A TV tower has a height of 100 m. The averag | vedclass

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(C) The range of the $TV$ tower is given by $d = \sqrt{2hR}$.The area covered by the tower is $A = \pi d^2 = \pi (2hR) = 2 \pi hR$.Given: $h = 100 \ m

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$4 	imes 10^6$

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(C) The range of the $TV$ tower is given by $d = \sqrt{2hR}$.The area covered by the tower is $A = \pi d^2 = \pi (2hR) = 2 \pi hR$.Given: $h = 100 \ m = 0.1 \ km$, $R = 6.4 	imes 10^6 \ m = 6400 \ km$, and population density $ho = 1000 \ 	ext{per} \ km^2$.Area $A = 2 	imes \pi 	imes 0.1 \ km 	imes 6400 \ km = 1280 \pi \ km^2$.Using $\pi \approx 3.14$, $A \approx 2 	imes 3.14 	imes 0.1 	imes 6400 = 4019.2 \ km^2 \approx 4000 \ km^2$.Population covered $= A 	imes ho = 4000 \ km^2 	imes 1000 \ 	ext{per} \ km^2 = 4 	imes 10^6$.

List $I$	List $II$
$A$. Attenuation	$I$. Combination of a receiver and transmitter.
$B$. Transducer	$II$. Process of retrieval of information from the carrier wave at receiver.
$C$. Demodulation	$III$. Converts one form of energy into another.
$D$. Repeater	$IV$. Loss of strength of a signal while propagating through a medium.

$A$ $TV$ tower has a height of $100 \ m$. The average population density around the tower is $1000 \ \text{per} \ km^2$. The radius of the earth is $6.4 \times 10^6 \ m$. The population covered by the tower is:

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