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**Hint :**In order to solve this question, we are going to take the expression for the frequency for a transistor- oscillator and then, by finding the ratio of the two frequencies $ {f_1} $ and $ {f_2} $ , and putting the changes in the values of inductance and the capacitance , the frequency $ {f_2} $ can be found in terms of $ {f_1} $ .

The frequency for a transistor – oscillator is given by

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

The ratio of two frequencies $ {f_1} $ and $ {f_2} $

$ \dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{\sqrt {{L_2}{C_2}} }}{{\sqrt {{L_1}{C_1}} }} $

**Complete Step By Step Answer:**

It is given in the question that a transistor- oscillator using a resonator circuit with an inductor $ L $ (of negligible resistance) and a capacitor $ C $ in series produce oscillations of frequency $ f $

We need to find the frequency when $ L $ is doubled and $ C $ is changed to $ 4C $

The frequency for a transistor – oscillator is given by

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

The ratio of two frequencies $ {f_1} $ and $ {f_2} $

$ \dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{\sqrt {{L_2}{C_2}} }}{{\sqrt {{L_1}{C_1}} }} $

Putting the values of the inductance and capacitance,

Here $ L $ is doubled and $ C $ is changed to $ 4C $

$ \dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{\sqrt {2L \times 4C} }}{{\sqrt {L \times C} }} \\

\Rightarrow \dfrac{{{f_1}}}{{{f_2}}} = \sqrt 8 = 2\sqrt 2 \\ $

Now, if we find the frequency $ {f_2} $ in terms of the frequency $ {f_1} $

We get $ {f_2} = \dfrac{{{f_1}}}{{2\sqrt 2 }} $

**Hence, option $ \left( C \right)\dfrac{f}{{2\sqrt 2 }} $ is the correct answer.**

**Note :**

A transistor can be operated as an oscillator for producing continuous undamped oscillations of any desired frequency if oscillatory and feedback circuits are properly connected to it. The frequency here is the resonant frequency that depends upon the values of inductance and the capacitance.

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