Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Communication

Question:

The line of sight distance between two antennas of heights 144 m and 324 m above earth surface is: 

(Given $\sqrt{20}$  = 4.4)

Options:

124.2 km

105.6 km

36.4 km

52.8 km

Correct Answer:

105.6 km

Explanation:

$\text{ Range R } = \sqrt{2Rh_T}+ \sqrt{2Rh_R} = \sqrt{2\times 6400\times 10^3\times 144}+ \sqrt{2\times 6400\times 10^3\times 324}$

$= 9600\sqrt{20} + 14400\sqrt{20} = 24000\sqrt{20}$

$= 24\times 4.4 Km = 105.6Km$