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Question

(A) The distance $d$ for line-of-sight communication between a transmitting antenna of height $h_T$ and a receiving antenna of height $h_R$ is given b

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(A) The distance $d$ for line-of-sight communication between a transmitting antenna of height $h_T$ and a receiving antenna of height $h_R$ is given by $d = \sqrt{2 R h_T} + \sqrt{2 R h_R}$.Given that the transmitting antenna is on the surface of the earth,$h_T = 0$.Therefore,the formula simplifies to $d = \sqrt{2 R h_R}$.Squaring both sides,we get $d^2 = 2 R h_R$.Solving for $h_R$,we get $h_R = \frac{d^2}{2 R}$.Given $d = 4\,km = 4000\,m$ and $R = 6400\,km = 6400000\,m$.Substituting the values: $h_R = \frac{(4000)^2}{2 	imes 6400000} = \frac{16000000}{12800000} = \frac{160}{128} = 1.25\,m$.We are given $h_R = x 	imes 10^{-2}\,m$.So,$1.25 = x 	imes 10^{-2} \implies x = 1.25 	imes 10^2 = 125$.

Column $A$	Column $B$
$A$. $D$-layer	$I$. $250-400 \ km$
$B$. $E$-layer	$II$. $170-190 \ km$
$C$. $F_1$-layer	$III$. $95-120 \ km$
$D$. $F_2$-layer	$IV$. $65-75 \ km$

$A$ transmitting antenna is kept on the surface of the earth. The minimum height of the receiving antenna required to receive the signal in line-of-sight at $4\,km$ distance from it is $x \times 10^{-2}\,m$. The value of $x$ is (Let radius of earth $R = 6400\,km$).

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