A radio can tune over the frequency range of a portion of MW broadcast band:

Solution:

The range of frequency (ν) of a radio is 800 kHz to 1200 kHz.

Lower tuning frequency, ν1 = 800 kHz = 800 × 103 Hz

Upper tuning frequency, ν2 = 1200 kHz = 1200× 103 Hz

Effective inductance of circuit L = 200 μH = 200 × 10−6 H

Capacitance of variable capacitor for ν1 is given as

$C_{1}=\frac{1}{\omega_{1}^{2} L}$

Where,

$\omega_{1}=$ Angular frequency for capacitor $C_{1}$

$=2 \pi v_{1}=2 \pi \times 800 \times 10^{3} \mathrm{rad} \mathrm{s}^{-1}$

$\therefore C_{1}=\frac{1}{\left(2 \pi \times 800 \times 10^{3}\right)^{2} \times 200 \times 10^{-6}}$

$=1.9809 \times 10^{-10} \mathrm{~F}=198.1 \mathrm{pF}$

Capacitance of variable capacitor for $v_{2}$,

$C_{2}=\frac{1}{\omega_{2}^{2} L}$

Where,

$\omega_{2}=$ Angular frequency for capacitor $C_{2}$

$=2 \pi v_{2}=2 \pi \times 1200 \times 10^{3} \mathrm{rad} \mathrm{s}^{-1}$

$\therefore C_{2}=\frac{1}{\left(2 \pi \times 1200 \times 10^{3}\right)^{2} \times 200 \times 10^{-6}}$

$=88.04 \mathrm{pF}$

Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF.