Modern Physics Ques 257
- 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)
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Answer:
Correct Answer: 257.(b)
Solution:
- 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) \\ & =\quad \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 \\ & = \quad 13.98-8.37=5.61 \\ & \text { or } \quad h _T=31.5 \approx 32 m \end{aligned} $
BETA
