The height of a transmitting antenna at the t | vedclass

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

Question

(D) The maximum line-of-sight distance $d$ between a transmitting antenna of height $h_T$ and a receiving antenna of height $h_R$ is given by the form

Vedclass · Accepted Answer

$192$

Vedclass · Answer

(D) The maximum line-of-sight distance $d$ between a transmitting antenna of height $h_T$ and a receiving antenna of height $h_R$ is given by the formula:$d = \sqrt{2 R h_T} + \sqrt{2 R h_R}$Given:$h_T = 25 \; m$$h_R = 49 \; m$$R = 64 	imes 10^{5} \; m$Substituting the values:$d = \sqrt{2 	imes 64 	imes 10^{5} 	imes 25} + \sqrt{2 	imes 64 	imes 10^{5} 	imes 49}$$d = \sqrt{2 	imes 64 	imes 10^{5}} 	imes (\sqrt{25} + \sqrt{49})$$d = \sqrt{128 	imes 10^{5}} 	imes (5 + 7)$$d = \sqrt{1280 	imes 10^{4}} 	imes 12$$d = \sqrt{256 	imes 5 	imes 10^{4}} 	imes 12$$d = (16 	imes 10^{2} \sqrt{5}) 	imes 12$$d = 192 \sqrt{5} 	imes 10^{2} \; m$Comparing this with $K \sqrt{5} 	imes 10^{2} \; m$,we get $K = 192$.

Similar Questions

$A$ carrier wave $v_{c}(t) = 160 \sin(2 \pi \times 10^{6} t) \text{ V}$ is made to vary between $V_{\max} = 200 \text{ V}$ and $V_{\min} = 120 \text{ V}$ by a message signal $v_{m}(t) = A_{m} \sin(2 \pi \times 10^{3} t) \text{ V}$. The peak voltage $A_{m}$ of the modulating signal is ..... .

When the receiving antenna is on the ground,the range of a transmitting antenna of height $980 \ m$ is (Radius of the earth $= 6400 \ km$) (in $km$)

The advantage of an optical fibre is:

$A$ message signal of peak voltage $12 \ V$ is used to amplitude modulate a carrier signal of frequency $1.2 \ MHz$. The amplitude of the side bands is (in $V$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module