Question:
The $TV$ signals have a bandwidth of $3.7\, MHz$. The number of $TV$ channels that can be accommodated with a band of $3700 \,GHz$ is
The $TV$ signals have a bandwidth of $3.7\, MHz$. The number of $TV$ channels that can be accommodated with a band of $3700 \,GHz$ is
Updated On: May 11, 2024
- $10^4$
- $10^5$
- $10^6$
- $10^7$
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The Correct Option is C
Solution and Explanation
Here,
Bandwidth of each channel= $3.7 MHz$
$ = 3.7 \times 10^6 \, Hz $
Total bandwidth $= 3700 \, GHz= 3700 \times 10^9 \, Hz$
Number of TV channels accomodated
$ = \frac{Total \, bandwidth}{Bandwidth\, per\, channel}= \frac{3700 \times 10^9}{3.7 \times 10^6} = 10^6$
Bandwidth of each channel= $3.7 MHz$
$ = 3.7 \times 10^6 \, Hz $
Total bandwidth $= 3700 \, GHz= 3700 \times 10^9 \, Hz$
Number of TV channels accomodated
$ = \frac{Total \, bandwidth}{Bandwidth\, per\, channel}= \frac{3700 \times 10^9}{3.7 \times 10^6} = 10^6$
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