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A transmitting antenna of height $20 \mathrm{~m}$ and the receiving antenna of height $h$ are separated by a distance of $40 \mathrm{~km}$ for satisfactory communication in line of sight (Los) mode. Then the value of $h$ is (Give, radius of earth is $6400 \mathrm{~km}$.)
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Verified Answer
The correct answer is:
45 m
Given, height of transmitting antenna $h_T=20 \mathrm{~m}$, distance of LOS, $d_m=40 \mathrm{~km}$ As maximum distance of line of sight mode
$d_m=\sqrt{2 R h_T}+\sqrt{2 R h_R}$
putting the given values, we get
$\begin{aligned}
40= & \sqrt{2 \times 6.4 \times 10^3 \times 20 \times 10^{-3}} \\
& +\sqrt{2 \times 6.4 \times 10^3 \times h_R \times 10^{-3}} \\
40 & =16+\sqrt{2 \times 6.4 \times 10^3 \times 10^{-3} \times h_R} \\
\Rightarrow 2 \times 6.4 \times h_T & =24 \times 24 \\
\Rightarrow \quad h_T & =45 \mathrm{~m}
\end{aligned}$
Hence, the correct option is (b).
$d_m=\sqrt{2 R h_T}+\sqrt{2 R h_R}$
putting the given values, we get
$\begin{aligned}
40= & \sqrt{2 \times 6.4 \times 10^3 \times 20 \times 10^{-3}} \\
& +\sqrt{2 \times 6.4 \times 10^3 \times h_R \times 10^{-3}} \\
40 & =16+\sqrt{2 \times 6.4 \times 10^3 \times 10^{-3} \times h_R} \\
\Rightarrow 2 \times 6.4 \times h_T & =24 \times 24 \\
\Rightarrow \quad h_T & =45 \mathrm{~m}
\end{aligned}$
Hence, the correct option is (b).
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