Question

The height of a television tower is $100m$. If radius of the Earth is $6.4\times {10}^6$ and average-population density surrounding the tower is $1000perk{m}^2$, then the population covered by the television transmission is

• A. $2.06\times {10}^6$
• B. $4.02\times {10}^6$
• C. $5.18\times {10}^9$
• D. $6.04\times {10}^9$

Answer 1

Answer: (B)

$4.02\times {10}^6$

$h=100m$, $R=6.4\times {10}^{6}m$
See figure given.
Average population density $=1000k{m}^{-2}={10}^{-3}{m}^{-2}$
Maximum range of TV transmission
$d=\sqrt{2hR}={(2\times 100\times 6.4\times {10}^6)}^{\frac{1}{2}}$
$\therefore$ Maximum area covered by T. V. transmission
$=\pi d^2=\pi \times 2hR=2\times 3.14\times 1006.4\times {10}^6$
$=4.02\times {10}^9{m}^2$
$\therefore$ The population covered by T.V. transmission
$=(4.02\times {10}^9)\times {10}^{-3}=4.02\times {10}^6$