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(A) The coverage distance $d$ of a $TV$ tower of height $h$ is given by $d = \sqrt{2Rh}$.Given the radius of the Earth $R = 6400 \, km$.The area cover

Vedclass · Accepted Answer

$150$

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(A) The coverage distance $d$ of a $TV$ tower of height $h$ is given by $d = \sqrt{2Rh}$.Given the radius of the Earth $R = 6400 \, km$.The area covered by the tower is $A = \pi d^2 = \pi (2Rh)$.Given the population $P = 6.03 	imes 10^5$ and population density $ho = 100 \, 	ext{people/km}^2$.The area $A$ is also given by $A = \frac{P}{ho} = \frac{6.03 	imes 10^5}{100} = 6030 \, 	ext{km}^2$.Equating the two expressions for area: $6030 = \pi 	imes 2 	imes 6400 	imes h$ (where $h$ is in $km$).$h = \frac{6030}{2 	imes \pi 	imes 6400} \, km$.$h = \frac{6030}{40212} \approx 0.150 \, km$.Converting to meters: $h = 0.150 	imes 1000 = 150 \, m$.

The required height of a $TV$ tower which can cover the population of $6.03$ lakh is $h$. If the average population density is $100$ per square $km$ and the radius of the Earth is $6400 \, km$,then the value of $h$ will be $... \, m$.

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