Question

A TV tower has a height of 100 m. How much
population is covered by the TV broadcast if the
average population density around the tower is
100 km⁻² (radius of the earth = 6.37 × 10⁶ m)
(a) 4 lakh (b) 4 billion
(c) 40,000 (d) 40 lakh

Answer 1

Answer:

A TV tower has a height of 100 m. How much

population is covered by the TV broadcast if the

average population density around the tower is

100 km⁻² (radius of the earth = 6.37 × 10⁶ m)

(a) 4 lakh✓✓

(b) 4 billion

(c) 40,000 (d) 40 lakh

Explanation:

$Solid angle subtended: w = 2\pi  (1 - cos  \theta)  =  2\pi  (1 - \dfrac{R}{R+h}) </p><p></p><p></p><p>Now, area covered: A =  w R^2  =  2\pi  R^2  (1- \dfrac{R}{R+h})  = 2\pi  R^2  (\dfrac{h}{R+h}) </p><p></p><p></p><p>For h << R   \implies A = 2\pi  h  R </p><p></p><p></p><p>Thus, net population covered: N  = Area \times population\ density  = 2\pi  hR \times </p><p></p><p></p><p>Given: = 100\ km^{-2}  =  100  \times 10^{-6} \ m^{-2}  =  10^{-4 }\ m^{-2}</p><p></p><p></p><p>N = 2(3.14) \times 100  \times 6.37 \times 10^6  \times 10^{-4}</p><p></p><p></p><p>N =  4 \times 10^5  = 4\ lakh</p><p></p><p>$