Question

The maximum distance between the transmitting and receiving TV towers is \[{\text{72 km}}\]. If the ratio of the heights of the TV transmitting tower to receiving tower is \[{\text{16:25}}\], the heights of the transmitting and receiving towers are
\[
  A.\;\;\;\;\;51.2{\text{ }}m,{\text{ }}82{\text{ }}m \\
  B.\;\;\;\;\;40{\text{ }}m,{\text{ }}80{\text{ }}m \\
  C.\;\;\;\;\;80{\text{ }}m,{\text{ }}125{\text{ }}m \\
  D.\;\;\;\;\;25{\text{ }}m,{\text{ }}75{\text{ }}m \\
  \]

Answer 1

Hint Use the relation for the maximum distance between transmitting and receiving TV towers which is given by
${d_m} = \sqrt {2{R_E}{h_T}} + \sqrt {2{R_E}{h_R}} $
Where, \[{{\text{d}}_{\text{m}}}\] = Maximum distance between the transmitting and receiving towers.
     \[{R_E}\] = Radius of the Earth = 6400 km.
     \[{{\text{h}}_{\text{T}}}\] = Height of the transmitting tower.
     \[{{\text{h}}_{\text{R}}}\] = Height of the receiving tower.

Complete step by step solution
We know that the maximum distance between transmitting and receiving TV towers is given by
${d_m} = \sqrt {2{R_E}{h_T}} + \sqrt {2{R_E}{h_R}} $……(i)
Where, \[{{\text{d}}_{\text{m}}}\] = Maximum distance between the transmitting and receiving towers.
     \[{R_E}\] = Radius of the Earth = 6400 km.
     \[{{\text{h}}_{\text{T}}}\] = Height of the transmitting tower.
     \[{{\text{h}}_{\text{R}}}\] = Height of the receiving tower.
Given: \[{{\text{d}}_{\text{m}}}\] \[{\text{ = 72 km}}\].
$\dfrac{{{h_T}}}{{{h_R}}} = \dfrac{{16}}{{25}}$
Cross multiplying, we get
${h_T} = \dfrac{{16}}{{25}}{h_R}$……(ii)
After putting the value of $d_m$ and $R_E$ in equation (i), we get
$72\times{10^3} = \sqrt {2\times6400{\text{ x 1}}{{\text{0}}^3}\times{h_T}} + \sqrt {2\times6400\times{{\text{10}}^3}\times{h_R}} $, (Multiplied with 103 to convert km to metre)
Now putting the value of $h_T$ in the above equation, we get
$72\times{10^3} = \sqrt {2\times6400{\text{ x 1}}{{\text{0}}^3}\times\dfrac{{16}}{{25}}\times{h_R}} + \sqrt {2\times6400{\times}{{\text{10}}^3}\times{h_R}} $
Taking $\sqrt {{h_R}} $ common from RHS, we get
$72\times{10^3} = \sqrt {{h_R}} \left( {\sqrt {2\times6400{\text{ x 1}}{{\text{0}}^3}\times\dfrac{{16}}{{25}}{\text{ }}} + \sqrt {2\times6400{\text{ x 1}}{{\text{0}}^3}} } \right)$
Or, $72\times{10^3} = 6439.875\times\sqrt {{h_R}} $
Or, $\sqrt {{h_R}} = 11.18$
Squaring both sides, we get
${h_R} = 125$m (approx.)
Now putting the value of $h_R$ in equation (ii), we get
${h_T} = \dfrac{{16}}{{25}}\times 125$
Or, ${h_T} = 80$m.

$\therefore $ Option C (80 m, 125 m) is the correct option.

Note: In the above question the value for radius of Earth was not given so try to use this type of important data. If a calculator is not provided better skip this question. Don’t try to square equation (i) to remove the roots because it won’t work as in the RHS you have to use the following formula
$({a^2} + {b^2}) = {a^2} + {b^2} + 2ab$
You will end up with a square root in the 2ab part.