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(C) The resonant frequency of an $LC$ circuit is given by $f_0 = \frac{1}{2\pi\sqrt{LC}}$.When a dielectric slab of constant $K$ is inserted between t

Vedclass · Accepted Answer

$f_0 / 4$

Vedclass · Answer

(C) The resonant frequency of an $LC$ circuit is given by $f_0 = \frac{1}{2\pi\sqrt{LC}}$.When a dielectric slab of constant $K$ is inserted between the plates of the capacitor,the new capacitance becomes $C' = KC$.The new resonant frequency $f'$ is given by $f' = \frac{1}{2\pi\sqrt{LC'}} = \frac{1}{2\pi\sqrt{L(KC)}} = \frac{1}{\sqrt{K}} 	imes \frac{1}{2\pi\sqrt{LC}}$.Substituting the given values,$K = 16$ and $f_0 = \frac{1}{2\pi\sqrt{LC}}$,we get $f' = \frac{f_0}{\sqrt{16}} = \frac{f_0}{4}$.Therefore,the correct option is $C$.

The resonant frequency of an $LC$ circuit is $f_0$. If a dielectric slab of constant $K = 16$ is inserted completely between the plates of the capacitor,then the new resonant frequency is:

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