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(B) The maximum distance $d_m$ for line-of-sight communication is given by the formula: $d_m = \sqrt{2 R h_T} + \sqrt{2 R h_R}$.Here, $h_T = 20 \,m =

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$45$

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(B) The maximum distance $d_m$ for line-of-sight communication is given by the formula: $d_m = \sqrt{2 R h_T} + \sqrt{2 R h_R}$.Here, $h_T = 20 \,m = 20 	imes 10^{-3} \,km$, $d_m = 40 \,km$, and $R = 6400 \,km$.Substituting the values:$40 = \sqrt{2 	imes 6400 	imes 20 	imes 10^{-3}} + \sqrt{2 	imes 6400 	imes h}$$40 = \sqrt{256} + \sqrt{12800 	imes h}$$40 = 16 + \sqrt{12800 	imes h}$$24 = \sqrt{12800 	imes h}$Squaring both sides:$576 = 12800 	imes h$$h = \frac{576}{12800} \,km = 0.045 \,km = 45 \,m$.Thus, the correct option is $B$.

$A$ transmitting antenna of height $20 \,m$ and a receiving antenna of height $h$ are separated by a distance of $40 \,km$ for satisfactory communication in line-of-sight $(LOS)$ mode. The value of $h$ is (Given: radius of Earth is $6400 \,km$). (in $\,m$)

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