The tuning circuit of a radio receiver has a resistance of \[50\text{ }\Omega \] , an inductor of \[10\text{ mH}\] and a variable capacitor. A \[1\text{ MHz}\] radio wave produces a potential difference of \[\text{0}\text{.1 mV}\] . The value of the capacitor to produce resonance is (take \[{{\pi }^{\text{2}}}=10\] )
(A) \[2.5\text{ pF}\]
(B) \[\text{5}\text{.0 pF}\]
(C) \[25\text{ pF}\]
(D) \[\text{50 pF}\]

Hint: A tuning circuit consists of an inductor and a capacitor, usually in parallel. Either the inductance of the inductor or the capacitance of the capacitor must be able to be varied. By varying either the capacitance or inductance the resonant frequency of the circuit is varied. When the resonant frequency matches an incoming signal, for example, the carrier wave of your favourite radio station, then the circuit is tuned to that frequency and the radio program can be heard.

Formula Used:
\[f\text{=}\dfrac{1}{2\pi \sqrt{LC}}\]

Complete step by step answer:
We have been given that,
Resistance present in the tuning circuit, \[(R)=50\text{ }\Omega \]
Value of Inductor in the circuit, \[(L)=10\text{ mH=10}\times \text{1}{{\text{0}}^{\text{-3}}}\text{H=1}{{\text{0}}^{\text{-2}}}H\] since we know that \[1\text{mH=1}{{\text{0}}^{\text{-3}}}\text{H}\]
Frequency of oscillation of the tuning circuit, \[(f)=1\text{ MHz=1}{{\text{0}}^{\text{6}}}Hz\] since we know that \[1\text{ MHz=1}{{\text{0}}^{\text{6}}}Hz\]
Potential difference produced by the circuit, \[\text{(V)=0}\text{.1 mV}\]
Let the value of capacitance of the circuit be \[\text{C farads}\] .
Now, a tuning circuit operates in the condition of resonance and we know that the expression for the resonant frequency of a circuit having an inductor and a capacitor is
\[f\text{=}\dfrac{1}{2\pi \sqrt{LC}}\] where the symbols have their meaning as discussed above.
Squaring both sides of this expression, we get \[{{f}^{2}}\text{=}\dfrac{1}{4{{\pi }^{2}}LC}\]
Rearranging the terms a little bit, we can say that \[C=\dfrac{1}{4{{\pi }^{2}}{{f}^{2}}L}\]
Substituting the values given to us in the above expression, we get
\[\begin{align}
  & C=\dfrac{1}{4{{\pi }^{2}}\times {{({{10}^{6}})}^{2}}\times ({{10}^{-2}})} \\
 & \Rightarrow C=\dfrac{1}{4\times 10\times {{10}^{12}}\times {{10}^{-2}}}(\because {{\pi }^{\text{2}}}=10) \\
 & \Rightarrow C=0.25\times {{10}^{-11}}F \\
 & \Rightarrow C=2.5pF(\because 1pF={{10}^{-12}}F) \\
\end{align}\]
Hence we can see that the correct option is (A).

Note: We have also been given the values of the resistance and the potential difference in the circuit when we didn’t need them at all in our solution. This is a tactic of the examiner, to give a lot of data so that the student gets intimidated and makes a mistake. Many students find the impedance of the circuit with the resistance and the inductance given. You should keep in mind that oscillations in a circuit only take place between the capacitor and the inductor.