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Hint: In this question use the direct formula that the maximum distance up to which the signals can be broad casted from this tower in LOS (line of sight) mode is ${D_{\max }} = \sqrt {2R{H_T}} + \sqrt {2R{H_R}} $, where ${H_T}$ is the height of the transmission antenna and ${H_R}$is the height of the receiving antenna. Direct substitution of values into the formula will help getting the right answer for this problem statement.

Given data:

Height of the transmission tower = 140 m

Let ${H_T} = 140$ m.

And the height of the receiving antenna = 40 m

Let ${H_R} = 40$ m.

Now the maximum distance up to which the signals can be broad casted from this tower in LOS (line of sight) mode is given by the formula which is given as

$ \Rightarrow {D_{\max }} = \sqrt {2R{H_T}} + \sqrt {2R{H_R}} $

Where r = radius of the earth = 6.4 $ \times {10^6}$m.

Now simply substitute the values in the above equation we have,

$ \Rightarrow {D_{\max }} = \sqrt {2 \times 6.4 \times {{10}^6} \times 140} + \sqrt {2 \times 6.4 \times {{10}^6} \times 40} $

Now simplify this we have, as $\sqrt {{{10}^6}} = 1000$

$ \Rightarrow {D_{\max }} = 1000\sqrt {2 \times 64 \times 14} + 1000\sqrt {2 \times 64 \times 4} $

Now again simplify this we have,

As square root of 64 is 8 and square root of 4 is 2 so we have,

$ \Rightarrow {D_{\max }} = 1000\left( {16\sqrt 7 + 16\sqrt 2 } \right)$

$ \Rightarrow {D_{\max }} = 16000\left( {\sqrt 7 + \sqrt 2 } \right)$

Now as we know that $\sqrt 7 = 2.645$ and $\sqrt 2 = 1.414$ so use this value in above equation we have,

$ \Rightarrow {D_{\max }} = 16000\left( {2.645 + 1.414} \right) = 16000\left( {4.06} \right) = 64959.4$meter.

$ \Rightarrow {D_{\max }} \simeq 65$Km, as 1Km = 1000 m.

So the maximum distance up to which the signals can be broad casted from this tower in LOS (line of sight) mode is 65 km.

So this is the required answer.

Note – The key point here was that the broadcast was made along the line of sight that is line of sight refers to a straight line along which any observer has unobstructed vision. Here the waves from the transmission antenna are travelling directly to the receiving end and are not subjected to any reflections due to the obstacles present in between the path of transmission and receiving antenna.

__Complete step-by-step solution__-Given data:

Height of the transmission tower = 140 m

Let ${H_T} = 140$ m.

And the height of the receiving antenna = 40 m

Let ${H_R} = 40$ m.

Now the maximum distance up to which the signals can be broad casted from this tower in LOS (line of sight) mode is given by the formula which is given as

$ \Rightarrow {D_{\max }} = \sqrt {2R{H_T}} + \sqrt {2R{H_R}} $

Where r = radius of the earth = 6.4 $ \times {10^6}$m.

Now simply substitute the values in the above equation we have,

$ \Rightarrow {D_{\max }} = \sqrt {2 \times 6.4 \times {{10}^6} \times 140} + \sqrt {2 \times 6.4 \times {{10}^6} \times 40} $

Now simplify this we have, as $\sqrt {{{10}^6}} = 1000$

$ \Rightarrow {D_{\max }} = 1000\sqrt {2 \times 64 \times 14} + 1000\sqrt {2 \times 64 \times 4} $

Now again simplify this we have,

As square root of 64 is 8 and square root of 4 is 2 so we have,

$ \Rightarrow {D_{\max }} = 1000\left( {16\sqrt 7 + 16\sqrt 2 } \right)$

$ \Rightarrow {D_{\max }} = 16000\left( {\sqrt 7 + \sqrt 2 } \right)$

Now as we know that $\sqrt 7 = 2.645$ and $\sqrt 2 = 1.414$ so use this value in above equation we have,

$ \Rightarrow {D_{\max }} = 16000\left( {2.645 + 1.414} \right) = 16000\left( {4.06} \right) = 64959.4$meter.

$ \Rightarrow {D_{\max }} \simeq 65$Km, as 1Km = 1000 m.

So the maximum distance up to which the signals can be broad casted from this tower in LOS (line of sight) mode is 65 km.

So this is the required answer.

__Hence option (D) is the correct answer.__Note – The key point here was that the broadcast was made along the line of sight that is line of sight refers to a straight line along which any observer has unobstructed vision. Here the waves from the transmission antenna are travelling directly to the receiving end and are not subjected to any reflections due to the obstacles present in between the path of transmission and receiving antenna.

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