Question

If the area to be covered for TV telecast is doubled, then the height of the transmitting antenna (TV tower) will have to be

• A. doubled
• B. halved
• C. quadrupled
• D. kept unchanged

Answer 1

The correct option is A

doubled

Explanation for the correct option:

Step 1: Given

The area to be covered for TV telecast is doubled.

Let $A_1$ be the initial area covered.
Thus, the area doubled is $A_2=2A_1$

Step 2: Formulas used

The range (i.e., the maximum distance between transmitting and receiving antenna) of an antenna is given as,

$d=\sqrt{2RH}$
Where $R$ is the radius of the earth, and $H$ is the height of the antenna.

The area covered will be, $A=\mathrm{\pi }{\left(\frac{d}{2}\right)}^2$

Step 3: Calculation of height

Let $H_2$ be the height required for double the initial area coverage and $d_2$ be the range because of it.

So,

$\begin{array}{cc}& {\mathrm{A}}_2=\mathrm{\pi }{\left(\frac{d_{\mathit{2}}}{2}\right)}^2\\ \Rightarrow & {\mathrm{A}}_2=\mathrm{\pi }\frac{2R{H}_2}{4}\end{array}$

And,

$\begin{array}{cc}& A_1=\mathrm{\pi }{\left(\frac{d_1}{2}\right)}^2\\ \Rightarrow & A_1=\frac{2\mathrm{\pi }R{H}_1}{4}\end{array}$
Where $H_1$ is the height of the antenna initially and $d_1$ is the consequent range.

But,

$\begin{array}{cc}& A_2=2A_1\\ \Rightarrow & \frac{2\mathrm{\pi }R{H}_2}{4}=\frac{2\times 2\mathrm{\pi }R{H}_1}{4}\\ \Rightarrow & H_2=2H_1\end{array}$,

Thus, if the area to be covered is to be doubled, then the height of the antenna also has to be doubled.

Hence, option A is correct.