The height of a TV tower is 150 m. If the po | vedclass

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(C) The area covered by the $TV$ tower is given by $A = \pi d_T^2$, where $d_T = \sqrt{2h_TR}$ is the range of the tower.$A = \pi (2h_TR) = 2 \pi h_T

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$829.3$

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(C) The area covered by the $TV$ tower is given by $A = \pi d_T^2$, where $d_T = \sqrt{2h_TR}$ is the range of the tower.$A = \pi (2h_TR) = 2 \pi h_T R$Given $h_T = 150 \ m$ and $R = 6.4 	imes 10^6 \ m$:$A = 2 	imes 3.14 	imes 150 	imes 6.4 	imes 10^6 \ m^2$$A = 6028.8 	imes 10^6 \ m^2 = 6028.8 \ km^2$Population density = $\frac{	ext{Total Population}}{	ext{Area}}$Population = $50 \ 	ext{lakhs} = 50 	imes 10^5 = 5 	imes 10^6$Population density = $\frac{5 	imes 10^6}{6028.8} \approx 829.3 \ km^{-2}$.

The height of a $TV$ tower is $150 \ m$. If the population density in the area covered by the $TV$ tower is $50 \ \text{lakhs}$, then the population density around the $TV$ tower is ....... $km^{-2}$. (Radius of the Earth is $6.4 \times 10^{6} \ m$)

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