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 .

Step-by-step solution

Maximum range of TV transmission d = 2 R h d = 2 × 6400 × h × 10 - 3 (here, h in m ) Maximum area of TV transmission Area = π d 2 = π × 2 × 6400 × h × 10 - 3   km 2 Now, 6 . 03 × 100000 = 100 × π × 2 × 6400 × 10 - 3   h h = 6 . 03 × 10 5 10 × π × 128 h = 150   m