Question

A transmitting antenna of height $h$ and the receiving antenna of height $\dfrac{3}{4}h$ are separated by a distance of $d$ for satisfactory communication in line-of-sight mode. Then, the value of $h$ is
[Given, the radius of the earth is $R$].

$\begin{aligned} & A.\dfrac{{{d}^{2}}}{2R}{{\left( 2\sqrt{2}-\sqrt{6} \right)}^{2}} \\\ & B.\dfrac{{{d}^{2}}}{4R}{{\left( 2\sqrt{2}-\sqrt{6} \right)}^{2}} \\\ & C.\dfrac{{{d}^{2}}}{R}{{\left( 2\sqrt{2}-\sqrt{6} \right)}^{2}} \\\ & D.\dfrac{{{d}^{2}}}{8R}{{\left( 2\sqrt{2}-\sqrt{6} \right)}^{2}} \\\ \end{aligned}$

Answer 1

The height of antenna can be calculated using the equation,
${{d}_{m}}=\sqrt{2R{{h}_{T}}}+\sqrt{2R{{h}_{R}}}$
Where ${{d}_{m}}$ be the maximum distance of line of sight between the antenna and the transmitting antenna. Then after solving and rearranging we will get the value of ${{d}_{m}}$ in terms of $h$. The denominator of the term is factorized after this and then squares the both sides of the equation. This will lead you to the answer.

Complete answer:
first of all let us look at what all are given in the question.
The height of the transmitting antenna has been given as the equation,
${{h}_{T}}=h$
And also the height of the receiving antenna is given by the equation,
${{h}_{R}}=\dfrac{3}{4}h$
As we know the radius of the earth is $R$.
The maximum line of sight distance between these two antenna is given by the equation,
${{d}_{m}}=\sqrt{2R{{h}_{T}}}+\sqrt{2R{{h}_{R}}}$
In the question, it is mentioned that the value of ${{d}_{m}}$is,
${{d}_{m}}=d$
Substituting all these values in it will give,

& d=\sqrt{2Rh}+\sqrt{2R\times \dfrac{3}{4}h} \\\ & d=\sqrt{2R}\times \left( \sqrt{h}+\sqrt{\dfrac{3h}{4}} \right) \\\ \end{aligned}$$ Taking the $$\sqrt{2R}$$ into the denominator, will give, $$\dfrac{d}{\sqrt{2R}}=\left( \sqrt{h}+\sqrt{\dfrac{3h}{4}} \right)$$ Now let us take the common terms out, $$\begin{aligned} & \dfrac{d}{\sqrt{2R}}=\sqrt{h}\left( 1+\sqrt{\dfrac{3}{4}} \right) \\\ & \dfrac{d}{\sqrt{2R}}=\sqrt{h}\left( \dfrac{2+\sqrt{3}}{2} \right) \\\ \end{aligned}$$ Let us take all the terms into the left hand side keeping $$\sqrt{h}$$only in right hand side will give, $$\dfrac{2d}{\sqrt{2R}\times \left( 2+\sqrt{3} \right)}=\sqrt{h}$$ Factoring the denominator will be given as, $$\begin{aligned} & \sqrt{h}=\dfrac{2d\left( 2-\sqrt{3} \right)}{\sqrt{2R}} \\\ & \sqrt{h}=\dfrac{2\sqrt{2}d\left( 2-\sqrt{3} \right)}{2\sqrt{R}} \\\ \end{aligned}$$ Cancelling the common terms in the equation, $$\sqrt{h}=\dfrac{\sqrt{2}d\left( 2-\sqrt{3} \right)}{\sqrt{R}}$$ $$\sqrt{h}=\dfrac{d\left( 2\sqrt{2}-\sqrt{6} \right)}{\sqrt{R}}$$ Squaring the both sides of equations will give the value of height of the antenna. $$h=\dfrac{{{d}^{2}}{{\left( 2\sqrt{2}-\sqrt{6} \right)}^{2}}}{R}$$ **So, the correct answer is “Option C”.** **Note:** A transmission antenna is a basic device used in radio technology. They are made up of a conductor that passes an electric current whose intensity is varying over time and converts it into radiofrequency radiation that can travel through space. A receiving antenna is having the opposite of the process done by the transmission antenna.