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JEE Main 20198 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 8 April
Shift: Evening Shift
Subject: Physics
Chapter: Communication System

JEE Main 2019 · 8 April · Evening Shift · Physics · Communication System

In a line of sight radio communication, a distance of about 50km is...

In a line of sight radio communication, a distance of about

50km

is kept between the transmitting and receiving antennas. If the height of the receiving antenna is

70m,

then the minimum height of the transmitting antenna should be: (Radius of the Earth

= 6 . 4 \times {10}^{6}m

)

Options

A $20m$
B $51m$
C $40m$
D $32m$

Correct answer

D. $32m$

Step-by-step solution

Range,

d = \sqrt{2R{h}_{T}} + \sqrt{2R{h}_{r}}

⇒ 50 = \sqrt{2 \times 6400 \times {h}_{T}} + \sqrt{2 \times 6400 \times \left(70 \times {10}^{- 3}\right)}

⇒ \sqrt{2 \times 6400 \times {h}_{T}} = 50 - 8\sqrt{14}

{h}_{T} = \dfrac{{\left(50 - 8\sqrt{14}\right)}^{2}}{2 \times 6400} = 0.031458km

{h}_{T} \approx 32m

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