Nsert Soln Class 12 Physics Chapter 15

CHAPTER-15

Question 15.1: * Answer (b) Answer:
10 MHz
For beyond-the-horizon communication, it is necessary for the signal waves to travel a large distance. 10 KHz signals cannot be radiated efficiently because of the antenna size. The high energy signal waves (1GHz − 1000 GHz) penetrate the ionosphere. 10 MHz frequencies get reflected easily from the ionosphere. Hence, signal waves of such frequencies are suitable for beyond-the-horizon communication.

Question 15.2: * Answer (d) Answer:
Space waves
Owing to its high frequency, an ultra high frequency (UHF) wave can neither travel along the trajectory of the ground nor can it get reflected by the ionosphere. The signals having UHF are propagated through line-of-sight communication, which is nothing but space wave propagation.

Question 15.3: * Answer (c) Answer:
A digital signal uses the binary (0 and 1) system for transferring message signals. Such a system cannot utilise the decimal system (which corresponds to analogue signals). Digital signals represent discontinuous values.

Question 15.4: * Answer
Line-of-sight communication means that there is no physical obstruction between the transmitter and the receiver. In such communications it is not necessary for the transmitting and receiving antennas to be at the same height.
Height of the given antenna, h = 81 m
Radius of earth, R = 6.4 × 106 m
For range, d = (2Rh)½, the service area of the antenna is given by the relation:
A = πd2
= π (2Rh)
= 3.14 × 2 × 6.4 × 106 × 81
= 3255.55 × 106 m2
= 3255.55
∼ 3256 km2

Question 15.5: * Answer
Amplitude of the carrier wave, Ac = 12 V
Modulation index, m = 75% = 0.75
Amplitude of the modulating wave = Am
Using the relation for modulation index:

Question 15.6: * Answer
It can be observed from the given modulating signal that the amplitude of the modulating signal, Am = 1 V
It is given that the carrier wave c (t) = 2 sin (8πt)
∴Amplitude of the carrier wave, Ac