,Hz$
Let dielectric constant $=K$
The natural frequency, $f_{N}=\frac{1}{2 \pi \sqrt{L C}}$
After filling the dielectric material of dielectric constant $K$
$f_{M}=\frac{1}{2 \pi \sqrt{K L C}} \,\,\,\left(\because C^{'}=K C\right)$
$\therefore \frac{f_{N}}{f_{M}}=\frac{1 / 2 \pi \sqrt{L C}}{1 / 2 \pi \sqrt{K L C}}=\sqrt{K}$
$\Rightarrow \frac{125 \times 10^{3}}{100 \times 10^{3}}=\sqrt{K}$
$\left[f_{M}=125\, kHz -25\, kHz =100\, kHz =100 \times 10^{3}\, Hz \right]$
$ \Rightarrow \frac{5}{4}=\sqrt{K} $
$ \Rightarrow K=\left(\frac{5}{4}\right)^{2}=\frac{25}{16}=1.562$