JEE Main · 2022 · 27 Jun 2022 · Shift 1 · EM Waves

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The height of a transmitting antenna at the top of a tower is 25 m and that of receiving antenna is 49 m. The maximum distance between them, for satisfactory communication in LOS (Line-Of-Sight) is

K 5 \times 1 0^{2}

m. The value of

K

is ______.
[Assume radius of Earth is

64 \times 1 0^{5}

m] (Calculate upto nearest integer value)

Answer: 192

Solution

The line-of-sight (LOS) distance for radio communication between two antennas at heights

h_{T}

and

h_{R}

above Earth's surface is:

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

where

R

is Earth's radius.

Given

h_{T} = 25

h_{R} = 49

m, and

R = 64 \times 1 0^{5}

d = 2 \times 64 \times 1 0^{5} (25 + 49) = 128 \times 1 0^{5} (5 + 7) = 128 \times 1 0^{5} \times 12

128 \times 1 0^{5} = 128 \times 1 0^{2.5} = 82 \times 1 0^{2.5} = 82 \times 1 0^{2} \times 10

Simplifying:

128 \times 1 0^{5} = 16 \times 1 0^{2} 5

d = 16 \times 1 0^{2} 5 \times 12 = 1925 \times 1 0^{2} m

Therefore,

K = 192

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 27 Jun 2022 · Shift 1 · EM Waves

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