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 $100k{{m}^{-2}}$.
(Radius of the earth $=6.37\times {{10}^{6}}m$)
$A)\text{ 4 lakh}$
$B)\text{ 4 thousand}$
$C)\text{ 40 thousand}$
$D)\text{ 40 lakh}$

Answer 1

Hint: This problem can be solved by using the direct formula for the area of coverage of broadcast of a tower in terms of its height and the earth’s radius when the height of the tower is negligible in comparison to the radius of the earth. After getting the area, we can multiply it with the population density to get the number of the population covered.
Formula used:
$A=2\pi h{{R}_{E}}$
$\text{Population density = }\dfrac{\text{Population}}{\text{Area}}$

Complete step by step solution:
We will use the direct formula for the area of coverage of a broadcasting tower and multiply the area to the population density to get the population covered by the tower.
The area of coverage $A$ of a broadcasting tower of height $h$ above the surface of the earth is given by
$A=2\pi h{{R}_{E}}$ --(1)
Where ${{R}_{E}}=6.37\times {{10}^{6}}km$ is the radius of the earth. This formula is valid for $h\ll {{R}_{E}}$.
Hence, now let us analyze the question.
The given height of the tower is $h=100m$.
The given radius of the earth is ${{R}_{E}}=6.37\times {{10}^{6}}km$.
Let the area of coverage of the tower be $A$.
Let the required population number which is covered by the broadcast of the tower be $N$.

The given population density is $\sigma =100k{{m}^{-2}}=100\times {{10}^{-6}}{{m}^{-2}}={{10}^{-4}}{{m}^{-2}}$ $\left( \because 1k{{m}^{-2}}={{10}^{-6}}{{m}^{-2}} \right)$
Hence, using (1), we get
$A=2\pi h{{R}_{E}}$
$\therefore A=2\pi \times 100\times 6.37\times {{10}^{6}}=4\times {{10}^{9}}{{m}^{2}}$ --(2)
Now, the relation between population density, area and population is
$\text{Population density = }\dfrac{\text{Population}}{\text{Area}}$ --(3)
Using (3), we get
$\sigma =\dfrac{N}{A}$
Using (2) in the above equation, we get
${{10}^{-4}}=\dfrac{N}{4\times {{10}^{9}}}$
$\therefore N=4\times {{10}^{9}}\times {{10}^{-4}}=4\times {{10}^{5}}=4\text{ lakh}$ $\left( \because {{10}^{5}}=1,00,000=1\text{ lakh} \right)$
Hence, the population covered by the broadcasting tower is $4\text{ lakh}$.
Hence, the correct option is $A)\text{ 4 lakh}$.

Note: The formula for the area covered by a broadcasting tower is an approximation which is valid when the height of the tower is negligible in comparison to the radius of the earth (which is almost always the case). However, the exact formula from which this approximation is made is
$A=2\pi {{R}_{E}}^{2}\left( \dfrac{h}{h+{{R}_{E}}} \right)$
This formula is derived by using the formula for the solid angle and the area of coverage of the solid angle which is formed by the tower upon the surface of the earth.