For VHF signal broadcasting,what is the maxim | vedclass

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(A) The distance to the horizon $d$ for an antenna of height $h$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the earth.The ser

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(A) The distance to the horizon $d$ for an antenna of height $h$ is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the earth.The service area $A$ covered by the antenna is the area of a circle with radius $d$,given by $A = \pi d^{2}$.Substituting the expression for $d$,we get $A = \pi (\sqrt{2Rh})^{2} = 2\pi Rh$.Given $R = 6400\, km = 6400 	imes 10^{3}\, m$ and $h = 30\, m$.Converting $R$ to $km$ to match the required unit for area: $R = 6400\, km$ and $h = 30\, m = 0.03\, km$.$A = 3.14 	imes 2 	imes 6400 	imes 0.03$.$A = 3.14 	imes 384 = 1205.76\, km^{2}$.Rounding off to the nearest integer,we get $A \simeq 1206\, km^{2}$.

For $VHF$ signal broadcasting,what is the maximum service area in $km^{2}$ that will be covered by an antenna tower of height $30\, m$,if the receiving antenna is placed at the ground? Let the radius of the earth be $6400\, km$. (Round off to the nearest integer) (Take $\pi$ as $3.14$)

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