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Question: The T.V transmission tower in Delhi has a height of 240 m. The distance up to which the broadcast ca...

The T.V transmission tower in Delhi has a height of 240 m. The distance up to which the broadcast can be received (taking the radius of earth to be 6.4 ×\times 106^6 m)
A) 100 km
B) 60 km
C) 55 km
D) 50 km

Explanation

Solution

The distance of broadcast depends on the height of the transmission antenna. When a space wave transforms from an antenna, travelling in a straight line, directly reaches the receiving antenna.
The distance up to which the T.V. signals can directly be received depending upon the height of T.V transmissions antenna. The distance of transmission can be finds with the help of formula dT=2RhT{d_T} = \sqrt {2R{h_T}}

Step by step solution:
To solve this question we have a formula which gives us the relation between maximum transmission distances to the height of the antenna.
dT=2RhT{d_T} = \sqrt {2R{h_T}}
In question given values are
Height of transmission antenna hT=240m{h_T} = 240m
Radius of earth R=6.4×106mR = 6.4 \times {10^6}m
The maximum distance on earth from the transmitter where a signal can be received.
dT=2RhT dT=2×6.4×106×240 dT=3072×106 dT=3072×(106)12 dT=55.42×106×12 dT=55.42×103m    \Rightarrow {d_T} = \sqrt {2R{h_T}} \\\ \Rightarrow {d_T} = \sqrt {2 \times 6.4 \times {{10}^6} \times 240} \\\ \Rightarrow {d_T} = \sqrt {3072 \times {{10}^6}} \\\ \Rightarrow {d_T} = \sqrt {3072} \times {\left( {{{10}^6}} \right)^{\dfrac{1}{2}}} \\\ \Rightarrow {d_T} = 55.42 \times {10^{6 \times \dfrac{1}{2}}} \\\ \Rightarrow {d_T} = 55.42 \times {10^3}m \\\ \\\ \\\
We find the approximate distance of transmission on ground is 55.42×103m55.42 \times {10^3}m .
Now we have to convert the distance in km
We know 1000m=1km1000m = 1km
Or 103m=1km{10^3}m = 1km
Hence we get our answer 55.42km55.42km

Here option C is correct.

Note: A space wave travels in a straight line from transmitting antenna to the receiving antenna.
Due to the curvature of the earth direct space waves get blocked at some point.
If the heights of transmitting and receiving antenna both are given in question then we directly add the transmission distance of both antennas.
Formula will become
dnet=dT+dR dnet=2RhT+2RhR  {d_{net}} = {d_T} + {d_R} \\\ {d_{net}} = \sqrt {2R{h_T}} + \sqrt {2R{h_R}} \\\
Where dnet{d_{net}} \Rightarrow total distance on earth from the transmitter where a signal can be received.
hT{h_T} \Rightarrow Height of transmitting antenna.
hR{h_R} \Rightarrow Height of receiving antenna.