Question

A TV tower has a height of \[70\,{\text{m}}\]. If the average population density around the tower is \[1000\,{\text{k}}{{\text{m}}^{ - {\text{2}}}}\], the population covered by the TV tower is:
(A) \[2.816 \times {10^6}\]
(B) \[2.86 \times {10^9}\]
(C) \[2.816 \times {10^3}\]
(D) \[2.816 \times {10^{12}}\]

Answer 1

Hint: First of all, we will convert all the units to the respective S.I units. We will find the expression for the distance up to which the tower can be viewed. After that calculating the total area over which the tower can be viewed, we can calculate the number of populations covered.

Complete step by step solution:
In the given question, we are supplied with the following data:
There is a TV tower whose height is \[70\,{\text{m}}\] .
The average density of the population around the tower is \[1000\,{\text{k}}{{\text{m}}^{ - {\text{2}}}}\] .
The radius of the Earth is \[6.4 \times {10^6}\,{\text{m}}\] .
We are asked to find the population covered by the tower.
To begin with, we will first convert the unit of population density from per square kilometre to per square metre. We can also easily find the distance up to which the transmission tower can be viewed.
Let us proceed to solve the problem.
The average density of the population around the tower is \[1000\,{\text{k}}{{\text{m}}^{ - {\text{2}}}}\] .
Now we will convert the units:
$1000\,{\text{k}}{{\text{m}}^{ - {\text{2}}}} \\
\Rightarrow 1000 \times {\left( {1 \times {{10}^3}\,{\text{m}}} \right)^{ - 2}} \\
\Rightarrow 1000 \times {10^{ - 6}}\,{{\text{m}}^{ - 2}} \\
\Rightarrow {10^{ - 3}}\,{{\text{m}}^{ - 2}} \\$
Now, we will calculate the distance up to which the tower is visible for the residents, which can be done as follows:
\[d = \sqrt {2hR} \] …… (1)
Where,
\[d\] indicates the distance up to which the tower is visible for the residents.
\[h\] indicates the height of the tower.
\[R\] indicates the radius of the earth.
Again, we know, the total area over which the transmission tower can be viewed with ease can be calculated as:
$A = \pi {d^2} \\
\Rightarrow A = \pi \times 2hR \\
\Rightarrow A = 2\pi hR \\$
Now, we can calculate the total population covered by the transmission tower is:
\[n = \rho \times 2\pi hR\] …… (2)
Where,
\[n\] indicates the total number of populations.
\[\rho \] indicates the average density of the population.
\[h\] indicates the height of the tower.
\[R\] indicates the radius of the earth.
Now, we substitute the required values in the equation (2) we get:
$n = \rho \times 2\pi hR \\
\Rightarrow n = {10^{ - 3}} \times 2 \times 3.14 \times 70 \times 6.4 \times {10^6} \\
\therefore n = 2.81 \times {10^6} \\$

Hence, the total number of populations covered by the tower is \[2.81 \times {10^6}\].The correct option is (A).

Note:While solving the problem remember that we have to include the radius of the earth too. This is because the earth is not flat rather it has curvature. So, an object situated high up, can be viewed by a limited number of people due to the curvature of the earth. A tower can be viewed by more and more if the height is increased to a greater value.