A TV tower has a height of 150 m. If the pop | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The range $d$ of a $TV$ tower of height $h$ is given by $d = \sqrt{2Rh}$.Given: $h = 150 \ m$, $R = 6.4 	imes 10^6 \ m$.$d = \sqrt{2 	imes 6.4 \

Vedclass · Accepted Answer

$60.288$

Vedclass · Answer

(A) The range $d$ of a $TV$ tower of height $h$ is given by $d = \sqrt{2Rh}$.Given: $h = 150 \ m$, $R = 6.4 	imes 10^6 \ m$.$d = \sqrt{2 	imes 6.4 	imes 10^6 	imes 150} = \sqrt{1920 	imes 10^6} = \sqrt{19.2 	imes 10^8} \approx 43.817 	imes 10^3 \ m = 43.817 \ km$.The area covered by the tower is $A = \pi d^2$.$A = 3.14 	imes (43.817)^2 \approx 3.14 	imes 1920 \approx 6028.8 \ km^2$.The population covered is $	ext{Population} = 	ext{Area} 	imes 	ext{Population Density}$.$	ext{Population} = 6028.8 \ km^2 	imes 10^3 \ km^{-2} = 6,028,800$.Converting to lakhs: $6,028,800 / 100,000 = 60.288 \ 	ext{lakh}$.

$A$ $TV$ tower has a height of $150 \ m$. If the population density around the $TV$ tower is $10^3 \ km^{-2}$, then the population covered by the tower is (Radius of the earth, $R = 6.4 \times 10^6 \ m$). (in $\text{lakh}$)

Similar Questions

$A$ signal $A \cos \omega t$ is transmitted using $v_0 \sin \omega_0 t$ as a carrier wave. The correct amplitude modulated $(AM)$ signal is:

$A$ transmitting antenna covers a range of $12.8 \ km$. Find the height of the antenna in $m$. (Radius of the Earth $= 6400 \ km$)

In the case of amplitude modulation,to avoid distortion,the modulation index $(\mu)$ should be:

For $100\%$ modulation,what should be the value of the minimum amplitude $(E_{min})$ of the modulated wave?

In amplitude modulation:

Vedclass Test Series

Exam Paper Generator

Online Exam Module