Correct Answer - Option 1 : 2000 kHz
Concept:
For a given carrier wave of frequency fc with modulation frequency fm, the bandwidth is calculated by
fupper = fc + fm
flower = fc – fm ----(1)
To avoid overlapping of bandwidths, next broadcast frequencies can be
f1 = fc ± 2fm
f2 = fc ± 3fm and so on
So, next immediate available broadcast frequency is f1 = fc + 2fm and \(f_1' = {f_c} - 2{f_m}\)
So,
f1 = fc + 2fm
Calculation:
Given,
Modulation frequency, fm = 250 kHz
fm is 10% of fc
i.e., fm = 10% of fc
\(250\;kHz = \frac{{10}}{{100}} \times {f_c}\)
2500 = fc
fc = 2500 kHz
And also that fm is 10% of fe, i.e fe = 2500 kHz
So, f1 = 2500 + (2 × 250)
f1 = 2500 + 500 = 3000 kHz
Now,
\(f_1' = {f_c} - 2{f_m}\)
\(f_1' = 2500 - \left( {2 \times 250} \right)\)
\(f_1' = 2500 - 500 = 2000\) kHz
Therefore, if another AM station approaches us for license, the lowest broadcast frequency which we can allot is 2000 kHz.