JEE Main · 2019 · 08 Apr 2019 · Shift 2 · EM Waves

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In a line of sight radio communication, a distance of about 50 km is kept between the transmitting and receiving antennas. If the height of the receiving antenna is 70 m, then the minimum height of the transmitting antenna should be: (Radius of the Earth

= 6.4 \times 1 0^{6} m

)

(A)32 m✓ Correct
(B)40 m
(C)51 m
(D)20 m

Answer: (A)

Solution

Line-of-sight range between two antennas is

d = 2 R h_{T} + 2 R h_{R}

, the sum of the two horizon distances. With

d = 50 \times 1 0^{3} m

R = 6.4 \times 1 0^{6} m

and

h_{R} = 70 m

, the receiver term is

2 \times 6.4 \times 1 0^{6} \times 70 \approx 29.9 \times 1 0^{3} m

. That leaves

2 R h_{T} = 50 \times 1 0^{3} - 29.9 \times 1 0^{3} \approx 20.1 \times 1 0^{3} m

for the transmitter, so

h_{T} = \frac{( 20.1 \times 1 0 ^{3} ) ^{2}}{2 \times 6.4 \times 1 0 ^{6}} \approx 32 m

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 08 Apr 2019 · Shift 2 · EM Waves

Solution

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