Modern Physics 7 Question 11

11. In a line of sight radio communication, a distance of about 50 $km$ is kept between the transmitting and receiving antennas. If the height of the receiving antenna is $70 m$, then the minimum height of the transmitting antenna should be

(Radius of the earth $=6.4 \times 10^{6} m$ )

(a) $20 m$

(b) $32 m$

(c) $40 m$

(d) $51 m$

(Main 2019, 8 April II)

Show Answer

Solution:

  1. Key Idea In line of sight communication, distance $d$ between transmitting antenna and receiving antenna is given by

$$ d=\sqrt{2 R h _T}+\sqrt{2 R h _R} $$

Here in figure, $h _R$ and $h _T$ is the height of receiving and transmitting antenna, respectively.

Given, $d=50 km=50 \times 10^{3} m$

$$ h _R=70 m, R=6.4 \times 10^{6} m $$

Then, distance between transmitting and

receiving antenna, i.e. $d=\sqrt{2 R h _T}+\sqrt{2 R h _R}$

$$ \begin{aligned} & 50 \times 10^{3}=\sqrt{2 R}\left(\sqrt{h _T}+\sqrt{h _R}\right) \\ & =\sqrt{2 \times 6.4 \times 10^{6}}\left(\sqrt{h _T}+\sqrt{70}\right) \\ & \Rightarrow \quad \sqrt{h _T} \approx \frac{50 \times 10^{3}}{3577.7}-8.37 \\ & =13.98-8.37=5.61 \\ & \text { or } \quad h _T=31.5 \approx 32 m \end{aligned} $$



NCERT Chapter Video Solution

Dual Pane