A radar has a power of 1 text{ kW} and operat | vedclass

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

Question

(A) The maximum line-of-sight distance $d_M$ for a transmitter at height $h$ above the Earth's surface is given by the formula:$d_M = \sqrt{2hR}$Given

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(A) The maximum line-of-sight distance $d_M$ for a transmitter at height $h$ above the Earth's surface is given by the formula:$d_M = \sqrt{2hR}$Given:Height $h = 500 	ext{ m}$Earth's radius $R = 6.4 	imes 10^{6} 	ext{ m}$Substituting the values:$d_M = \sqrt{2 	imes 500 	imes 6.4 	imes 10^{6}}$$d_M = \sqrt{1000 	imes 6.4 	imes 10^{6}}$$d_M = \sqrt{6.4 	imes 10^{9}}$$d_M = \sqrt{6400 	imes 10^{6}}$$d_M = 80 	imes 10^{3} 	ext{ m}$$d_M = 80 	ext{ km}$Thus,the radar can detect objects up to a maximum distance of $80 	ext{ km}$.

Similar Questions

If a $TV$ transmission antenna covers a radius of $128 \, km$,what should be the height of the antenna in $m$?

The waves which are reflected back to the earth by the ionosphere are:

The heights of transmitting and receiving antennas are respectively $\frac{1}{20000}$ and $\frac{1}{80000}$ times the radius of the earth. The maximum distance between these two antennas for satisfactory communication in line of sight mode is (Radius of the earth $= 6.4 \times 10^6 \ m$) (in $km$)

If $E_c$ and $E_m$ are peak values of carrier and modulating signals,respectively,then for $100 \%$ modulation,

Vedclass Test Series

Exam Paper Generator

Online Exam Module