JEE Main 2022PhysicsCommunication SystemEasyNumerical

JEE Main 2022 — Communication System Question with Solution

JEE Main 2022 (25 Jul Shift 1)

Question

The required height of a TV tower which can cover the population of 6.03 lakh is h. If the average population density is 100 per square km and the radius of earth is 6400 km, then the value of h will be _____ m.

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Show full solutionCorrect answer: 150
Correct answer
150

Step-by-step explanation

Maximum range of TV transmission d=2Rh

d=2×6400×h×10-3 (here, h in m)

Maximum area of TV transmission

Area =πd2

=π×2×6400×h×10-3 km2

Now, 6.03×100000=100×π×2×6400×10-3 h

h=6.03×10510×π×128

h=150 m

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About this question

This is a previous-year question from JEE Main 2022, covering the Communication System chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.