Question

To double the covering range of TV transmitter tower, its height should be multiplied by:-
A. $\dfrac{1}{{\sqrt 2 }}$
B. $4$
C. $\sqrt 2$
D. $2$

Answer 1

To solve the question, all you need to do is apply the formula of the covering range of a TV transmitter by considering a tower of height $h$ on the Earth’s surface and then using the properties of triangle and circle, find a relation between the covering distance $d$ and the radius of earth, ${R_e}$ and the height of TV transmitter $h$ .

Complete step by step answer:
We will proceed with the solution exactly as explained in the hint section of the solution to the asked question. To reach at the answer, we first need to assume a TV transmitter tower of height $h$ on the surface of earth where, we can assert the radius of earth as ${R_e}$ and the covering range can be taken as the tangent drawn to the earth from the highest point of the TV transmitter or TV tower. After that, we can easily use the properties of the circle and right-angled triangle to achieve a relation between the covering range, height of the TV transmitter and the radius of earth.
Let’s have a look at the diagram of how it would look if we assume the TV transmitter of height $h$ on the surface of earth.

In the diagram, we can see that the covering range is nothing but the tangent drawn to the circle, from the highest point of the TV transmitter. Since any tangent, at the point of contact with the circle, becomes perpendicular to the radius of the circle. Hence, we can see that the situation becomes a right-angled triangle, where the sum of height of the TV transmitter and radius of earth is the hypotenuse of the triangle.
Let’s have a look at what is what:
The covering range, $d$ , can be taken as one of the base or the perpendicular.
The radius of the earth, perpendicular to the covering range, can be taken as one of the base or the perpendicular.
The TV transmitter plus the radius of earth can be taken as the hypotenuse of the triangle.
Hence, we can write:
${d^2} + {R_e}^2 = {\left( {h + {R_e}} \right)^2}$
Upon simplifying, we get:
${d^2} + R_e^2 = R_e^2 + {h^2} + 2h{R_e}$
Upon transposing and cancelling out terms, we get:
$d = \sqrt {{h^2} + 2h{R_e}}$
We already know that ${R_e}$ is the radius of earth which has a value of almost $6400{\kern 1pt} km$ and ${R_e} > > > h$
Hence, we can write the above-mentioned equation as:
$d = \sqrt {2h{R_e}}$
Let us assume the new height of the TV transmitter to be $h'$
Now, we can write:
$d' = \sqrt {2h'{R_e}}$
But it is given to us that: $d' = 2d$
Substituting, we get:
$2d = 2\sqrt {2h{R_e}} = \sqrt {2h'{R_e}}$
Upon solving, we get:
$h' = 4h$
Hence, we can see that the correct option is the option (B) since the value matches to the one that we found out.

Note: Since the covering range of a TV tower or TV transmitter is a very generalized and useful result, you are supposed to learn it to achieve quick and easy marks by solving such question. Do not make the mistake of neglecting the value of ${h^2}$ when it is told in the question that the height of the TV tower is comparable to the radius of the planet.