The maximum capacitance of a variable capacitor is $33\; pF$. What should be the self-inductance to be connected to this capacitor for the natural frequency of the LC circuit to be $810 \; KHz$. Corresponding to A.M. broadcast band of Radio Nepal?

Given, Capacitance $(C) = 33\; pF = 33 * 10^{-12}\;F$

Let, $L$ be the self-inductance connected with the capacitance, so that the natural frequency of this LC circuit is $810\; KHz$,

$i.e\; \; \; \;  f = 810 \; * 10^3 \; Hz $

Now, we have 

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

$L =$ $ \frac{1}{4 \; \pi^2 \; (810 \; * \;10^3)^2 \; * \; 33 \; * \; 10^{-12}}$  $= 1.17 * 10^{-3}\; H$

Thus, Inductance connected $(L) = 1.17 * 10^{-3}\;H$

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