The number of AM broadcast stations that can be accommodated in a $300 k H z$ band width for the highest modulating frequency $15 k H z$ will be

10

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5

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7

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12

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Solution

The correct option is A $10$
Number of channels $= \frac{B W}{2 \times H i g h e s t m o d u l a t i n g f r e q u e n c y} = \frac{300 \times 10^{3}}{2 \times 15 \times 10^{3}} = 10$

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