Question

A T.V. tower has a height of 100m. How much population is covered by the T.V. broadcast if the average population density around the tower is 1000 per sq.km. Radius of the earth $6.37\times {10}^{6}m$. By how much height of the tower be increased to double its coverage range?

Answer 1

Here, h=100m;
R=$6.37\times {10}^{6}m$;
Population density, $\rho =1000{km}^{-2}$
$1000\times {10}^{-6}{m}^{-6}{m}^{-2}={10}^{-3}{m}^{-2}$
Population covered = $\rho \pi d^2=\rho \pi \left(2hR\right)$
$={10}^{-3}\times \frac{22}{7}\times 2\times 100\times 6.37\times {10}^6$
$40\times {10}^5=40lakhs$
Now, $d^{\prime }=\sqrt{{2h}^{\prime }R}=2d=2\sqrt{2hR}$
or $h^{\prime }=4h=4\times 100=400m$
Increase in height of tower = $h^{\prime }-h$
=400-100 = 300m.