Question

The required height of a TV tower which can cover the population of $603$ 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 ______$m$

Answer 1

Correct Answer: 150

To determine the required height of a TV tower that can cover a population of $603$ lakh with a population density of $100$ per square km, we need to calculate the area that must be covered and subsequently find the corresponding height of the tower.
1. Calculate the coverage area needed:
The total population to be covered is $603$ lakh, which is $60300000$ people. With a density of $100$ people per square km, the required area (\(A\)) is:
  \(A=\frac{60300000}{100}=603000\) square km.
2. Determine the coverage radius:
The area (\(A\)) is related to the coverage radius (\(R\)) by the formula for the area of a circle: \(A=\pi R^2\). Solving for \(R\), we get:
  \(R=\sqrt{\frac{603000}{\pi}}\approx 437.38\, km\).
3. Use the Earth curvature formula:
The distance to the horizon from a height \(h\) and Earth’s radius \(R_e=6400\, km\) can be given by:
  \(d=\sqrt{2R_eh+h^2}\).
Since \(d\) must equal the coverage radius (437.38 km), we approximate as \(d\approx\sqrt{2R_eh}\) because \(h\ll R_e\):
  \(437.38\approx\sqrt{2\times6400\times h}\).
4. Solve for \(h\):
Square both sides to eliminate the square root:
  \(437.38^2=2\times6400\times h\).
Calculate \(h\) as follows:
  \(h=\frac{437.38^2}{2\times6400}\approx 149.555625\, m.\)
This value falls within the given range (150, 150).
Thus, the height of the TV tower required is approximately 150 meters.