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(B) $1$. Calculate the maximum line-of-sight distance for space wave communication: $d_m = \sqrt{2Rh} = \sqrt{2 	imes 6.4 	imes 10^6 	imes 300} = \

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Sky wave propagation

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(B) $1$. Calculate the maximum line-of-sight distance for space wave communication: $d_m = \sqrt{2Rh} = \sqrt{2 	imes 6.4 	imes 10^6 	imes 300} = \sqrt{3840 	imes 10^6} \approx 61.97\, km \approx 62\, km$.$2$. Since the distance between the transmitter and receiver is $100\, km$,which is greater than $d_m$ $(100\, km > 62\, km)$,space wave propagation is not possible.$3$. Calculate the critical frequency of the ionosphere: $f_c = 9(N_{\max})^{1/2} = 9 	imes (10^{12})^{1/2} = 9 	imes 10^6\, Hz = 9\, MHz$.$4$. Since the signal frequency $(5\, MHz)$ is less than the critical frequency $(9\, MHz)$,the signal can be reflected by the ionosphere.$5$. Therefore,the signal is received via sky wave propagation.

List-$I$	List-$II$
$A$. $AM$ Broadcast	$I$. $88-108\,MHz$
$B$. $FM$ Broadcast	$II$. $540-1600\,kHz$
$C$. Television	$III$. $3.7-4.2\,GHz$
$D$. Satellite Communication	$IV$. $54-890\,MHz$

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The waves used for line-of-sight $(LOS)$ communication are

For an amplitude-modulated wave, the audio frequency is $500 \, \text{cycles/second}$. What is the actual carrier frequency in $\text{cycles/sec}$?

Match List-$I$ with List-$II$:

List-$I$ List-$II$

$A$. $AM$ Broadcast $I$. $88-108\,MHz$

$B$. $FM$ Broadcast $II$. $540-1600\,kHz$

$C$. Television $III$. $3.7-4.2\,GHz$

$D$. Satellite Communication $IV$. $54-890\,MHz$

Choose the correct answer from the options given below:

Only $2 \%$ of the optical source frequency is the available channel bandwidth for an optical communication system operating at $1000 \,nm$. If an audio signal requires a bandwidth of $8 \,kHz$,how many channels can be accommodated for transmission?

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