# NCERT Solutions For Class 12 Physics Chapter 15 Communication Systems

NCERT Solutions for Class 12 Physics Chapter 15 Communication Systems provided here is extremely helpful in revising complete syllabus and getting a strong base on it. NCERT 12th Class Physics Communication Systems all questions are solved with detailed explanation available for students. In this article we had given of NCERT solutions for Communication Systems Class 12 Physics step by step solution for each and every question of the chapter. These solutions will also help you with your homework. Best teachers across the India created NCERT solutions for Class 12 Physics Chapter 15 Communication Systems according to curriculum and pattern of syllabus as per guidelines of NCERT (CBSE) Books.

Table of Contents

## 12th Class NCERT Physics Chapter 15 Communication Systems

 Name of Organization NCERT Name of Class 12th Class Name of Subject Physics Name of Chapter Chapter 15 Name of Content Communication Systems Name of Category NCERT Solutions Official site http://ncert.nic.in/

### Chapter 15 Communication Systems covers numerous Questions and answers from all topics and sub-topics which are given below

 Section Name Topic Name 15 Communication Systems 15.1 Introduction 15.2 Elements of a Communication System 15.3 Basic Terminology Used in Electronic Communication Systems 15.4 Bandwidth of Signals 15.5 Bandwidth of Transmission Medium 15.6 Propagation of Electromagnetic Waves 15.7 Modulation and its Necessity 15.8 Amplitude Modulation 15.9 Production of Amplitude Modulated Wave 15.10 Detection of Amplitude Modulated Wave

Question 15.1:
Which of the following frequencies will be suitable for beyond-the-horizon communication
using sky waves?
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz
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:
Frequencies in the UHF range normally propagate by means of:
(a) Ground waves.
(b) Sky waves.
(c) Surface waves.
(d) Space waves.
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:
Digital signals
(i) Do not provide a continuous set of values,
(ii) Represent values as discrete steps,
(iii) Can utilize binary system, and
(iv) Can utilize decimal as well as binary systems.
Which of the above statements are true?
(a) (i) and (ii) only
(b) (ii) and (iii) only
(c) (i), (ii) and (iii) but not (iv)
(d) All of (i), (ii), (iii) and (iv).
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:
Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication? A TV transmitting antenna is 81m tall. How much service area can it cover if the receiving antenna is at the ground level?
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 × 10^6 m
For range, d = 2Rh, the service area of the antenna is given by the relation:
A = πd2
= π (2Rh)
= 3.14 × 2 × 6.4 × 10^6 × 81
= 3255.55 × 10^6 m^2
= 3255.55