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A TV transmission tower has a height of $240 \mathrm{~m}$. Signals broadcast from this tower will be received by LOS communication at a distance of (assume the radius of earth to be $\left(6.4 \times 10^6 \mathrm{~m}\right)$
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The correct answers are:
$24 \mathrm{~km}$
,
$55 \mathrm{~km}$
,
$50 \mathrm{~km}$
$24 \mathrm{~km}$
,
$55 \mathrm{~km}$
,
$50 \mathrm{~km}$
As given that,
height of tower $\mathrm{h}_{\mathrm{T}}=240 \mathrm{~m}$
Radius of earth $\mathrm{R}=6.4 \times 10^6 \mathrm{~m}$
For LOS (line of sight) communication range
$$
\mathrm{d}_{\mathrm{T}}=\sqrt{2 \mathrm{Rh}_{\mathrm{T}}}
$$
So, the maximum distance on earth from the transmitter upto which a signal can be received is $\mathrm{d}_{\mathrm{T}}=\sqrt{2 \mathrm{Rh}}$
Putting all these values in Eq. (i),
we get $\mathrm{d}_{\mathrm{T}}=\sqrt{2 \mathrm{Rh}}=\sqrt{2 \times 6.4 \times 10^6 \times 240}$
$$
=55.4 \times 10^3 \mathrm{~m}=55.4 \mathrm{~km}
$$
So, the range of $55.4 \mathrm{~km}$ covers the distance $24 \mathrm{~km}, 55 \mathrm{~km}$ and $50 \mathrm{~km}$.
height of tower $\mathrm{h}_{\mathrm{T}}=240 \mathrm{~m}$
Radius of earth $\mathrm{R}=6.4 \times 10^6 \mathrm{~m}$
For LOS (line of sight) communication range
$$
\mathrm{d}_{\mathrm{T}}=\sqrt{2 \mathrm{Rh}_{\mathrm{T}}}
$$
So, the maximum distance on earth from the transmitter upto which a signal can be received is $\mathrm{d}_{\mathrm{T}}=\sqrt{2 \mathrm{Rh}}$
Putting all these values in Eq. (i),
we get $\mathrm{d}_{\mathrm{T}}=\sqrt{2 \mathrm{Rh}}=\sqrt{2 \times 6.4 \times 10^6 \times 240}$
$$
=55.4 \times 10^3 \mathrm{~m}=55.4 \mathrm{~km}
$$
So, the range of $55.4 \mathrm{~km}$ covers the distance $24 \mathrm{~km}, 55 \mathrm{~km}$ and $50 \mathrm{~km}$.
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