Given,
L=200μH=200×10−6H
Frequency range of a portion of MW broadcast band 800kHz(sayfmrn) to 1200kHz(sayfmax ).
So, frequency of free oscillations f should lie between this range.
⇒fmin<f<fmax
As frequency of free oscillations
f=12π√LC
⇒C=14π2f2L
The range of C for given value of L
⇒C∝1fx
If f=fman=800kHz
Cmax=14π2×(800×107)2×200×10−6
Cmax=197.7×10−12F=197.7pF
If f=fmax=1200kHz
Cmin=14π2×(1200×10−2)2×200×10−6
Cmin=87.8×10−12F=87.8pF
Hence, Cmin<C<Cmax
Final Answer: The variable capacitor should have a range of about 88pF to 198pF.