Question

A TV tower has a height of $100m$. How much population is covered by the TV broadcast if the average population density around the tower is 1000 $k{m}^{-2}$ ? (radius of earth = 6.4 $\times$${10}^6$ $m$).

• A. $4\times {10}^5$
• B. $5\times {10}^5$
• C. $4\times {10}^6$
• D. $8\times {10}^7$

Answer 1

The correct option is C $4\times {10}^6$

Here, $h=100m$, $R_e=6.4\times {10}^{6}m$
Average population density= $1000k{m}^{-2}={10}^{-3}{m}^{\mathrm{-2}}$
TV transmission range, $d=\sqrt{2h{R}_e}$
Therefore, area of TV transmission range
$A=\pi d^2=2\pi h{R}_e=2\times 3.14\times 100\times 6.4\times {10}^6=4\times {10}^9{m}^2$
Therefore, population covered by TV broadcast= $4\times {10}^9\times {10}^{-3}=4\times {10}^6$