Q 12-08-131JEE MainJEE Main 2022 (25 Jul, Shift 1)Medium
The required height of a TV tower which can cover a population of $6.03$ lakh is $h$. If the average population density is $100$ per square km and the radius of earth is $6400\ \text{km}$, then the value of $h$ will be ______ m.
Numerical value type. Enter your answer.
Answer: 150
Area to cover $= \dfrac{6.03\times10^5}{100} = 6030\ \text{km}^2 = \pi d^2 = \pi(2hR)$.
$$h = \frac{6030}{2\pi\times6400}\ \text{km}\approx0.15\ \text{km} = 150\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics