Question:
A TV tower has a height 150 m. What is the total population covered by the TV tower, if the population density around the TV tower is $10^3\, km^{-2}?$ Radius of the earth is $6.4\times10^{6}m$
A TV tower has a height 150 m. What is the total population covered by the TV tower, if the population density around the TV tower is $10^3\, km^{-2}?$ Radius of the earth is $6.4\times10^{6}m$
Updated On: May 15, 2024
- 60.288 lakh
- 40.192 lakh
- 100 lakh
- 20.22 lakh
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The Correct Option is A
Solution and Explanation
$h =150\, m,R = 6.4 \times 10^{6}\, m$
Average population density $ = 10^3\, km^{-2}$
$\hspace25mm = (10^3) (10^3) m^{-2} = 10^{-3}\, m^{-2}$
Distance upto which the transmission could be viewed
$\hspace30mm d = \sqrt{2hR}$
Total area over which transmission could be viewer
$\hspace30mm = \pi d^2 = 2 \pi h R$
Population covered $= 10^{-3} \times 2 \pi h R $
$\hspace40mm = 10^{-3} \times 2 \times 3.14 \times 150 \times 6.4 \times 10^6 $
$\hspace40mm = 60.288\, lakh$
Average population density $ = 10^3\, km^{-2}$
$\hspace25mm = (10^3) (10^3) m^{-2} = 10^{-3}\, m^{-2}$
Distance upto which the transmission could be viewed
$\hspace30mm d = \sqrt{2hR}$
Total area over which transmission could be viewer
$\hspace30mm = \pi d^2 = 2 \pi h R$
Population covered $= 10^{-3} \times 2 \pi h R $
$\hspace40mm = 10^{-3} \times 2 \times 3.14 \times 150 \times 6.4 \times 10^6 $
$\hspace40mm = 60.288\, lakh$
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