Question 4: How are EM waves produced?

Step-by-step solution provided.

Question 9: What are radio waves? Give its two uses.

Step-by-step solution provided.

Question 17: What is meant by noise?

Step-by-step solution provided.

Question 23: Explain the necessity of a carrier wave in communication.

Step-by-step solution provided.

Maharashtra State Board Class 11 (FYJC / HSC) Physics Solutions Chapter 13 Electromagnetic Waves and Communication System - Digest Answers | Maharashtra Board Solutions

Maharashtra State Board 11th Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

1. Choose the correct option.

Question 1 Maharashtra Board Solution

The EM wave emitted by the Sun and responsible for heating the Earth’s atmosphere due to the greenhouse effect is (A) Infra-red radiation (B) X-ray (C) Microwave (D) Visible light

Solution & Step-by-Step Answer:

(A) Infra-red radiation

Question 2 Maharashtra Board Solution

Earth’s atmosphere is richest in (A) UV (B) IR (C) X-ray (D) Microwaves

Solution & Step-by-Step Answer:

(B) IR

Question 3 Maharashtra Board Solution

How does the frequency of a beam of ultraviolet light change when it travels from air into glass? (A) depends on the values of p and e (B) increases (C) decreases (D) remains same

Solution & Step-by-Step Answer:

(D) remains same

Question 4 Maharashtra Board Solution

The direction of EM wave is given by (A) × (B) . (C) along (D) along

Solution & Step-by-Step Answer:

(A) ×

Question 5 Maharashtra Board Solution

The maximum distance upto which TV transmission from a TV tower of height h can be received is proportional to (A) h½ (B) h (C) h3/2 (D) h²

Solution & Step-by-Step Answer:

(A) h½

Question 6 Maharashtra Board Solution

The waves used by artificial satellites for communication purposes are (A) Microwave (B) AM radio waves (C) FM radio waves (D) X-rays

Solution & Step-by-Step Answer:

(A) Microwave

Question 7 Maharashtra Board Solution

If a TV telecast is to cover a radius of 640 km, what should be the height of transmitting antenna? (A) 32000 m (B) 53000 m (C) 42000 m (D) 55000 m

Solution & Step-by-Step Answer:

(A) 32000 m

2. Answer briefly.

Question 1 Maharashtra Board Solution

State two characteristics of an EM wave.

Solution & Step-by-Step Answer:

i. The electric and magnetic fields, and are always perpendicular to each other and also to the direction of propagation of the EM wave. Thus, the EM waves are transverse waves.

ii. The cross product ( × ) gives the direction in which the EM wave travels. ( × ) also gives the energy carried by EM wave.

Question 2 Maharashtra Board Solution

Why are microwaves used in radar?

Solution & Step-by-Step Answer:

Microwaves are used in radar systems for identifying the location of distant objects like ships, aeroplanes etc.

Question 3 Maharashtra Board Solution

What are EM waves?

Solution & Step-by-Step Answer:

Waves that are caused by the acceleration of charged particles and consist of electric and magnetic fields vibrating sinusoidally at right angles to each other and to the direction of propagation are called EM waves or EM radiation.

Question 4 Maharashtra Board Solution

How are EM waves produced?

Solution & Step-by-Step Answer:

Question 5 Maharashtra Board Solution

Can we produce a pure electric or magnetic wave in space? Why?

Solution & Step-by-Step Answer:

No. In vacuum, an electric field cannot directly induce another electric field so a “pure” electric field wave cannot exist and same can be said for a “pure” magnetic wave.

Question 6 Maharashtra Board Solution

Does an ordinary electric lamp emit EM waves?

Solution & Step-by-Step Answer:

Yes, ordinary electric lamp emits EM waves.

Question 7 Maharashtra Board Solution

Why light waves travel in vacuum whereas sound wave cannot?

Solution & Step-by-Step Answer:

Light waves are electromagnetic waves which can travel in vacuum whereas sound waves travel due to the vibration of particles of medium. Without any particles present (like in a vacuum) no vibrations can be produced. Hence, the sound wave cannot travel through the vacuum.

Question 8 Maharashtra Board Solution

What are ultraviolet rays? Give two uses.

Solution & Step-by-Step Answer:

Production:

Uses:

Question 9 Maharashtra Board Solution

What are radio waves? Give its two uses.

Solution & Step-by-Step Answer:

Uses:

Question 10 Maharashtra Board Solution

Name the most harmful radiation entering the Earth’s atmosphere from the outer space.

Solution & Step-by-Step Answer:

Ultraviolet radiation.

Question 11 Maharashtra Board Solution

Give reasons for the following: i. Long distance radio broadcast uses short wave bands. ii. Satellites are used for long distance TV transmission.

Solution & Step-by-Step Answer:

i. Long distance radio broadcast uses short wave bands because electromagnetic waves only in the frequency range of short wave bands only are reflected by the ionosphere.

ii. a. It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere.
b. Hence, satellites are helpful in long distance TV transmission.

Question 12 Maharashtra Board Solution

Name the three basic units of any communication system.

Solution & Step-by-Step Answer:

Three basic (essential) elements of every communication system are transmitter, communication channel and receiver.

Question 13 Maharashtra Board Solution

What is a carrier wave?

Solution & Step-by-Step Answer:

The high frequency waves on which the signals to be transmitted are superimposed are called carrier waves.

Question 14 Maharashtra Board Solution

Why high frequency carrier waves are used for transmission of audio signals?

Solution & Step-by-Step Answer:

An audio signal has low frequency (<20 kHz) and low frequency signals cannot be transmitted over large distances. Because of this, a high frequency carrier waves are used for transmission.

Question 15 Maharashtra Board Solution

What is modulation?

Solution & Step-by-Step Answer:

The signals in communication system (e.g. music, speech etc.) are low frequency signals and cannot be transmitted over large distances. In order to transmit the signal to large distances, it is superimposed on a high frequency wave (called carrier wave). This process is called modulation.

Question 16 Maharashtra Board Solution

What is meant by amplitude modulation?

Solution & Step-by-Step Answer:

When the amplitude of carrier wave is varied in accordance with the modulating signal, the process is called amplitude modulation.

Question 17 Maharashtra Board Solution

What is meant by noise?

Solution & Step-by-Step Answer:

Question 18 Maharashtra Board Solution

What is meant by bandwidth?

Solution & Step-by-Step Answer:

The bandwidth of an electronic circuit is the range of frequencies over which it operates efficiently.

Question 19 Maharashtra Board Solution

What is demodulation?

Solution & Step-by-Step Answer:

The process of regaining signal from a modulated wave is called demodulation. This is the reverse process of modulation.

Question 20 Maharashtra Board Solution

What type of modulation is required for television broadcast?

Solution & Step-by-Step Answer:

Amplitude modulation is required for television broadcast.

Question 21 Maharashtra Board Solution

How does the effective power radiated by an antenna vary with wavelength?

Solution & Step-by-Step Answer:

Question 22 Maharashtra Board Solution

Why should broadcasting programs use different frequencies?

Solution & Step-by-Step Answer:

If broadcasting programs run on same frequency, then the information carried by these waves will get mixed up with each other. Hence, different broadcasting programs should run on different frequencies.

Question 23 Maharashtra Board Solution

Explain the necessity of a carrier wave in communication.

Solution & Step-by-Step Answer:

Question 24 Maharashtra Board Solution

Why does amplitude modulation give noisy reception?

Solution & Step-by-Step Answer:

i. In amplitude modulation, carrier is varied in accordance with the message signal.

ii. The higher the amplitude, the greater is magnitude of the signal. So even if due to any reason, the magnitude of the signal changes, it will lead to variation in the amplitude of the signal. So its easy for noise to disturb the amplitude modulated signal.

Question 25 Maharashtra Board Solution

Explain why is modulation needed.

Solution & Step-by-Step Answer:

Modulation helps in avoiding mixing up of signals from different transmitters as different carrier wave frequencies can be allotted to different transmitters. Without the use of these waves, the audio signals, if transmitted directly by different transmitters, would get mixed up.

3. Solve the numerical problem.

Question 1 Maharashtra Board Solution

Calculate the frequency in MHz of a radio wave of wavelength 250 m. Remember that the speed of all EM waves in vacuum is 3.0 × 108 m/s.

Solution & Step-by-Step Answer:

Given: λ = 250 m, c = 3 × 108 m/s To find: Frequency (v) Formula: c = v8 Calculation: From formula, v = = = 1.2 × 106 Hz = 1.2 MHz

Question 2 Maharashtra Board Solution

Calculate the wavelength in nm of an X-ray wave of frequency 2.0 × 1018 Hz.

Solution & Step-by-Step Answer:

Given: c = 3 × 108, v = 2 × 1018 Hz To find: Wavelength (λ) Formula: c = vλ Calculation. From formula, λ = = = 1.5 × 10-10 = 0.15 nm

Question 3 Maharashtra Board Solution

The speed of light is 3 × 108 m/s. Calculate the frequency of red light of wavelength of 6.5 × 10-7 m.

Solution & Step-by-Step Answer:

Given: c = 3 × 108 m/s, λ = 6.5 × 10-7 m To find: Frequency (v) Formula: c = vλ Calculation: From formula, v = = = 4.6 × 1014 Hz

Question 4 Maharashtra Board Solution

Calculate the wavelength of a microwave of frequency 8.0 GHz.

Solution & Step-by-Step Answer:

Given: v = 8 GHz = 8 × 109 Hz, c = 3 × 108 m/s To find: Wavelength (λ) Formula: c = vλ Calculation: From formula, λ = = = 3.75 × 10-2 = 3.75 cm

Question 5 Maharashtra Board Solution

In a EM wave the electric field oscillates sinusoidally at a frequency of 2 × 1010 What is the wavelength of the wave?

Solution & Step-by-Step Answer:

Given: v = 2 × 1010 Hz, c = 3 × 108 m To find: Wavelength (λ) Formula: c = vλ Calculation: From formula, λ = = = 1.5 × 10-2

Question 6 Maharashtra Board Solution

The amplitude of the magnetic field part of a harmonic EM wave in vacuum is B0 = 5 X 10-7 T. What is the amplitude of the electric field part of the wave?

Solution & Step-by-Step Answer:

Given: B0 = 5 × 10-7 T, c = 3 × 108 To find: Amplitude of electric field (E0) Formula: c = Calculation /From formula, E0 = c × B0 = 3 × 108 × 5 × 10-7 = 150 V/m

Question 7 Maharashtra Board Solution

A TV tower has a height of 200 m. How much population is covered by TV transmission if the average population density around the tower is 1000/km²? (Radius of the Earth = 6.4 × 106 m)

Solution & Step-by-Step Answer:

Given: h = 200 m, Population density (n) = 1000/km² = 1000 × 10-6/m² = 10-3/m² R = 6.4 ×106 m To find: Population covered Formulae: i. A = πd² = π()² = 2πRh ii. Population covered = nA Calculation /From formula (i), A = 2πRh = 2 × 3.142 × 6.4 × 106 × 200 ≈ 8 × 109 m² From formula (ii), Population covered = nA = 10-3 × 8 × 109 = 8 × 106

Question 8 Maharashtra Board Solution

Height of a TV tower is 600 m at a given place. Calculate its coverage range if the radius of the Earth is 6400 km. What should be the height to get the double coverage area?

Solution & Step-by-Step Answer:

Given: h = 600 m, R = 6.4 × 106 m To find: Range (d) Height to get the double coverage (h’) Formula: d = Calculation: From formula, d = = 87.6 × 10³ = 87.6 km Now, for A’ = 2A π(d’)² = 2 (πd²) ∴ (d’)² = 2d² From formula, h’ = = = 2 × h ……….. (∵ h = ) = 2 × 600 =1200 m

Question 9 Maharashtra Board Solution

A transmitting antenna at the top of a tower has a height 32 m and that of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in line of sight mode? Given radius of Earth is 6.4 × 106 m.

Solution & Step-by-Step Answer:

Given: ht = 32 m, hr = 50 m, R = 6.4 × 106 m To find: Maximum distance or range (d) Formula: d = Calculation: From formula, dt = = = 20.238 × 10³ m = 20.238 km dr = = = 25.298 × 10³ m = 25.298 km Now, d = dt + dr = 20.238 + 25.298 = 45.536 km

11th Physics Digest Chapter 13 Electromagnetic Waves and Communication System Intext Questions and Answers

Can you recall? (Textbookpage no. 229)

Question 1 Maharashtra Board Solution

i. What is a wave?

Solution & Step-by-Step Answer:

Wave is an oscillatory disturbance which travels through a medium without change in its form.

ii. What is the difference between longitudinal and transverse waves?
Answer:
a. Transverse wave: A wave in which particles of the medium vibrate in a direction perpendicular to the direction of propagation of wave is called transverse wave.
b. Longitudinal wave: A wave in which particles of the medium vibrate in a direction parallel to the direction of propagation of wave is called longitudinal wave.

iii. What are electric and magnetic fields and what are their sources?
Answer:
a. Electric field is the force experienced by a test charge in presence of the given charge at the given distance from it.
b. A magnetic field is produced around a magnet or around a current carrying conductor.

iv. By which mechanism heat is lost by hot bodies?
Answer:
Hot bodies lose the heat in the form of radiation.

Question 2 Maharashtra Board Solution

What are Lenz’s law, Ampere’s law and Faraday’s law?

Solution & Step-by-Step Answer:

Lenz’s law: Whereas, Lenz’s law states that, the direction of the induced emf is such that the change is opposed.

Ampere’s law:
Ampere’s law describes the relation between the induced magnetic field associated with a loop and the current flowing through the loop.

Faraday’s law:
Faraday’s law states that, time varying magnetic field induces an electromotive force (emf) and an electric field.

Internet my friend. (Tpxtboakpage no. 240)

https//www.iiap.res.in/centers/iao
[Students are expected to visit the above mentioned website and collect more information about different EM wave propagations used by astronomical observatories.]

Chapter 13 Electromagnetic Waves and Communication System Solutions