Question

A TV tower has a height of $100m$. How much population is covered by the TV broadcast, if the average population density around the tower is $100k{m}^{-2}$.

(radius of the Earth $=6.37\times {10}^{6}m$)

• A. $4lakh$
• B. $4thousand$
• C. $40thousand$
• D. $40lakh$

Answer 1

The correct option is A $4lakh$

Solid angle subtended: $w=2\pi (1-cos\theta )=2\pi (1-\frac{R}{R+h})$

Now, area covered: $A=w{R}^2=2\pi R^2(1-\frac{R}{R+h})=2\pi R^2\left(\frac{h}{R+h}\right)$

For $h<<R⟹A=2\pi hR$

Thus, net population covered: $N=Area\times populationdensity=2\pi hR\times \rho$

Given: $\rho =100k{m}^{-2}=100\times {10}^{-6}{m}^{-2}={10}^{-4}{m}^{-2}$

$N=2\left(3.14\right)\times 100\times 6.37\times {10}^6\times {10}^{-4}$

$⟹N=4\times {10}^5=4lakh$