Question

The capacitor of an oscillatory LC circuit is enclosed in a container. When the container is empty, the frequency is $10\text{kHz}$. When the container is filled with a gas, the frequency changes by $50\text{Hz}$. The dielectric constant of the gas is -

• A. $2.01$
• B. $1.01$
• C. $3.01$
• D. $5.01$

Answer 1

The correct option is B $1.01$

We know that, frequency of oscillations of charge in the LC circuit is,

$f=\frac{1}{2\pi \sqrt{LC}}$

$\Rightarrow f\propto \frac{1}{\sqrt{C}}$

$\therefore \frac{f_2}{f_1}=\sqrt{\frac{C_1}{C_2}}....\left(1\right)$

Also, capacitance,

$C=\frac{K{ϵ}_{0}A}{d}$

$\Rightarrow \frac{C_1}{C_2}=\frac{K_1}{K_2}=\frac{1}{K}....\left(2\right)$

Here, $K$ is the dielectric constant of the gas.

From equations $\left(1\right)$ and $\left(2\right)$,

$\frac{f_2}{f_1}=\sqrt{\frac{1}{K}}$

As the dielectric medium in the capacitor changes from vacuum to gas, the capacitance will increase, so the frequency will decrease, $i.e.f_2<f_1$.

$\therefore \frac{f_2}{f_1}=\frac{{10}^4-50}{{10}^4}=\sqrt{\frac{1}{K}}$

$\Rightarrow K=1.01$

Hence, option $\left(B\right)$ is the correct answer.