A TV tower has a height of 5 , m in a region | vedclass

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

Question

(B) Given: Height of the $TV$ tower, $h = 5 \, m = 5 	imes 10^{-3} \, km$. Population density, $n = \frac{1000}{\pi} \, 	ext{people/km}^2$. Radius o

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(B) Given: Height of the $TV$ tower, $h = 5 \, m = 5 	imes 10^{-3} \, km$. Population density, $n = \frac{1000}{\pi} \, 	ext{people/km}^2$. Radius of the Earth, $R_e \approx 6400 \, km$.The maximum range $(d)$ of the transmission is given by the formula: $d = \sqrt{2 h R_e}$.The area covered by the transmission is $A = \pi d^2 = \pi (2 h R_e)$.The population covered $(P_c)$ is given by: $P_c = n 	imes A$.Substituting the values:$P_c = \left( \frac{1000}{\pi} ight) 	imes (\pi 	imes 2 	imes h 	imes R_e)$$P_c = 1000 	imes 2 	imes (5 	imes 10^{-3} \, km) 	imes (6400 \, km)$$P_c = 1000 	imes 10 	imes 10^{-3} 	imes 6400$$P_c = 10 	imes 6400 = 64000$.Since the question asks for the number of people in thousands, the answer is $64$ thousand.

$A$. Optical Fiber Communication	$P$. Ultrasound
$B$. Radar	$Q$. Infrared Light
$C$. Sonar	$R$. Microwaves
$D$. Mobile Phones	$S$. Radio Waves

$A$ $TV$ tower has a height of $5 \, m$ in a region of population density $\frac{1000}{\pi}$ per square kilometre. The number of people that can receive the transmission is nearly, (in thousands)

Similar Questions

$A$ signal of frequency $3 \text{ kHz}$ is amplitude modulated on a carrier wave of frequency $2.5 \text{ MHz}$. The upper and lower sideband frequencies in the resultant signal are

What is the modulation index of the $AM$ wave shown in the figure?

The size of the antenna required for a carrier wave of frequency $3 \text{ MHz}$ is: (in $\text{ m}$)

The signal obtained at the output of a microphone is of which type?

Vedclass Test Series

Exam Paper Generator

Online Exam Module