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 $1000k{m}^{-2}$?

• A. $4.0\times {10}^6$
• B. $8.0\times {10}^6$
• C. $4.0\times {10}^4$
• D. $8.0\times {10}^4$

Answer 1

The correct option is A $4.0\times {10}^6$

$h=100m$, $R_e=6.4\times {10}^{6}m,P_{density}={10}^{-3}{m}^{-2},P_{cov}=P_{density}\times Area$
=${10}^{-3}\times \pi \times d^2$ ......(1)
where,
$d=\sqrt{2h{R}_e}$
so, $d^2=2h{R}_e$
putting in eq.(1).

$P_{cov}={10}^{-3}\times \pi \times 2\times h{R}_e=4\times {10}^6$