What should be the height of transmitting antenna and the population covered if the televition telecast is to cover a radius of 150 km? The average population density around the tower is 200 / km2 and the value of Re = 6.5 * 106 m.

Option 1 - <p>Height = 1800m&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>Population Covered = 1413 × 10<sup>8</sup></p>
Option 2 - <p>Height = 1241m</p> <p>Population Covered&nbsp; = 7 × 10<sup>5</sup></p>
Option 3 - <p>Height = 1600m</p> <p>Population Covered = 2 × 10<sup>5</sup></p>
Option 4 - <p>Height = 1731m</p> <p>Population Covered = 1413 × 10<sup>5</sup></p>
2 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
V
6 months ago
Correct Option - 4
Detailed Solution:

d = 150 km

d = 2 h R

h = d 2 2 R = ( 1 5 0 * 1 0 3 ) 2 2 * 6 . 5 * 1 0 6 = 1 7 3 1 m .

Population covered

π r 2 * σ = 3 . 1 4 * 1 5 0 2 * 2 0 0 0 = 1 4 1 3 * 1 0 5

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According to question, we can write

  d = 2 h R d 2 d 1 = h 2 h 1 h 2 = ( d 2 d 1 ) 2 h 1 = ( 2 1 ) 2 × 1 2 5 = 5 0 0 m

Increment in height of tower = h2 – h1 = 500 – 125 = 375 m

Low pass filter will allow low frequency signal to pass while high pass filter allow high frequency to pass through

µ = A? /A? = 0.5 {A? = 20 volt, A? = 40 volt}
m (t) = A? sin ω? t {ω? = 2π×10? }
c (t) = A? sin ω? t {ω? = 2π×10×10³}
C? (t) = (A? + A? sin ω? t) sin ω? t ⇒ A? {1+ µsin ω? t} sin ω? t

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Physics Ncert Solutions Class 12th 2023

Physics Ncert Solutions Class 12th 2023

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