The tuning circuit of a radio receiver has a resistance of $$50\,\Omega ,$$  an inductor of $$10\,mH$$  and a variable capacitor. A $$1\,MHz$$  radio wave produces a potential difference of $$0.1\,mV.$$  The values of the capacitor to produce resonance is (Take $${\pi ^2} = 10$$  )

Solution :
$$\eqalign{ & L = 10\,mHz = {10^{ - 2}}\,Hz \cr & f = 1\,MHz = {10^6}\,Hz \cr & f = \frac{1}{{2\pi \sqrt {LC} }} \cr & \Rightarrow {f^2} = \frac{1}{{4{\pi ^2}LC}} \cr & \Rightarrow C = \frac{1}{{4{\pi ^2}{f^2}L}} = \frac{1}{{4 \times 10 \times {{10}^{ - 2}} \times {{10}^{12}}}} \cr & = \frac{{{{10}^{ - 12}}}}{4} = 2.5\,pF \cr} $$